303. Integrals II.

by Virgil Nicula, Jul 26, 2011, 6:37 PM

$\blacksquare$ PP1. Prove that $\left(\forall\right)a>0$ exists the inequality $g(a)\equiv \int_0^{\frac {\pi}{2}}\left|a\cdot\sin 2x-\cos^2x\right|\mathrm{dx}\ \ge\ \frac 12$ .

Proof. With the substitution $t=\tan x$ obtain $g(a)=\int_0^{\infty}\frac {|2at-1|}{\left(t^2+1\right)^2}\mathrm{dx}$ $\implies$ $g(a)=\arctan\frac {1}{2a}+a-\frac {\pi}{4}$ . Prove easily that $g'(a)\ \mathrm{\ .s.s.\ }(2a-1)$

for any $a>0$ , i.e. $\min_{a>0}\ g(a)=g\left(\frac 12\right)=\frac 12$ . Remark. $g(a)=\int_0^{\frac {\pi}{2}}\cos x|2a\sin x-\cos x|\ \mathrm{dx}\le\sqrt {4a^2+1}\cdot\int_0^{\frac {\pi}{2}}\cos x\ \mathrm{dx}=\sqrt {4a^2+1}$ .

Hence $\left(\forall\right) a>0$ , $\boxed {\ \frac 12\ \le\ \int_0^{\frac {\pi}{2}}\left|a\cdot\sin 2x-\cos^2x\right|\mathrm{dx}\ \le\ \sqrt {4a^2+1}\ }$ .

$\blacksquare$ PP2. Evaluate $I\equiv \int_0^{2\pi} \left|x^2-\pi ^ 2 -\sin ^ 2 x\right|\ \mathrm{dx}$ .

Proof. $f(x)=x^2-\pi^2-\sin^2x\ ,\ x\in [0,2\pi ]$ $\implies$ $f''(x)=4\sin^2x\ge 0$ ; $f'(x)=2x-\sin 2x\nearrow$ (strict increasing) and

$f'(0)=0\implies f'(x)\ge 0$ ; $f\nearrow$ and $f(\pi )=0\implies$ $f(x)\ .s.s.\ (x-\pi )$ , i.e. $f(x)$ has same sign with $(x-\pi )$ , what means

$(*)\ \left\{\begin{array}{ccc} x\in [0,\pi ] & \implies & f(x)\le 0\\\\ x\in(\pi,2\pi ] & \implies & f(x)\ge 0\end{array}\right\|$ . Let $F\in\int \left(x^2-\pi ^ 2 -\sin ^ 2 x\right)\ \mathrm{dx}$ , where $F(x)=\frac {x^3}{3}-\pi^2x-\frac x2+\frac {\sin 2x}{4}\implies$

$I=\int_0^{2\pi}f(x)\mathrm{dx}=$ $-\int_0^{\pi}f(x)\mathrm{dx}+\int_{\pi}^{2\pi}f(x)\mathrm{dx}=$ $F(0)+F(2\pi )-2F(\pi )\implies$ $\boxed{\ I=2\pi^3\ }\ .$

Remark. I"ll prove elementary the relations $(*)$ . If $0\le x\le\pi\implies \sin^2x\ge 0\ge x^2-\pi^2$ . If $\pi\le x\le 2\pi$ , then

$0\le x-\pi\le\pi\implies$ $0\le -\sin x=\sin (x-\pi )\le x-\pi\implies\sin^2x\le (x-\pi )^2\le (x-\pi )(x+\pi )=x^2-\pi^2$ .

$\blacksquare$ PP3. Ascertain the relation $\mathrm {R}(a,b,c) = 0$ so that $K\equiv \int_{ - \infty}^{\infty}\frac {x^2 + a}{x^4 + bx^2 + c^2}\ \mathrm{dx} = \pi$ , where $\{a,b,c\}\subset (0,\infty )$ .

Proof. $\blacktriangleright\ \boxed {\ I\equiv\int_0^{\infty}\frac {x^2 + c}{x^4 + bx^2 + c^2}\ \mathrm{dx}\ } =$ $\int_0^{\infty}\frac {1 + \frac {c}{x^2}}{x^2 + \frac {c^2}{x^2} + b}\ \mathrm {dx} =$ $\int_0^{\infty}\frac {\left(x - \frac cx\right)'}{\left(x - \frac cx\right)^2 + 2c + b}\ \mathrm {dx}$ . Thus, $I = \left|\frac {1}{\sqrt {2c + b}}\arctan\frac {x^2 - c}{x\sqrt {2c + b}}\right|_0^{\infty} =$

$\frac {\pi}{\sqrt {b + 2c}}$ $\implies$ $\boxed {\ I = \frac {\pi}{\sqrt {b + 2c}}\ }$ . $\blacktriangleright\ \boxed {\ J\equiv\int_0^{\infty}\frac {x^2 - c}{x^4 + bx^2 + c^2}\ \mathrm{dx} \ } =$ $\int_0^{\infty}\frac {1 - \frac {c}{x^2}}{x^2 + \frac {c^2}{x^2} + b}\ \mathrm {dx} =$ $\int_0^{\infty}\frac {\left(x + \frac cx\right)'}{\left(x + \frac cx\right)^2 - 2c + b}\ \mathrm {dx}$ . If $b = 2c$ , then

$J = - \left|\frac {x}{x^2 + c}\right|_0^{\infty} = 0$ . If $b > 2c$ , then $J = \left|\frac {1}{\sqrt {b - 2c}}\arctan\frac {x^2 + c}{x\sqrt {b - 2c}}\right|_0^{\infty} = 0$ . If $b < 2c$ , then $J = \left|\frac {1}{2\sqrt {2c - b}}\ln\frac {x^2 - x\sqrt {2c - b} + c}{x^2 + x\sqrt {2c - b} + c}\right|_0^{\infty} = 0$ .

Therefore, for any $b > 0$ , $c > 0$ we have $\boxed {\ J = 0\ }$ . $\blacktriangleright\ \boxed {\ U\equiv\int_0^{\infty}\frac {x^2}{x^4 + bx^2 + c^2}\ \mathrm{dx}\ } = \frac 12\cdot (I + J) = \frac 12\cdot I$ and

$\boxed{V\equiv\int_0^{\infty}\frac {1}{x^4+bx^2+c^2}\ \mathrm{dx}=\frac {1}{2c}\cdot (I-J)=\frac {1}{2c}\cdot I}$ . Therefore, $K=2\cdot \int_{0}^{\infty}\frac {x^2 + a}{x^4 + bx^2 + c^2}\ \mathrm{dx} = 2(U + aV) =$ $\left(1 + \frac ac\right)\cdot I$ $\implies$

$\boxed {\boxed {\ K = \left(1 + \frac ac\right)\cdot \frac {\pi}{\sqrt {b + 2c}}\ }}$ . In conclusion, $\boxed {\ K = \pi\ \Longleftrightarrow\ \left(1 + \frac ac\right)^2 = b + 2c\ }$ .

Remark. Show easily that if $b > 2c$ , then my relation $\boxed {\ \left(1 + \frac ac\right)^2 = b + 2c\ }$ is equivalently with "Misan's relation, i.e.

$\boxed {\ \frac {1}{\sqrt 2} \;\; \frac {1}{\sqrt {b^2 - 4c^2}}\;\;\frac {1}{2c}\left[ \sqrt {b + \sqrt {b^2 - 4c^2}}\cdot\left(\sqrt {b^2 - 4c^2} + (2a - b)\right) + \sqrt {b - \sqrt {b^2 - 4c^2}}\cdot\left(\sqrt {b^2 - 4c^2} - (2a - b)\right)\right] = 1\ }$ .

$\blacksquare$ PP4. Let $a,\ b$ be postive constant numbers. Evaluate $\int_0^{\frac {\pi}{2}} \frac {a\cos x - b\sin x}{b\sin x + a\cos x}\ \mathrm {dx}$ .

Proof. Suppose w.l.o.g. $a\ne 0\ \ \vee\ \ b\ne 0$ , i.e. $a^2 + b^2\ne 0$ . Denote $f(x) = b\sin x + a\cos x$ . I look for the constant $A$ , $B$ so that for any $x\in\mathbb R$ ,

$a\cos x - b\sin x = Af(x) + Bf'(x)$ , i.e. $a\cos x - b\sin x =$ $A(b\sin x + a\cos x) + B(b\cos x - a\sin x)$ . Obtain easily that $A = \frac {a^2 - b^2}{a^2 + b^2}$

and $B = \frac {2ab}{a^2 + b^2}$ . Thus, $\frac {a\cos x - b\sin x}{b\sin x + a\cos x}\ \mathrm {dx} =$ $\int\frac {Af(x) + Bf'(x)}{f(x)}\ \mathrm {dx} =$ $Ax + B\ln|f(x)| + \mathrm C$ , i.e. $\int\frac {a\cos x - b\sin x}{b\sin x + a\cos x}\ \mathrm {dx} =$

$\frac {a^2 - b^2}{a^2 + b^2}\cdot x + \frac {2ab}{a^2 + b^2}\cdot\ln|b\sin x + a\cos x| + \mathrm C$ . In conclusion, $\int_0^{\frac {\pi}{2}}\frac {a\cos x - b\sin x}{b\sin x + a\cos x}\ \mathrm {dx} =$ $\frac {a^2 - b^2}{a^2 + b^2}\cdot\frac {\pi}{2} + \frac {2ab}{a^2 + b^2}\cdot\ln\left|\frac ab\right|$ .

$\blacksquare$ PP5. Compute $I = \int_{0}^{\frac {\pi}{2}} \frac {1}{1 + \tan^{\alpha}{x}}\ \mathrm {dx}$ , where $\alpha\in\mathbb R^*$ .

Proof. You can apply the well-known integral identity $\boxed {\ \int_a^bf(x)\ \mathrm {dx} = \int_a^bf(a + b - x)\ \mathrm {dx}\ }$ ("keeping of the interval").

Therefore, $I=\int_{0}^{\frac {\pi}{2}} \frac {1}{1 + \tan^{\alpha}{x}}\ \mathrm {dx} =$ $\int_{0}^{\frac {\pi}{2}} \frac {1}{1 + \tan^{\alpha}\left(\frac {\pi}{2} - x\right)}\ \mathrm {dx} =$ $\int_{0}^{\frac {\pi}{2}} \frac {1}{1+\cot^{\alpha}{x}}\ \mathrm {dx}=$ $\int_{0}^{\frac {\pi}{2}} \frac {\tan^{\alpha} x}{1 + \tan^{\alpha}x}\ \mathrm {dx}$ .

In conclusion, $2\cdot I = \int_{0}^{\frac {\pi}{2}} \frac {1}{1 + \tan^{\alpha}{x}}\ \mathrm {dx}+ \int_{0}^{\frac {\pi}{2}} \frac {\tan^{\alpha} x}{1 + \tan^{\alpha}x}\ \mathrm {dx}=\int_0^{\frac {\pi}{2}}\ \mathrm {dx}=\frac {\pi}{2}$ $\implies$ $I = \frac {\pi}{4}$ . See and here .

$\blacksquare$ PP6. Given are $\{m,n\}\subset\mathbb Z$ so that $m + n$ is an odd integer. Find $\min_{\begin{array}{c} \{a,b\}\subset\mathbb R \\ \ a + b = 1\end{array}}\int_{0}^{\pi}(a\sin mx + b\sin nx)^{2}\ \mathrm {dx}$ .

Proof. $I(a,b)\equiv\int_{0}^{\pi}(a\sin mx + b\sin nx)^{2}\ \mathrm {dx} =$ $\int_{0}^{\pi}\left[a\sin m(\pi - x) + b\sin n(\pi - x)\right]^{2}\ \mathrm {dx} =$

$\int_{0}^{\pi}\left[( - 1)^{m + 1}a\sin mx + ( - 1)^{n + 1}b\sin nx\right]^{2}\ \mathrm {dx}$ . Therefore, $2I(a,b) =$

$\int_0^{\pi}\left\{(a\sin mx + b\sin nx)^{2} + \left[( - 1)^{m + 1}a\sin mx + ( - 1)^{n + 1}b\sin nx\right]^{2}\right\}\ \mathrm {dx}$ $\implies$

$I(a,b) = \int_0^{\pi}\left(a^2\sin^2mx + b^2\sin^2nx\right)\ \mathrm {dx} =$ $\frac 12\cdot\int_0^{\pi}\left[\left(a^2 + b^2\right) - \left(a^2\cos 2mx + b^2\cos 2nx\right)\right]\ \mathrm {dx}$ .

Thus, $I(a,b) = \frac {\pi}{2}\cdot \left(a^2 + b^2\right)\ge\frac {\pi}{4}$ because $a^2 + b^2\ge \frac 12\cdot (a + b)^2 = \frac 12$ . We have equality iff $a = b = \frac 12$ .

$\blacksquare$ PP7. Prove that integral $I=\int_{0}^{\infty}\frac {\ln x}{x^2 + a^2}\ \mathrm {dx}$ converges and calculate it.

Proof. Using the substitution $x: =ax$ obtain $I=\frac 1a\cdot\left(J+\ln a \cdot\int_0^{\infty}\frac {1}{x^2+1}\ \mathrm {dx}\right)$ , where $J=\int_0^{\infty}\frac {\ln x}{x^2+1}\ \mathrm {dx}$ .

Since $\int_0^{\infty}\frac {1}{x^2+1}\ \mathrm {dx}=\frac {\pi}{2}$ and using the substitution $x: =\frac 1x$ obtain $J=-J$ , i.e. $J=0$ , means that $I=\frac {\pi}{2}\cdot\frac {\ln a}{a}$ .

$\blacksquare$ PP8. Calculate $I= \int_0^{\pi}\frac {x\sin x}{1+\cos^2x}\ \mathrm{dx}$ .

Proof. $I\stackrel{(x:=\pi -x)}{\ =\ } \int_0^{\pi}\frac {(\pi - x)\sin x}{1 + \cos^{2}x}\ \mathrm{dx}$ $\implies$ $I=\frac {\pi}{2}\cdot\int_0^{\pi}\frac {\sin x}{1 + \cos^2x}\ \mathrm {dx}=$ $\left|\frac {\pi}{2}\cdot \arctan (\cos x)\right|_{\pi}^0$ $\implies$ $I = \frac {\pi^2}{4}$ .

$\blacksquare$ PP9. Let $I_n(x)=\int_1^x (\ln t)^ndt\ (x>0)$ for $n\in\mathbb N^*$ . Prove by mathematical induction that $I_n(x)$ is expressed by $I_n(x)=xf_n(\ln x)+C_n\ (n\geq 1)$

in terms of some polynomial $f_n(y)$ with degree $n$ and some constant number $C_n$ . Express the constant term of $f_n(y)$ in terms of $n$ .

Proof. Prove easily that $\boxed {I_{k+1}(x)=x\ln^{k+1}x-(k+1)\cdot I_k(x)\ ,\ k\in\mathbb N}$ . Divide by $(k+1)!$ and denote

$t=\ln x\ ,\ J_k(x)=\frac {1}{k!}\cdot I_k(x)\ ,\ k\in\mathbb N$ . Thus, $\boxed {J_{k+1}(x)=x\cdot\frac {t^{k+1}}{(k+1)!}-J_k(x)\ ,\ k\in\mathbb N}$ a.s.o.

$\blacksquare$ PP10. $\boxed {I_n\equiv\int_0^{\infty}\frac {1}{\left(x^2 + 1^2\right)\left(x^2 + 2^2\right)\ldots\left(x^2 + n^2\right)}\ \mathrm {dx} = \pi\cdot\sum_{k = 1}^n\frac {( - 1)^{k - 1}\cdot k}{(n - k)!\cdot(n + k)!}\ }$ .

Proof. Denote $x^2 = y$ . Thus, $\prod_{k = 1}^n\frac {1}{\left(x^2 + k^2\right)} = \prod_{k = 1}^n\frac {1}{\left(y + k^2\right)} =$ $\sum_{k = 1}^n\frac {A_k}{y + k^2}$ $\implies$ $1 = \sum_{k = 1}^n\left[A_k\cdot\prod_{j = 1\ ,\ j\ne k}^n(y + j^2)\right]$ for any $y\in\mathbb R$ .

Particularly, for $y = - k^2$ obtain : $1 = A_k\cdot\prod_{j = 1\ ,\ j\ne k}^n(j^2 - k^2) =$ $A_k\cdot\prod_{j = 1}^{k - 1}(j - k)\cdot\prod_{j = k + 1}^n(j - k)\cdot\prod_{j = 1\ ,\ j\ne k}^n(j + k)$ $\implies$

$1=A_k\cdot ( - 1)^{k - 1}(k - 1)!\cdot(n - k)!\cdot\frac {(n + k)!}{k!\cdot 2k}$ $\implies$ $A_k = \frac {( - 1)^k\cdot 2k^2}{(n - k)!(n + k)!}$ . Therefore, $I_n = \int_0^{\infty}\prod_{k = 1}^n\frac {1}{x^2 + k^2}\ \mathrm {dx} =$

$\sum_{k = 1}^nA_k\int_0^{\infty}\frac {1}{x^2 + k^2}\ \mathrm {dx} =$ $\frac {\pi}{2}\cdot\sum_{k = 1}^n\frac {A_k}{k}$ . In conclusion, $\boxed {I_n = \pi\cdot\sum_{k = 1}^n\frac {( - 1)^{k - 1}\cdot k}{(n - k)!(n + k)!}}$ .

Remark. Denote $f(x)=\int_0^x\frac {t}{\left(t^2 + 1^2\right)\left(t^2 + 2^2\right)\ldots\left(t^2 + n^2\right)}\ \mathrm {dt}$ . Observe that

$\lim_{x\to 0}\frac {f(x)}{x^2}\stackrel{(l'H)}{=}$ $\lim_{x\to 0}\frac {f'(x)}{2x}=$ $\lim_{x\to 0}\frac {1}{2\left(x^2 + 1^2\right)\left(x^2 + 2^2\right)\ldots\left(x^2 + n^2\right)}=\frac{1}{2 (n!)^2}$ .

Can prove analogously that $\boxed {\int_0^{\infty}\frac {1}{\left(x^4 + 1^2\right)\left(x^4 + 2^2\right)\ldots\left(x^4 + n^2\right)}\ \mathrm {dx} = \frac {\pi}{2}\cdot\sum_{k = 1}^n\frac {( - 1)^{k - 1}\cdot\sqrt {2k}}{(n - k)!\cdot(n + k)!}\ }$ . See here.

$\blacksquare$ PP11. Ascertain $\lim_{x\to\infty}\left[x^3\cdot\int_{-\frac 1x}^{\frac 1x}\frac {\ln (1+t^2)}{1+e^t}\ \mathrm {dt}\right]$ .

Proof 1. If $f(t)=\frac {\ln (1 + t^2)}{1 + e^t}$ then we have $\int_{-\frac {1}{x}}^{\frac {1}{x}}f(t)\ \mathrm{dt}=$ $\int_{ - \frac {1}{x}}^{\frac {1}{x}} \frac {f(t) + f( - t)}{2} \ \mathrm{dt}=$ $\frac {1}{2} \int_{ - \frac {1}{x}}^{\frac {1}{x}} \ln (1 + t^2) \ \mathrm{dt}=$ $\arctan t - t + \frac {t}{2} \ln (1 + t^2) \Big|_{ - \frac {1}{x}}^{\frac {1}{x}}=$

$\frac { - 2 + 2x \ \mbox{arccot} x + \ln \left( 1 + \frac {1}{x^2} \right) }{x}$ . Then the our limit becomes $\lim_{x \to\infty} \left\{ x^2 \left[ - 2 + 2x \ \mbox{arccot} x + \ln \left( 1 + \frac {1}{x^2} \right) \right] \right\}$ . We can easily prove that

$\lim_{x\to \infty} \ln \left( 1 + \frac {1}{x^2} \right)^{x^2} = 1$ . Also $\lim_{x \to\infty}x^2(- 2 + 2x \, \mbox{arccot} x) = \lim_{x \to\infty} \frac { - 2 + 2x \, \mbox{arccot} x}{\frac {1}{x^2}} \stackrel{\mathbf{H}}{ = } \ldots \stackrel{\mathbf{H}}{ = } \ldots = - \frac {2}{3}$ . And whole limit yields $\frac 13$ .

Proof 2. $\lim_{x\to\infty}\left[x^3\cdot\int_{-\frac 1x}^{\frac 1x}\frac {\ln (1+t^2)}{1+e^t}\ \mathrm {dt}\right]=$ $\lim_{x\to\infty}\frac {\int_{-\frac 1x}^{\frac 1x}\frac {\ln\left(1+t^2\right)}{1+e^t}\ \mathrm{dt}}{\frac {1}{x^3}}\stackrel{\left(y=\frac 1x\right)}{=}$ $\lim_{y\to 0} \frac {\int_{-y}^{y}\frac {\ln\left(1+t^2\right)}{1+e^t}\ \mathrm{dt}}{y^3}\stackrel{(l'H)}{=}$

$\lim_{y\to 0}\frac {\frac {\ln\left(1+y^2\right)}{1+e^y}+\frac {\ln\left(1+y^2\right)}{1+e^{-y}}}{3y^2}=$ $\lim_{y\to 0}\frac {\ln\left(1+y^2\right)}{3y^2}\stackrel{(z=y^2)}{=}$ $\lim_{z\to 0}\frac {\ln\left(1+z\right)}{3z}=\frac 13$ .

$\blacksquare$ PP12. Find the recurrent relations for $A_n = \int (\sin x + \cos x)^n\ \mathrm {dx}$ and $a_n = \int_0^{\frac {\pi}{2}}(\sin x + \cos x)^n\ \mathrm {dx}$ , where $n\in \mathcal N$ .

Proof. I"ll use the identity $(\sin x + \cos x)^2 + (\cos x - \sin x)^2 = 2$ . Thus, for $n\in\mathcal N$ ,

$A_{n + 2} = \int (\sin x + \cos x)^n\left[2 - (\cos x - \sin x)^2\right]\ \mathrm {dx}\implies$ $\boxed{A_{n + 2} = 2A_n - H_n}$ ,

where $H_n = \int (\sin x + \cos x)^n(\cos x - \sin x)^2\ \mathrm {dx}$ . Apply the integration by parts :

$\overline {\underline {\left\|\begin{array}{ccc} u(x) = \cos x - \sin x & \implies & u'(x) = - (\sin x + \cos x) \\ \\ v'(x) = (\sin x + \cos x)^n(\cos x - \sin x) & \implies & v(x) = \frac {1}{n + 1}\cdot (\sin x + \cos x)^{n + 1}\end{array}\right\|}}$

Therefore, $H_n = \frac {1}{n + 1}\cdot (\sin x + \cos x)^{n + 1}(\cos x - \sin x) + \frac {1}{n + 1}\cdot A_{n + 2}$ and

$\boxed {\begin{array}{c} A_0 = x + \mathcal C\ \ ,\ \ A_1 = \sin x - \cos x + \mathcal C \\ \\ (n + 2)\cdot A_{n + 2} = 2(n + 1)\cdot A_n - (\sin x + \cos x)^{n + 1}(\cos x - \sin x)\ ,\ n\in\mathcal N\end{array}}$ .

Analogously, $\boxed {\begin{array}{c} a_0 = \frac {\pi}{2}\ \ ,\ \ a_1 = 2 \\ \\ (n + 2)\cdot a_{n + 2} = 2(n + 1)\cdot a_n + 2\end{array}}$ .

$\blacksquare$ PP13. Find the recurrent relations for $A_n = \int (1 + \sin x + \cos x)^n\ \mathrm {dx}$ and $a_n = \int_0^{\frac {\pi}{2}}(1 + \sin x + \cos x)^n\ \mathrm {dx}$ , where $n\in \mathcal N$ .

Proof. I"ll use the identity $(1 + \sin x + \cos x)^2 = 2(1 + \sin x + \cos x + \sin x\cos x)$ .

For $n\in\mathcal N$ , $A_{n + 2} = 2\cdot \int (1 + \sin x + \cos x)^n(1 + \cos x + \sin x + \sin x\cos x)\ \mathrm {dx}\implies$

$\boxed{A_{n + 2} = 2A_{n + 1} + H_n}$ , where $H_n = \int (1 + \sin x + \cos x)^n\sin 2x\ \mathrm {dx}\implies$

$H_n = \int (1 + \sin x + \cos x)^n\left[1 - (\cos x - \sin x)^2\right]\ \mathrm {dx}\implies$ $\boxed {H_n = A_n - L_n}$ , where

$L_n = \int (1 + \sin x + \cos x)^n(\cos x - \sin x)^2\ \mathrm {dx}$ . Apply the integration by parts :

$\overline {\underline {\left\|\begin{array}{ccc} u(x) = \cos x - \sin x & \implies & u'(x) = - (\sin x + \cos x) \\ \\ v'(x) = (1 + \sin x + \cos x)^n(\cos x - \sin x) & \implies & v(x) = \frac {1}{n + 1}\cdot (1 + \sin x + \cos x)^{n + 1}\end{array}\right\|}}$

Therefore, $L_n = \frac {1}{n + 1}\cdot (1 + \sin x + \cos x)^{n + 1}(\cos x - \sin x) + \frac {1}{n + 1}\cdot \left(A_{n + 2} - A_{n + 1}\right)$ .

$\boxed {\begin{array}{c} A_0 = x + \mathcal C\ \ ,\ \ A_1 = x + \sin x - \cos x + \mathcal C \\ \\ (n + 2)\cdot A_{n + 2} = (2n + 3)\cdot A_{n + 1} + (n + 1)\cdot A_n - (1 + \sin x + \cos x)^{n + 1}(\cos x - \sin x)\ ,\ n\in\mathcal N\end{array}}$ .

Analogously, $\boxed {\begin{array}{c} a_0 = \frac {\pi}{2}\ \ ,\ \ a_1 = 2 + \frac {\pi}{2} \\ \\ (n + 2)\cdot a_{n + 2} = (2n+ 3)\cdot a_{n + 1} + (n+1)\cdot a_n + 2^{n + 2}\ ,\ n\in\mathcal N\end{array}}$ .

$\blacksquare$ PP14. Ascertain $\lim_{a\to\infty}\frac 1a\cdot\int_0^{\infty}\frac {x^2+ax+1}{x^4+1}\cdot\arctan x\ \mathrm {dx}$ .

Proof. $I(a)=\int_0^\infty \frac{x^2+ax+1}{x^4+1} \, \arctan x \, \mathrm{dx}$ after the substitution $x\to \frac{1}{x}$ becomes $\int_0^\infty \frac{x^2+ax+1}{x^4+1} \, \arctan\left(\frac{1}{x}\right)\, \mathrm{dx}$ .

So $2\, I(a)=\int_0^\infty \frac{x^2+ax+1}{x^4+1} \, \underbrace{\left[ \arctan x+\arctan\left(\frac{1}{x}\right)\right]}_{=\frac{\pi}{2}} \, \mathrm{dx}$ $\Longrightarrow I(a)=\frac{\pi}{4} \left[\underbrace{\int_0^\infty \frac{x^2+1}{x^4+1} \mathrm{dx}}_{\frac{\pi}{\sqrt{2}}}+\underbrace{\int_0^\infty \frac{ax}{x^4+1} \mathrm{dx}}_{a\, \frac{\pi}{4}} \right] \Rightarrow \frac{I(a)}{a}\to \left(\frac{\pi}{4}\right)^2$ .

$\blacksquare$ PP15. $I_2\equiv\int_0^1 \frac {x^3}{\sqrt {x^2 + 1}}\ \mathrm {dx} =$ $\int_0^1 \left(x\sqrt {x^2 + 1} - \frac {x}{\sqrt {x^2 + 1}}\right)\ \mathrm {dx} =$ $\left[\frac {1}{3}(x^2 + 1)^{\frac {3}{2}} - \sqrt {x^2 + 1}\right]_0^1$ $\implies\boxed {I_2 = \frac {2 - \sqrt {2}}{3}}\ .$

Proof. Kunny, I like the your nice proof because it is without integration by parts and without substitutions, i.e. it is a integration by "binocular" ! Thank you !

Can show analogously that $I_n\equiv\boxed {\int_0^1\frac {x^{2n - 1}}{\sqrt {x^n + 1}}\ \mathrm {dx} = \frac {2(2 - \sqrt 2)}{3n}}\ !$ . Indeed, $\frac {x^{2n - 1}}{\sqrt {x^n + 1}} =$ $x^{n - 1}\left(\sqrt {x^n + 1} - \frac {1}{\sqrt {x^n + 1}}\right) =$

$\frac 1n\left[\left(x^n + 1\right)^{\frac 12} - \left(x^n + 1\right)^{ - \frac 12}\right]\left(x^n + 1\right)'$ $\implies$ $\frac 1n\left[\frac 23\left(x^n + 1\right)^{\frac 32} - 2\left(x^n + 1\right)^{\frac 12}\right]_0^1$ $\implies$ $I_n = \frac 1n\left[\frac 23\left(2\sqrt 2 - 1\right) - 2\left(\sqrt 2 - 1\right)\right]$ $\implies$

$I_n\equiv\boxed {\int_0^1\frac {x^{2n - 1}}{\sqrt {x^n + 1}}\ \mathrm {dx} = \frac {2(2 - \sqrt 2)}{3n}}$ . Remark. Can show analogously more generally that $I_{n,p}\equiv$ $\boxed {\int_0^1\frac {x^{2n - 1}}{\sqrt [p] {x^n + 1}}\ \mathrm {dx} =\frac {p\left(2 - \sqrt [p] {2^{p-1}}\right)}{n(p-1)(2p-1)}}\ !$ .

Indeed, can use the substitution $\left\{\begin{array}{c} t = \phi (x) = x^n + 1\\\\ \phi '(x)=nx^{n-1}\end{array}\right\|$ , i.e. $\int_0^1\frac {x^{2n - 1}}{\sqrt [p] {x^n + 1}}\ \mathrm {dx} =$ $\frac 1n\int_0^1\frac {\left(x^n + 1\right) - 1}{\sqrt [p] {x^n + 1}}\left(x^n + 1\right)'\ \mathrm {dx} =$ $\frac 1n\int_1^2\frac {t - 1}{\sqrt [p] t}\ \mathrm {dt}$ a.s.o.

$\blacksquare$ PP16. Ascertain a recurrence for $a_n = \int_0^{\infty }\frac {x^{n - 1}\cdot\arctan x}{\left(x^2 + 1\right)^n}\ \mathrm {dx}$ , where $n\in \mathcal N\ .$

$\blacksquare$ PP17. Find $I=\int_{0}^{\frac{\pi}{4}}\left(\frac{\cos x}{\sin x+\cos x}\right)^{2}\ \mathrm{dx}$ and $J=\int_{0}^{\frac{\pi}{4}}\left(\frac{\sin x+\cos x}{\cos x}\right)^{2}\ \mathrm{dx}$ .

Proof. I"ll use $t=\phi (x)=\frac{\pi}{4}-x\ :\ J=2\int_{0}^{\frac{\pi}{4}}\frac{1+\sin 2x}{1+\cos 2x}\ \mathrm{dx}=$ $2\int_{0}^{\frac{\pi}{4}}\frac{1+\cos 2x}{1+\sin 2x}\ \mathrm{dx}=$ $4\int_{0}^{\frac{\pi}{4}}\frac{\cos^{2}x}{(\sin x+\cos x)^{2}}\ \mathrm{dx}$ $\Longrightarrow \boxed{\ J=4\cdot I\ }$ .

Thus, $J=\int_0^{\frac {\pi}{4}}\frac {1+\sin 2x}{\cos^2x}\ \mathrm{dx}=$ $\int_0^{\frac {\pi}{4}}\left(\frac {1}{\cos^2x}+2\tan x\right)\ \mathrm{dx}=$ $\left|\left(\tan x-2\ln\cos x\right)\right|_0^{\frac {\pi}{4}}\implies$ $J=1+\ln 2$ and $I=\frac 14\cdot (1+\ln 2)$ .

$\blacksquare$ PP18. Prove that $:\ \int_{0}^{2n\pi}\max\{\sin x\ ,\ \arcsin (\sin x)\}\ \mathrm{dx}=$ $\frac{n(\pi^{2}-8)}{4}\ ;\ \lim_{n\to\infty}\left[\int_{a}^{b}(x-a)^{n}(x-b)^{n}\ \mathrm{dx}\right]^{\frac{1}{n}}=$ $\left(\frac{b-a}{2}\right)^{2}$ ,

where $a< b\ ;\ \lim_{n\to\infty}\int_{0}^{1}\frac{1}{x^{n}+x+1}\ \mathrm{dx}=$ $\ln 2\ ;\ \lim_{n\to\infty}\int_{0}^{1}\max\{x_{n},1-x\}\ \mathrm{dx}=1$ .

Proof. Soon.

$\blacksquare$ PP19. Compute $A=\int_{1}^{\infty }\frac{\ln (1+x)-\ln 2}{1+x^2}\ \mathrm{dx}$ .

Lemma. $\boxed{I=\int^1_0 \frac{\ln (x+1)}{x^2+1}\ \mathrm{dx}=\frac {\pi\ln2}{8}}\ (*)$ . Indeed, $I\equiv\int^1_0 \frac{\ln (x+1)}{x^2+1} \ \mathrm{dx}=$ $\int^1_0 \ln(1+x)(\arctan x)'\ \mathrm{dx}=$

$\int^{\frac{\pi}{4}}_0 \ln (1+\tan x)\ dx=$ $\int^{\frac{\pi}{4}}_0 \ln \left[1+\tan \left(\frac{\pi}{4} -x\right)\right] \ \mathrm{dx}=$ $\int^{\frac{\pi}{4}}_0\ln \frac{2}{1+\tan x}\ \mathrm{dx}=$ $\frac{\pi}{4}\ln 2 -I$ $\Longrightarrow I=\frac{\pi\ln 2}{8}$ .

Proof of the proposed problem. Using the substitution $x:=\frac 1x$ obtain that $A=\int_0^1\frac{\ln (1+x)-\ln x-\ln 2}{1+x^2}\ \mathrm{dx}\stackrel{(*)}{=}$

$\frac {\pi\ln2}{8}-\frac {\pi\ln2}{4}-\int_0^1\frac {\ln x}{1+x^2}\ \mathrm{dx}=$ $-\frac {\pi\ln2}{8}-J$ , where $J=\int_0^1\frac {\ln x}{1+x^2}\ \mathrm{dx}=$ $\left|\left(\ln x\cdot\arctan x\right)\right|_0^1-\int_0^1\frac {\arctan x}{x}\ \mathrm{dx}$ , where

$\lim_{x\searrow 0}\ln x\cdot\arctan x=0$ and $\int_0^1\frac {\arctan x}{x}\ \mathrm{dx}=K=\sum_{k=0}^{\infty}\frac {(-1)^k}{(2k+1)^2}$ - the Catalan's constant (<== click). Hence $\boxed{I=-\frac {\pi\ln2}{8}-K}$ .

Remark. $\int_{0}^{1} \frac{\arctan x}{x} \ \mathrm{dx} = \int_{0}^{1} \frac{1}{x}\cdot \Big( x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \ldots \Big)\ \mathrm{dx}$ $= \int_{0}^{1} \Big( 1 - \frac{x^{2}}{3} + \frac{x^{4}}{5} - \ldots \Big) \ \mathrm{dx} =$

$x - \frac{x^{3}}{9} + \frac{x^{5}}{25} - \ldots \Big|_{0}^{1}$ $= 1 - \frac{1}{9} + \frac{1}{25} - \ldots = \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^{2}}=K$ .

$\blacksquare$ PP20. $I=\int\frac {3e^x + 5\sin x + 10\cos x}{e^x + 4\sin x + 3\cos x}\ \mathrm {dx}$ and $J=\int\frac {\sin x}{e^x - \sin x - \cos x}\ \mathrm {dx}$ .

Proof. Denote $f(x)=e^x + 4\sin x + 3\cos x$ and look for real numbers $m$ , $n$ so that $3e^x + 5\sin x + 10\cos x=$ $mf(x)+nf'(x)=$

$m\left(e^x + 4\sin x + 3\cos x\right)+n\left(e^x+4\cos x-3\sin x\right)$ , i.e. $\left\{\begin{array}{c} m+n=3\\\ 4m-3n=5\\\ 3m+4n=10\end{array}\right\|\iff$ $m=2$ and $n=1$ . Therefore,

$\boxed{I=2x+\ln\left|e^x + 4\sin x + 3\cos x\right|+\mathbb C}$ . Denote $g(x)=e^x-\sin x-\cos x$ and look for real numbers $m$ , $n$

so that $\sin x =$ $mg(x)+ng'(x)=$ $m\left(e^x-\sin x-\cos x\right)+n\left(e^x-\cos x+\sin x\right)$ , i.e. $\left\{\begin{array}{c} m+n=0\\\ -m+n=1\end{array}\right\|\iff$

$m=-\frac 12$ and $n=\frac 12$ . Therefore, $\boxed{J=-\frac 12\cdot x+\frac 12\cdot\ln\left|e^x-\sin x-\cos x\right|+\mathbb C}$

$\blacksquare$ PP21. Ascertain a recurence relation for $a_n\equiv\int_0^{\frac {\pi}{2}}\cos^nx\cos nx\ \mathrm{dx}$ , where $n\in\mathbb N^*$ .

Proof. Denote $\boxed {a_n\equiv\int_0^{\frac {\pi}{2}}\cos^nx\cos nx\ \mathrm{dx}}$ . Then $2a_n = \int_0^{\frac {\pi}{2}}\left[\cos(n + 1)x + \cos(n - 1)x\right]\cos^{n - 1}x\ \mathrm{dx}$ . Thus,

$2a_n = a_{n - 1} + \int_0^{\frac {\pi}{2}}\cos^{n - 1}x(\cos x\cos nx - \sin x\sin nx)\ \mathrm{dx}$ $\implies$ $2a_n = a_{n - 1} + a_n - A$ , where $A = \int_0^{\frac {\pi}{2}}\sin nx\sin x\cos^{n - 1}x\ \mathrm {dx}$ .

Now I"ll apply the integration by parts to the definite integral $A$ : $\boxed {\ \begin{array}{ccc} u(x) = \sin nx & \implies & u'(x) = n\cos nx \\ \\ v'(x) = \cos^{n - 1}x\sin x & \implies & v(x) = - \frac 1n\cdot\cos^nx\end{array}\ }$ $\implies$ $A = a_n$ .

Therefore, $a_0 = \frac {\pi}{2}$ and for any $n\in \mathcal N^*$ , $a_{n} = \frac 12\cdot a_{n - 1}$ . In conclusion, $\boxed {\ a_n = \frac {\pi}{2^{n + 1}}\ ,\ n\in \mathcal N\ }$ .

PP22. Evaluate the definite integral $\int_0^{\frac {\pi^2}4}\frac 1{1+\sin\sqrt t+\cos\sqrt t}\ \mathrm{dt}$ .

Proof. Let $I$ be the value of the integral. Substitute $t:= t^2$ to get $I = \int_{0}^{\frac{\pi}{2}}\frac{2t}{1+\sin t+\cos t}\ \mathrm{dt}$ . Substituting $t:=\frac{\pi}{2}-t$ gives

$I = \int_{0}^{\frac{\pi}{2}}\frac{2\left(\dfrac{\pi}{2}-t\right)}{1+\sin\left(\frac{\pi}{2}-t\right)+\cos\left(\frac{\pi}{2}-t\right)}\ \mathrm{dt}$ $= \int_{0}^{\dfrac{\pi}{2}}\dfrac{\pi-2t}{1+\sin t+\cos t}\ \mathrm{dt}$ . Adding the two expressions and dividing by $2$ gives

$I = \frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\sin t+\cos t}\ \mathrm{dt}$ . Substituting $t: = \tan\frac{t}{2}$ gives $I = \frac{\pi}{2}\int_{0}^{1}\frac{1}{1+\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}} \cdot \dfrac{2}{1+t^2}\ \mathrm{dt}$ $\implies$ $\boxed{I=\frac{\pi\ln 2}{2}}$ .

PP23. Find the minimum value of $\int_0^1 \frac{|t-x|}{t+1}\ \mathrm{dt}$ , where $x\in\mathbb R$ .

Proof. Denote $F(t)\in \int\frac{|t-x|}{t+1}\ \mathrm{dt}$ for which $F(0)=0$ , i.e. $F(t)=t-(x+1)\ln|t+1|$ . Thus,

$f(x)\equiv\int_0^1 \frac{|t-x|}{t+1}\ \mathrm{dt}\implies$ $f(x)=\left\{\begin{array}{ccc} F(1)-F(0) & \mathrm{if} & x\le 0\\\\ F(0)+F(1)-2F(x) & \mathrm{if} & 0\le x\le 1\\\\ F(0)-F(1) & \mathrm{if} & x\ge 1\end{array}\right\|\iff$

$f(x)=\left\{\begin{array}{ccc} 1-(x+1)\ln 2\ \ \searrow & \mathrm{if} & x\le 0\\\\ 1-(x+1)\ln 2-2x+2(x+1)\ln (x+1)\ \ \searrow\ \nearrow & \mathrm{if} & 0\le x\le 1\\\\ (x+1)\ln 2-1\ \ \nearrow & \mathrm{if} & x\ge 1\end{array}\right\|$ ,

where $\left\{\begin{array}{c} f(0-0)=f(0+0)=1-\ln 2\\\\ f(1-0)=f(1+0)=2\ln 2-1\end{array}\right|$ . Therefore, $x_{\mathrm{min}}\in (0,1)$ .

For any $x\in (0,1)$ we have $f'(x)=\ln\frac {(x+1)^2}{2}$ and $f'(x)=$ $0\iff$ $x=\sqrt 2-1$ .

In conclusion, $x_{\mathrm{min}}=\sqrt 2-1$ and $\min_{x\in\mathbb R}f(x)=f\left(x_{\mathrm{min}}\right)=f\left(\sqrt 2-1\right)=3-2\sqrt 2$ .

PP24. Ascertain $\frac AB$ , where $A=\int_{0}^{1}({1-x^{n}})^{m}\ \mathrm{dx}\ \ ,\ \ B=\int_{0}^{1}({1-x^{n}})^{m+1}\ \mathrm{dx}$ and $\{m,n\}\subset \mathbb N^*$ .

Proof. $\left\{\begin{array}{ccc} u(x)=\left(1-x^n\right)^{m+1} & \implies & u'(x)=-n(m+1)x^{n-1}\left(1-x^n\right)^m\\\\ v'(x)=1 & \implies & v(x)=x\end{array}\right|$ $\implies$

$B=\int_{0}^{1}({1-x^{n}})^{m+1}\ \mathrm{dx}=$ $\int_0^1x\left(1-x^n\right)^{m+1}\ \mathrm{dx}+n(m+1)\cdot\int x^{n}\left(1-x^n\right)^m\ \mathrm{dx}=$

$n(m+1)\cdot\int \left[1-\left(1-x^{n}\right)\right]\left(1-x^n\right)^m\ \mathrm{dx}\implies$ $B=n(m+1)\cdot (A-B)\implies \frac AB=1+\frac {1}{n(m+1)}$ .

PP25. Ascertain $\lim_{n\to\infty}f(n)$ , where $f(n)= \int_0^{\pi} e^{x}|\sin nx|\ \mathrm{dx}\ ,\ n\in\mathbb N^*$ .

Riemann-Lebesgue lemma. Let $f$ be a continuous function on $0 \le a < b$ . If $g$ is a continuous and

$T$ -periodic on $[0,\infty)$ .Then $\lim_{n \to \infty} \int^{b}_{a} f(x) g(nx) \ \mathrm{dx} = \frac{1}{T} \cdot\int^{T}_{0} g(x) \ \mathrm{dx}\cdot \int_{a}^{b} f(x) \ \mathrm{dx}$ .

Proof

Proof 1. Can apply the upper Riemann-Lebesgue lemma : $\lim_{n \to \infty} \int_{0}^{\pi} e^{x} |\sin nx|\ \mathrm{dx} = \frac{1}{\pi} \int_{0}^{\pi} |\sin x|\ \mathrm{dx}\cdot \int_{0}^{\pi} e^{x}\ \mathrm{dx}$ $= \frac{2(e^{\pi}-1)}{\pi}$ .

Proof 2. . $f(n) = \int_{0}^\pi e^x | \sin nx |\mathrm{dx}=$ $\sum_{k=0}^{n-1} \int_{\frac {k\pi}{n} }^{\frac {(k+1)\pi}{n} } (-1)^k e^x \sin nx \ \mathrm{dx}=$ $\sum_{k=0}^{n-1} (-1)^k \left[ \frac{e^x}{1+n^2} (\sin nx - n \cos nx)\right]_{\frac {k\pi}{n} }^{\frac {(k+1)\pi}{n} }=$

$\sum_{k=0}^{n-1} (-1)^k \frac{e^{\frac {k\pi}{n} }}{1+n^2} \left(-ne^{\frac {\pi}{n} } \cos (k+1)\pi + n\cos k\pi\right)=$ $\sum_{k=0}^{n-1} \frac{n e^{\frac {k\pi}{n} }}{1+n^2} (-1)^k (\cos k\pi)(e^{\frac {\pi}{n} }+1)=$ $\frac{n(e^{\frac {\pi}{n} } + 1)}{1+n^2} \sum_{k=0}^{n-1} e^{\frac {k\pi}{n} } (-1)^{2k}=$

$\frac{n (e^{\frac {\pi}{n} } + 1)}{1+n^2} \frac{e^\pi - 1}{e^{\frac {\pi}{n} } - 1}\implies$ $\lim_{n\to\infty} f(n)=$ $\frac{e^\pi - 1}{\pi}\cdot \lim_{n\to\infty} \frac{n^2}{1+n^2}\cdot \lim_{n\to\infty} (e^{\frac {\pi}{n} } + 1) \cdot \lim_{n\to\infty} \frac{\frac {\pi}{n} }{e^{\frac {\pi}{n} } - 1}=$ $\frac{2(e^\pi - 1)}{\pi}\implies$ $\underline{\overline{\left\|\ \lim_{n \to\infty} f(n) = \frac{2(e^\pi-1)}{\pi}\ \right\|}}$ .

PP26. Find the non-constant function $f(x)$ such that $f(x)=x^2-\int_0^1 [f(t)+x]^2\ \mathrm{dt}$ .

Proof. $f(x)=x^{2}-\int_{0}^{1}[f(t)+x]^{2}\ \mathrm{dt}$ $\Longrightarrow f(x)=x^{2}-\int_{0}^{1}f^{2}(t)\ \mathrm{dt} -2x\int_{0}^{1}f(t)\ \mathrm{dt} - x^{2} \int_{0}^{1}\ \mathrm{dt}$ $\Longrightarrow$ $f(x)=-\int_{0}^{1}f^{2}(t)\ \mathrm{dt} -2x\int_{0}^{1}f(t)\ \mathrm{dt}\Longrightarrow$ exist $\{a,b\}\subset\mathbb R$

so that $f(x)=ax+b$ , i.e. $ax+b=x^{2}-\int_{0}^{1}(at+b+x)^{2}\ \mathrm{dt}\iff$ $ax+b=x^2-\frac {a^2}{3}-a(b+x)-(b+x)^2\iff$ $ax+b=-\frac {a^2}{3}-ab-ax-b^2-2bx\iff$

$\left\{\begin{array}{c} a=-a-2b\\\\ b=-\frac {a^2}{3}-ab-b^2\end{array}\right\|\iff$ $\left\{\begin{array}{c} b=-a\\\\ a=\frac {a^2}{3}\end{array}\right\|\iff$ $\left\{\begin{array}{c} b=-a\\\\ a^2=3a\end{array}\right\|\iff$ $a=-b=3$ . In conclusion, $\boxed{f(x)=3x-3}$ .

PP27. For large $n$ , show that $\int_{0}^{1}\frac{nx^{n-1}}{1+x^2}dx\approx \frac{1}{2}$ .

Proof 1. $J_n=\int_0^1 \frac{nx^{n-1}}{1+x^2}\ \mathrm{dx}=$ $\int_0^1\frac {1}{x^2+1}\cdot \left(x^n\right)'\ \mathrm{dx}=$ $\left|\frac{x^n}{1+x^2}\right|_0^1 +$ $\int_0^1 x^n\cdot \frac{2x}{\left(x^2+1\right)^2}\ \mathrm{dx}=$ $\frac 12+2\cdot \int_0^1 \frac{x^{n+1}}{(1+x^2)^2}\ \mathrm{dx}\ge \frac 12$ . On the other hand,

$J_n=\int_0^1 \frac{nx^{n-1}}{1+x^2}\ \mathrm{dx}=$ $n\cdot \int_0^1 \frac{x}{1+x^2}\cdot x^{n-2}\ \mathrm{dx}$ $\leq \frac{n}{2}\cdot \int_0^1 x^{n-2}\ \mathrm{dx}=$ $\left|\frac{n}{2}\cdot \frac{x^{n-1}}{n-1}\right|_0^1=\frac {n}{2(n-1)}$ . In conclusion, $\frac 12\le J_n\le \frac {n}{2(n-1)}$ $\implies J_n\rightarrow\frac 12$ .

Proof 2. Define $I_n=\int_0^1\frac {x^{n-1}}{1+x^2}\ \text{d}x$ , where $n\ge 1$ . Note that: $I_{n+1}-I_n=\int_0^1\frac {x^{n-1}(x-1)}{1+x^2}\ \text{d}x\le 0$ , so this means that $\left(I_n\right)_{n\ge 1}$ is decreasing.

On the other hand, one can easily derive the relation $\boxed{\, I_{n+2}+I_n=\frac 1n\, }$ and given the monotonicity of $\left(I_n\right)_{n\ge 1}$ we can now obtain the following chain

$\frac 1{n}=I_{n+2}+I_n\, \le\, 2I_n\, \le\, I_n+I_{n-2}=\frac 1{n-2}\implies$ $\boxed{\, \frac 12\, \le\, nI_n\, \le\, \frac n{2(n-2)}\, }$ . Now, by squeeze theorem it follows that $\lim_{n\to\infty}\, nI_n=\frac 12$ .

PP28. Let $f:[0,1]\rightarrow [0,\infty )$ be a function $f\in\mathbb C^1$ . Prove that the sequence $a_n=\int_0^1\frac {f(x)}{1+x^n}\ \mathrm{dx}\ ,\ n\in\mathbb N^*$ is convergent $(a_n\rightarrow l\in \mathbb R)$ and $\lim_{n\to\infty}n\left(l-a_n\right)=f(1)\ln2$ .

Proof. $0\le \int_0^1f(x)\ \mathrm{dx} -\int_0^1\frac {f(x)}{1+x^n}\ \mathrm{dx}=$ $\int_0^1\frac {x^n}{1+x^n}\cdot f(x)\ \mathrm{dx}\le$ $A\cdot \int_0^1x^n\ \mathrm{dx}=$ $\frac {A}{n+1}\ ,\ (\forall )n\in\mathbb N^*$ , where $A=\sup_{0\le x\le 1}f(x)$ . Thus, $\lim_{n\to\infty}\int_0^1\frac {f(x)}{1+x^n}\ \mathrm{dx}=$

$\int_0^1\frac {f(x)}{1+x^n}\ \mathrm{dx}=l$ . So, $n\left(l-a_n\right)=\int_0^1xf(x)\cdot \frac {nx^{n-1}}{1+x^n}\ \mathrm {dx}=$ $\left |xf(x)\cdot \ln\left(1+x^n\right)\right|_0^1-\int_0^1\left[f(x)+xf'(x)\right]\cdot \ln\left(1+x^n\right)\ \mathrm{dx}=$ $f(1)\ln 2-b_n$ ,

where $b_n=\int_0^1\left[f(x)+xf'(x)\right]\cdot \ln\left(1+x^n\right)\ \mathrm{dx} \le$ $\int_0^1\left[f(x)+xf'(x)\right]\cdot x^n\ \mathrm{dx}\le$ $A\int_0^1x^n\ \mathrm{dx}+B\cdot \int_0^1x^{n+1}\ \mathrm{dx}=\frac {A}{n+1}+\frac {B}{n+1}\rightarrow 0$ ,

where $B=\sup_{0\le x\le 1}f'(x)$ .In conclusion, $b_n\rightarrow 0$ and $\lim_{n\to\infty}n\left(l-a_n\right)=f(1)\ln 2$ .

PP29 (easy and nice). Evaluate $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{2+\sin x}{1+\cos x}\ dx$

Proof. Denote $I=\int\frac{2+\sin x}{1+\cos x}\ \mathrm{dx}=2K+L$ , where $K=\int\frac {1}{1+\cos x}\ \mathrm{dx}$ and $L=\int\frac {\sin x}{1+\cos x}\ \mathrm{dx}$ . Observe that $L=-\int\frac {(1+\cos x)'}{1+\cos x}\ \mathrm {dx}$

and $K=\int\frac {1-\cos x}{\sin^2x}\ \mathrm{dx}=$ $\int\frac {1}{\sin^2x}\ \mathrm{dx}-\int\frac {(\sin x)'}{\sin^2x}\ \mathrm{dx}\implies$ $L=-\ln (1+\cos x)+\mathbb C$ and $K=-\cot x+\frac {1}{\sin x}+\mathbb C$ .

In conclusion, $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{2+\sin x}{1+\cos x}\ \mathrm{dx}=$ $\left| -\ln (1+\cos x)-2\cot x+\frac {2}{\sin x}\right|_{\frac {\pi}{3}}^{\frac {\pi}{2}}=$ $\ln\frac 32+\frac {2}{\sqrt 3}+2-\frac {4}{\sqrt 3}=$ $\ln\frac 32+2\left(1-\frac{1}{\sqrt{3}}\right)$ .

See here , here and here

This post has been edited 180 times. Last edited by Virgil Nicula, Nov 21, 2015, 7:17 AM

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June 2013

379. Geometrical extremity I.

378. Some metrical relations.

377. Some remarkable properties of the symmedian.

May 2013

376. Trigonometrical identities.

374. Some proofs of the Law of Cosines.

373. Extension of Chebyshev's inequality.

April 2013

372. The distancies : OI , OI_a , ON a.s.o.

371. Some problems from NMO (final stage) - 2013, Romania.

February 2013

370. Some metrical relations in a triangle.

January 2013

369. Mixtilinear circles of ABC.

364. Some interesting inequalities I

December 2012

363. Geometry problems with perpendicularities.

November 2012

362. Fermat's problem and other metrical problems.

October 2012

361. Arhimedes theorem of the broken chord in a circle.

360. Some problems for the middle school.

359. Ptolemy's theorem and generalized Ptolemy's relation.

358. Some properties of the remarkable lines in a triangle.

September 2012

357. Morley theorem.

356. Order relation between s & (2R+r) in an acute triangle.

August 2012

355. Some problems with polynomials or equations I

354. Gradul II - Univ. B.B. Cluj, aug. 1998 Prof I sau II.

353. Some problems of the analytical geometry.

July 2012

352. Some definite integrals.

351. Problems with areas.

350. Another (easy) integrals.

349. Another interesting limits.

348. Subiectele de la BAC 2012 /Mate-Info.

June 2012

347. Two isogonal lines in a triangle.

346. Some easy integrals.

345. Characteristic function of a set.

344. Some trigonometry problems.

343. Problems with collinear points and concurrent lines.

May 2012

342. Exponential or logarithmic equations/inequations.

April 2012

341. Some nice and interesting "slicing" problems.

340. Geometry problems for middle school.

339. Derivatives. Rolle's function.

338. ROU National Math Olympiad 2012.

February 2012

337. Nice metrical problems.

January 2012

336. Some problems of the analytical geometry.

335. Some difficult geometric/trigonometric inequalities.

334. 2011 Japan Mathematical Olympiad Finals Problem 1.

333. Problems with the common tangent for two circles.

332. Some metrical problems from the contests I.

December 2011

331. Some nice problems with polynomials/equations I.

November 2011

330. A triangle and an its transversal through a point.

329. Some problems with complex numbers.

328. A general identities with areas in a triangle ABC.

October 2011

327. Some geometrical/algebraic extremum problems.

326. Some problems from trigonometry.

325. Without the l'Hospital's rules.

324. Some determinants.

323. Easy, nice and interesting problems for high school.

322. Incenter, Gergonne, Nagel a.s.o. of ABC.

321. Harmonical quadrilateral.

320. Some problems with metrical relations.

September 2011

319. Some problems with induction.

318. IRAN - 2011 (geometrie).

317. Some interesting geometry problems for middle school.

316. Some properties of isogonal rays in an angle of ABC.

August 2011

315. Indian Team Selection Test 2010 ST1 P1 and extension.

314. USAMTS 2009 Round 2 #5.

313. For middle school - some nice problems.

312. Very nice (famous trigonometrical problems) !

311. Some problems from USAMO and other contests.

310. Some properties of the tangential quadrilateral.

309. Beautiful problem ! IRAN Selection Test for IMO, 2004.

308. Some nice and easy properties in a triangle.

July 2011

307. Very nice proofs !

306. Some easy and nice "slicing" problems.

305. Position of the roots w.r.t. a given real number.

304. Integrals ("brothers") III.

303. Integrals II.

302. Some problems with sequencies of the "roots" .

301. Some trigonometrico-geometrical inequalities.

300. Lagrange's multiplier. Examples and problems.

299. A class of the recurrent sequencies.

298. Some nice problems from Crux Mathematicorum.

297. Nice inequalities with concrete numbers.

296. Some nice Vittasko's problems.

295. M1 3th subject, 2c - School Graduate, ROMANIA.

294. The Octav Halunga's inequality in a triangle.

June 2011

293. A difficult trigonometric equation (third point !).

292. Length of the angled bisector and some means.

291. Some integrals.

290. Characterize "OP perpendicular on OA" in ABC a.s.o.

289. Nice identities in an algebra/ geometry/trigonometry.

288. An inequality with two A-cevians in a triangle ABC.

287. Some metrical problems with or without trigonometry.

286. ABC is right triangle iff s=2R+r and other similar pp.

285. Nice and easy inequalities. For example, IMAC - 2009.

284. Ellipse (definition).

283. Some nice and easy problems.

282. Factorize of polynomials.

May 2011

281. Identity with R in an acute triangle (ILL - 1974).

280. Common roots of two polynomial equations.

279. O.M. Holland , O.M. Ireland 1988 and TST USA 2000, etc.

278. In memoria profesorului Alexandru Lupas.

277. A nice and difficult relation with Lemoine's axis.

276. Chinese TST 2009 and Second Round 2006.

275. Some interesting algebraic equations and systems.

274. An inequality in a triangle for its interior point.

273. The area of the triangle IOH. When I belongs to OH ?

272. Outside of triangle. Extension of Fermat's point.

271. Relations concerning Fermat points.

April 2011

270. Morrocan M.O. 2010 and Croatia TST 2009.

269. Some nice and easy properties in a trapezoid.

268. Saraghin contest, 2011.

267. Germany 2002, Russia 2003, IMO 2003, Aus.-Polish 1992.

266. Rectangles erected on sides of a triangle. Concurrence.

265. Really nice problem !

264. Identities with complex numbers.

263. Newton's sums. Applications.

262. IMO Shortlist 2005, Geometry 6th.

261. IMO 2009, problem 2 and JBMO 2007, problems 1 & 2.

260. Some nice, simple identities and their applications.

259. ABC equilateral triangle ==> DEF equilateral triangle.

258. Some difficult and nice problems with complex numbers.

257. Seeking roots of 2th equation with complex coef.

256. An extremum problems with complex numbers.

March 2011

255. A line which is parallelly with OH.

254. An interesting C. Mateescu's geometrical inequality.

253. An old algebraic inequality.

252. OI=f(A,R) in certain conditions.

251. Problems of Analytical Geometry.

250. An application of the Euler's relation.

249. An usual and nice metrical relations in quadrilateral.

248. Pagina instructiva.

247. An remarkable inequality with median.

246. R1-R2=OI (nice).

245. An inequality in an acute triangle.

244. Some properties of cyclic orthodiagonal quadrilaterals.

243. Nice ! (a+b) is constant.

242. A general geometrical problems.

241. Areas in a triangle.

240. Some interesting geometrical inequalities.

239. Algebraic marathon.

238. Some metrical problems.

237. Some problems with ... problems.

236. A locus with the metrical relation.

235. Proofs of some geometry problems with complex numbers.

234. Some simple and nice problems from contests.

233. An interesting identity and an easy & nice inequality.

February 2011

232. Some geometrical inequalities (own).

Pagina libera

231. A conditioned minimum value.

230. Some usual synthetical properties in a triangle.

229. Concurs online MATEFBC, ed. 5 -subiect 2.

228. OJM - Romania, 2010 problem 3.

227. IMO ShortList 2003 (problem 5).

226. Some inequalities from Calculus (Real Analysis).

225. An interesting problems from Kunny.

224. Some nice and easy inequalities in a triangle.

223. An inequality with complex numbers-RMO shortlist 2010.

222. An extension of probl. 2 from IMO 2009.

221. Some nice and simple "slicing" problems.

220. Some nice geometry problems from contest.

January 2011

219. A very nice geometrical inequality.

218. Some problems from the Suiss IMO Selection Team 2006.

217. A nice and interesting Murray S. Klamkin's problem.

216. Nice ! Find the length of hypotenuse (Belarus - 2010)

215. Limit of the unique root for polynominal equations.

214. A range of some functions.

213. Another classical "slicing" problems.

212. A simple applications of the Casey's theorem.

211. Pretty hard problem !

210. Tangential circle of the triangle ABC.

209. Nice problem, nice proof !

208. An interesting geometrical inequality.

207. Some interesting problems of Toshio Seimiya, Japan.

206. Four problems with extremum in a quadrilateral.

205. A "slicing" problem with 11 methods.

204. Saraghin 2010 (three geometry problems).

203. An application of the Pascal's permutation theorem.

December 2010

202. Concurrent lines in a right triangle.

201. Applications of Desarques' and Pappus' theorems.

200. Some nice applications of the inversion.

199. Some properties of symmedian. Two extensions.

198. Generalized Carnot's lemma.

197. Triangles outside of a triangle and concurrence.

195. Semicircle. Nice application of harmonical division.

194. The Lalescu's sequence. Properties.

193. An inequality with n! .

192. An easy and nice problem of Valentin Vornicu.

191. Six nice problems with the incircle of a triangle.

190. Very nice ! Coculescu's Contest seniors 2010.

189. Pascal's theorem and some nice and easy applications.

188. Some problems from vectorial geometry.

187. Distancies.

186. Without the l'Hospital's rules.

185. The Durrande's relation.

184. An interesting geometry problem from RMO 2010.

183. A remarkable characterization of equilateral triangle.

182. Another two nice problems with complex numbers (own).

181. Two special inequalities from CRUX (own).

180. Own proposed problem from CRUX (with complex numbers).

November 2010

179. The range of |z| , |z^2+a|=|bz+c|. Particular cases.

178. Some nice and simple geometry problems.

177. A nice proof of one identity in a triangle.

176. Another nice problems for middle school.

175. A Geometric Proof (without trig.) of Heron's Formula.

174. Some interesting properties of modulus (middle school).

173. An interesting trigonometric equations.

172. The Zelinsky's problem.

171. Another "slicing" proposed problems (middle school).

October 2010

170. A nice "slicing" proposed problem for middle school.

169. Some metrical problems in a circle (III).

168. Some properties of the Lemoine's point.

167. Nice problems - OIM, 1995 and ... Toshio Seimyia.

166. Some constant geometrical expressions.

165. Some metrical problems (5) in a circle (II).

164. Some metrical problems (5) in a circle.

163. Applications of Ptolemy's inequality. Ex. - China,1998.

162. Generalization of proposed problem from Mongolia, 2000.

161. A difficult inequality (barycentrical coordinates)

160. A very nice "slicing" problem.

159. Multiple derivatives in a point.

158. Very nice and difficult limits.

157. Steinbart's theorem. Applications.

156. Some problems with perpendicularities.

155. A constant ratio in a triangle ABC.

154. Difficult inequality (without derivatives eventually).

153. Geometrical minimum/maximum.

152. Concurrence, cyclically, metrics, symmedian a.s.o.

151. Miscellaneous problems for the admission at math.

150. Some (5) problems with fixed points.

149. Some (8) problems with collinearity or concurrence.

148. A special class of systems.

147. A problem with two polynominals.

146. Synthetic, metric, trigonometric and analitycal proofs.

145. Spain MO P5 - Point on median.

144. A conditioned concurrency in a triangle.

143. Assess of a ratio (nice and difficult problem).

142. Inscribed/circumscribed quadrilateral w.r.t. ABC.

141. N is interior <==> 1/3<a/b<3 (CRUX Mathematicorum).

140. A synthetical property <==> linear/angled relation.

139. A caracterization of a right angle in a triangle.

138. Generalized "slicing" problems (own).

137. Stewart's relation.

136. A geometrical interpretation of a system (own).

135. Three strong geometrical inequalities.

September 2010

134. Some problems with projections/perpendiculars (2).

133. Some problems with projections/perpendiculars (1).

132. Asupra unei inegalitati intr-un triunghi.

131. A stronger Form of the Steiner-Lehmus Theorem (own).

130. A triangle and a fixed point.

129. BX=CY and three properties.

128. Value of IG and two inequalities in neighbourhood of 0.

127. Pagina educativa.

126. Routh's theorem.

125. I like very much this Darij Grinberg's message.

124. A nice and easy problem with a right-angled triangle.

123. My extension and its proof.

122. An identity for an equilateral triangle. Extension.

121. Viete's relations with cos A , cos B , cos C.

120. Two equal neighbouring angles/sides in a quadrilateral.

119. Inspired by a Miculita's problem.

118. An usual remarkable geometrical locus.

117. Some applications of harmonic division/quadrilateral.

116. The length of AE (nine methods !).

115. An difficult equation with one radical.

114. A property of tangential triangle for ABC.

113. Range for a function.

112. The equivalence relation ".s.s."

111. Simple, but interesting ...

110. German TST 2004 - Exam V , Problem 1

109. Two equal areas ==> find A.

108. Old, short and nice proof of Euler's relation.

107. XY parallel BC and X,Y are isogonal/izotomic points.

106. OLM - Bucuresti (geometrie).

105. A problem with three circles.

104. Concursul Revistei de Matematica din Galati (RMG).

103. Metrical usual relations in triangle. Applications.

102. Characterization of a metrical relation in a triangle.

101. O problema cu numere complexe.

100. Bobiller's theorem.

99. Some simple and nice geometry problems.

98. Symmedian & median (metrical relations).

97. Problem of butterfly.

August 2010

96. Integrals.

95. Three circles in an angle.

94. Equations of angle bisectors (interior and exterior).

93. The Appolonius' construct problems.

92. CA , CB tangent to parabola - OIMU 2008 Problem 4.

91. JBMO 2010, Problem 3.

90. Nice and difficult inequality in ABC (own).

89. Similarity (parallel/antiparallel).

88. A nice geometric locus conditioned metrically.

87. An interesting vectorial identity.

86. A very nice maxmin.

85. Problem "slicing" : 60-24-12.

84. IRAN national math olympiad - 2010

83. A metrical characterization of concurrence.

82. AA' , BN , CM concur in Gergonne point.

81. Sawayama and Thebault’s theorem.

80. Three concurrent perpendiculars (the Carnot's lemma).

79. A quadrilateral with many perpendicularities.

78. Jakobi's theorem.

77. Two equivalent perpendicularities.

76. A parallelogram and a circle (nice !).

75. Extended Steinbart Theorem.

74. Inegalitatea Erdos-Mordell.

73. A nice problems with a square and a perpendicularity.

July 2010

72. Colinearity in a triangle - H , P , M .

70. Colinearity with pole/polar - H\in PQ .

71. An angle questions (synthetic/trigonometric proofs).

69*1. Slicing problem 3.

69. Position of x1, x2 w.r.t. a given real number k.

68. Some nice applications of "pole/polar".

67. Remarkable algebraic/geometrical inequalities (1).

66. Remarkable perpendicularities in a triangle.

65. Parallel to BC through I .

64. Problems of the type "instant" (at most one minute).

63. A nice problem with radical center (livetolove212).

62. Nice and easy problem with a right triangle.

61. IMO Shortlist 2009 (very nice problem and its proof).

60. Mediteranean M.C. 2010 Problem 3.

59. IMO 2010, problem 4.

58. A cevian and two incircles.

57. Opinii si comentarii relativ la inv. rom. II

56. A extremum problem with areas.

55. Two important circles - mixtilinear incircle/excircle.

54. Cinci grade de libertate.

53. An remarkable concurrency.

52. Harmonical division.

51*1. JBTST-3/2010 Aplicatii la proprietatea fluturelui.

51. A nice problem with perpendicularity from Jaime.

50. IMAC 2010 seniors, the first day.

49. The midpoint of a cevian in an acute triangle.

June 2010

48. An inequality in a triangle : h_a+max{b,c} ...

47. Interesting and difficult identities in a triangle.

46. Outside of a quadrilateral.

45. \cos B+\cos C=1 <==> nice property.

44. Relatia IA=IO sau IA=IM intr-un triunghi ABC etc.

43. A "slicing" problem with a parameter.

42. IMAC 2010 Problema 2.

41. ONM 2010 Calarasi Problema 3.

May 2010

40. Lema lui Haruki.

39. A nice and remarkable problem with Brocard's point.

38. Minimum area of triangle touching with semicircle.

37. Own extension of the Steiner-Lehmus' problem.

36. ONM (juniori) - 2010 Iasi, problema 4.

35. Nice problem from Arqady (Eduard_Tur).

34. IMO Shortlist 2002 geometry problem 7.

April 2010

32. Linkuri diverse.

31.Some nice metrical problems.

29. Inegalitate integrala de la Kunny.

28. A fost odată ...

27.Problem of Darij Grinberg (I - centroid of triangle DEF).

26. Outside of a triangle.

25. Geometric equations.

24. A fine and cool inequalities.

23. A very nice exponential inequality (own - from CRUX).

22. Some interesting limits (sequence/function).

21 bis. Some problems with complex numbers.

21. Probleme de geometrie (Yetty & I).

20. O ineg.cu suma de AG/AI .

19. Own proposed problems in CRUX Mathematicorum - Canada.

18. O problema a lui Toshio Seimiya.

17. O inegalitate "tare" intr-un triunghi.

16. Length of the (interior) angle bisector (10 proofs).

15. Inequalities stronger than Panaitopol's.

14. Opinii si comentarii relativ la inv. rom.

13. Inegalitatea de la Sorin Borodi.

12. Inegalitatea I. Maftei & S. Radulescu (2).

11. Inegalitatea I. Maftei & S. Radulescu (1).

10. Walker's inequality and its extension.

9. Inegalitatea HO+HA+HB+HC >= 3R .

8. The first problem Shortlist ONM 2010 Romania.

7. Some interesting "slicing" problems.

6. Interesting problems from JBMO.

5. Theoretical preliminary for inequalities.

4. Remarkable geometrical inequalities (2).

Submit

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blog
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long blog
by MrMustache, Nov 11, 2020, 4:52 PM
372554 views!
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wow... i am lost.

369302 views!

-piphi
by piphi, Jun 10, 2020, 11:44 PM
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by CSPAL, May 27, 2020, 4:17 PM
impressive
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by OlympusHero, May 14, 2020, 8:43 PM
Nice blog.
The element of Mathematics is dense here.
I love the style of your posts.
Thanks for writing the blog! I can learn Geometry from here too.
by epsilonist, Jun 27, 2011, 6:00 AM
Thank you very much for this blog.
by Affine, May 7, 2011, 6:20 PM
You seem to have solved every existing problem.
by Goutham, May 2, 2011, 10:59 AM
I love your posts, man (those which I understand)! It's interesting how you punish integrals with recurrence relations and sometimes make seemingly unrelated integrals evaluate each other!
by Caspar, Apr 11, 2011, 2:36 PM
Thank you a lot dear Virgil for an interesting and instructive blog.
by vittasko, Mar 12, 2011, 9:42 AM
Interesting Blog
by ShahinBJK, Mar 12, 2011, 5:07 AM
Interesant, d-le Nicula.
by silent_control, Mar 3, 2011, 6:00 PM
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by Goutham, Jan 30, 2011, 2:06 PM
15001th view.
by fries4guys, Jan 20, 2011, 4:57 PM
Thanks to all !
by Virgil Nicula, Dec 11, 2010, 8:32 PM
El_Ectric:
You can either copy some parts from the blog post, paste it at "Search Blogs" (which is lower on the page, after "Blog Stats") and find the whole post.

You're welcome.
by Thomas_, Dec 8, 2010, 6:42 PM
Thanks Thomas_.
by El_Ectric, Dec 8, 2010, 4:21 PM
El_Ectric: Try to open this blog with Internet Explorer (browser). If it still doesn't work, just copy the post you're interested in, and than paste it into Word Document, or almost any kind of reading documents and you'll have all the details.

Nice blog!
by Thomas_, Nov 24, 2010, 4:59 PM
Dear Virgil !

Congratulations for this blog !
Babis stergiou - Greece
by stergiu, Nov 12, 2010, 6:15 AM

72 shouts

My (math)blog

303. Integrals II.

by Virgil Nicula, Jul 26, 2011, 6:37 PM

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