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

by Virgil Nicula, Jul 5, 2011, 7:41 PM

PP. Denote $I_n=\int_1^2(x-1)^n(x-2)^n\ \mathrm{dx}$ , where $n\in\mathbb N$ . Prove that $(4n+2)\cdot I_n+n\cdot I_{n-1}=0$ .

Proof. I"ll use the partial integration :

$\left\{\begin{array}{ccc} u(x)=(x-1)^n(x-2)^n & \implies & u'(x)=n(x-1)^{n-1}(x-2)^{n-1}(2x-3)\\\\ v'(x)=1 & \implies & v(x)=x\end{array}\right\|$

$I_n=\left\|x(x-1)^n(x-2)^n\right\|_1^2-n\cdot $ $\int_1^2x(2x-3)(x-1)^{n-1}(x-2)^{n-1}\ \mathrm {dx}$ $\implies$ $I_n=-n\cdot\int_1^2\left(2x^2-3x\right)(x-1)^{n-1}(x-2)^{n-1}\ \mathrm {dx}$ .

Observe that $2x^2-3x=$ $2\left(x^2-3x+2\right)+3x-4=$ $2(x-1)(x-2)+\frac 32\cdot (2x-3)+\frac 12$ . Therefore,

$I_n=-n\cdot\left[2\cdot\int_1^2(x-1)^n(x-2)^n\ \mathrm{dx}+\frac 32\cdot\int_1^2(2x-3)(x-1)^{n-1}(x-2)^{n-1}\ \mathrm{dx}+\frac 12\cdot\int_1^2(x-1)^{n-1}(x-2)^{n-1}\ \mathrm{dx}\right]$ $\implies$

$I_n=-2n\cdot I_n-\frac {3n}{2}\cdot \int_1^2\left(x^2-3x+2\right)^{n-1}\left(x^2-3x+2\right)'\mathrm{dx}-$ $\frac n2\cdot I_{n-1}$ $\implies$ $(4n+2)\cdot I_n+n\cdot I_{n-1}=$

$-\frac {3n}{2}\cdot\left|\frac{\left(x^2-3x+2\right)^n}n\right|_1^2=0$ $\implies$ $(4n+2)\cdot I_n+n\cdot I_{n-1}=0$ $\implies $ $\boxed{\ I_n=-\frac {n}{2(2n+1)}\cdot I_{n-1}\ }$ .

Remark. Using the recurence relation $I_n=-\frac {n}{2(2n+1)}\cdot I_{n-1}\ ,\ n\in\mathbb N$ , where $I_0=1$ obtain that

$\prod_{k=1}^{n}I_{k}=\prod_{k=1}^n\frac {-k}{2(2k+1)}\cdot I_{k-1}\implies$ $I_n=\frac {(-1)^n\cdot n!}{2^{n}\cdot 3\cdot 5\cdot 7\cdot\ldots\cdot (2n-1)(2n+1)}\implies$ $\boxed {\ I_n=\frac {(-1)^n\left(n!\right)^2}{(2n+1)!}\ }$ .

An easy extension. Let $I_n=\int_a^b(x-a)^n(x-b)^n\ \mathrm{dx}$ , where $a< b$ and $n\in\mathbb N$ . Prove that $\left\{\begin{array}{c} I_n=-\frac {n(a-b)^2}{2(2n+1)}\cdot I_{n-1}\ ,\ n\ge 1\\\\ I_n=\frac {(-1)^n(n!)^2(a-b)^{2n}}{(2n+1)!}\end{array}\right\|$

and $|a-b|<2\implies$ $\lim_{n\to\infty}I_n=0$ because $\lim_{n\to\infty}\left|\frac {I_{n+1}}{I_n}\right|=\lim_{n\to\infty}\frac {(n+1)^2(a-b)^2}{(2n+2)(2n+3)}=\left(\frac {a-b}{2}\right)^2<1$ .

Remark. Prove easily that $\lim_{n\to\infty}\ \sqrt [n]{\int_a^b(x-a)^n(x-b)^n\ \mathrm{dx}}=\left(\frac {b-a}{2}\right)^2$ , where $0<a<b$ .
This post has been edited 45 times. Last edited by Virgil Nicula, Nov 21, 2015, 7:59 AM

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  • impressive :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

72 shouts
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