CMI Entrance 19#3

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CMI Entrance 19#3

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definite integrals

Recursion of integrals

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#1 h Oct 31, 2019, 6:07 PM • 1 Y

Y by Adventure10

Evaluate $\int_{ 0 }^{ \infty } ( 1 + x^2 )^{-( m + 1 )} \mathrm{d}x$ where $m \in \mathbb{N}$

This post has been edited 1 time. Last edited by bubu_2001, Jun 29, 2021, 2:58 PM

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#2 h Oct 31, 2019, 6:28 PM • 2 Y

Y by Adventure10, Mango247

bubu_2001 wrote:

Evaluate $\int_{ 0 }^{ \infty } ( 1 + x^2 )^{-( m + 1 )} dx$ where $m \in \mathbb{N}$

Put $x=\tan(\theta)$ Substitution possible as $\tan(\theta)$ is increasing bijective function, and we will use beta- gamma function.

$\int_{0}^{\frac{\pi}{2}}(1+\tan^2(\theta))^{-m-1}\sec^2(\theta)d\theta$

= $\int_{0}^{\frac{\pi}{2}}\cos^{2m}(\theta)d\theta$

Now pure bash by $gamma$ function using the formulae

$\int_{0}^{\frac{\pi}{2}}\sin^{2m-1}(x)\cos^{2n-1}(x)dx =\frac{1}{2}\beta(m,n)=\frac{\gamma(m)\gamma(n)}{\gamma{(m+n)}}$
And $\gamma\bigg({\frac{1}{2}}\bigg)=\sqrt{\pi}$

and get the answer as

$\frac{\pi(2m-1)(2m-3)(2m-5).....(1)}{2(m!)}$

This post has been edited 9 times. Last edited by Vietjung, Oct 31, 2019, 7:17 PM

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#3 h Nov 15, 2019, 7:37 AM • 2 Y

Y by ayan_mathematics_king, Adventure10

I have an alternate solution. I haven't used gamma function.

By absolutely identical substituition, arrive at the integral $I_m=\int_{0}^{\frac{\pi}{2}}\cos^{2m}(\theta)d\theta$ . Rewrite the integral as $\int_{0}^{\frac{\pi}{2}}\cos^{2m-1}(\theta)\cos\theta d\theta$ . Now integrate by parts:
$\displaystyle I_m=A+ (2m-1)\int_{0}^{\frac{\pi}{2}}\cos^{2m-2}(\theta)\sin^{2}\theta d\theta$ where $A$ is the definite integral from $0$ to $\frac{\pi}{2}$ of the function $\cos^{2m-1}\theta \sin\theta$ . Calculate it to get the answer as $0$ and thus the original integral becomes
$\displaystyle I_m=(2m-1)\int_{0}^{\frac{\pi}{2}}\cos^{2m-2}(\theta)\sin^{2}\theta d\theta$ $\displaystyle I_m=(2m-1)\int_{0}^{\frac{\pi}{2}}\cos^{2m-2}(\theta)(1-cos^{2}\theta) d\theta$ $\displaystyle I_m=(2m-1)\int_{0}^{\frac{\pi}{2}}\cos^{2m-2}(\theta)-\cos^{2m}\theta d\theta$ $\displaystyle I_m=(2m-1)\int_{0}^{\frac{\pi}{2}}\cos^{2m-2}(\theta) d\theta -(2m-1)\int_{0}^{\frac{\pi}{2}}\cos^{2m}\theta d\theta$

This leads to the recursion: $I_m=(2m-1)I_m-(2m-1)I_{m-1}$ . Simplify this to obtain :
$\boxed{\frac{I_m}{I_{m-1}}=\frac{2m-1}{2m}}. \square$ Routine substitutuion leads to the cancelletion, quite similar to the telescopic technique for summation. We get the final answer as : $I_m=\frac{\pi}{2}.\frac{(2m)!}{2^{2m}(m!)^2}$ .

PS

This post has been edited 7 times. Last edited by The_Maitreyo1, Jan 16, 2020, 1:35 PM

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#4 h Nov 15, 2019, 4:05 PM • 2 Y

Y by Adventure10, Mango247

See also https://artofproblemsolving.com/community/u209000h1943826p13407143

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#5 h Feb 29, 2020, 3:07 PM

Y by

See https://artofproblemsolving.com/community/q7h2013361p14120030

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#6 h Feb 29, 2020, 5:37 PM • 3 Y

Y by ashrith9sagar_1, tmleyncodes, Mango247

We can do it using the Residue Theorem too. The problem offers itself for complexification. First of all observe that $\int_{0}^{\infty}\frac{dx}{(1+x^2)^{m+1}}=\frac{1}{2}\int_{-\infty}^{\infty}\frac{dx}{(1+x^2)^{m+1}}$ Now turn our attention to the complex contour integral $\int_{\Gamma}\frac{dz}{(1+z^2)^{m+1}}$ where $\Gamma$ is the rectifiable closed path from $-R$ to $R$ through a straight line segment and from $R$ back to $-R$ on the upper semicircle centered at the origin and has radius $R$ . Wlog we might assume that $R>1$ in order the pole $z=i$ would locate in the semicircle. Now applying the Residue Theorem we have
$2\pi i \text{Res(i)}=\int_{\Gamma}\frac{dz}{(1+z^2)^{m+1}}=\int_{-R}^{R}\frac{dx}{(1+x^2)^{m+1}}+\int_{\gamma_R}\frac{dz}{(1+z^2)^{m+1}}\ \ \ \ \ \ \ (*)$ where $\gamma_R$ is the upper semicircle part of $\Gamma$ oriented counter clockwise. We need two results here:

a.) $\text{Res(i)}=-\frac{i}{2^{2m+1}}\binom{2m}{m}.$
In order to work out the residue at $z=i$ we rewrite the function into $\frac{1}{(1+z^2)^{m+1}}=\frac{1}{(z-i)^{m+1}(z+i)^{m+1}}.$ What we need that is the $m^{th}$ coefficient of the Taylor expansion of $\frac{1}{(z+i)^{m+1}}$ around $z=i.$ That is given by $\left[\frac{d^m}{dz^m}\frac{1}{(z+i)^{m+1}}\right]_{z=i}=-\frac{i}{2^{2m+1}}\binom{2m}{m}.$

b.) $\int_{\gamma_R}\frac{dz}{(1+z^2)^{m+1}}\rightarrow 0\ \ \text{if}\ \ R\rightarrow \infty$

Let parametrize $\gamma_R$ in the usual way setting $\gamma_R(t)=Re^{it},\ \ 0\le t\le \pi$ then we have $dz=Rie^{it}dt$ and
$\left|\int_{\gamma_R}\frac{dz}{(1+z^2)^{m+1}}\right|=\left|\int\limits_{0}^{\pi}\frac{Rie^{it}dt}{(1+R^2e^{2it})^{m+1}}\right|\le\frac{\pi R}{(R^2-1)^{m+1}}\rightarrow 0\ \ \text{if}\ \ R\rightarrow\infty$ We used here the ML inequality and the facts that $|Rie^{it}|=R$ , $|1+R^2e^{2it}|=\sqrt{1+2R\cos 2t+R^4}\ge \sqrt{(1-R^2)^2}=R^2-1.$

By virtue of a.), b.) and (*) we have $\int\limits_{0}^{\infty}\frac{dx}{(1+x^2)^{m+1}}=\frac{1}{2}2\pi i \left[ -\frac{i}{2^{2m+1}}\binom{2m}{m}\right]=\frac{\pi}{2^{2m+1}}\binom{2m}{m}.$

This post has been edited 1 time. Last edited by Leonmedes, Feb 29, 2020, 5:40 PM
Reason: typo

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R-sk

#7 h Feb 12, 2021, 4:50 PM • 3 Y

Y by Mango247, Mango247, Mango247

Pretty easy integral I have done it from college mathematics journal so we just need to substitute x=tantheta then on applying wallis product we get
required answer

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#8 h Apr 19, 2021, 3:37 PM

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Alternatively, the substitution $u = (1+x^2)^{-1}$ , which gives $du = -2x(1+x^2)^{-2}\,dx$ , yields
$\int_0^\infty(1+x^2)^{-(m+1)}\,dx = \int_0^1 \frac 12u^{m-1}\sqrt{\frac 1u - 1}\,du = \frac12\int_0^1u^{m-3/2}(1-u)^{-1/2} = \frac{\Gamma(m-\frac12)\Gamma(\frac12)}{2\Gamma(m)}.$ Now use $\Gamma(z+1) = z\Gamma(z)$ , $\Gamma(m) = (m-1)!$ , and $\Gamma(\tfrac12) = \sqrt\pi$ to finish.

This post has been edited 2 times. Last edited by djmathman, May 12, 2022, 5:17 PM
Reason: typo

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ayan_mathematics_king

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ayan_mathematics_king

#9 h Jul 22, 2021, 6:14 AM • 1 Y

Y by amar_04

Please find the mistake in my solution:
We have $I_m=\int_0^{\frac{\pi}{2}} {\cos(u)}^{2m} du =\int_0^{\frac{\pi}{2}} {\cos(u)}^{2(m-1)}\cdot \cos^{2} u du =\int_0^{\frac{\pi}{2}} {\cos(u)}^{2(m-1)} - \int_0^{\frac{\pi}{2}} {\cos(u)}^{2(m-1)} \sin^2 u du=I_{m-1}- J_{m-1}$
Where $J_{m-1}=\int_0^{\frac{\pi}{2}} {\cos(u)}^{2(m-1)} \sin^2 u du=-\frac{1}{2m-1}$ .
So, $I_m-I_{m-1}= \frac{1}{2m-1}$
Adding these gives us,
$I_m=\frac{\pi}{2} + \frac 13 +\frac 15+....+\frac{1}{2m-1}$ since $I_1=\frac{\pi}{2}$

@before Ohh ya thanks...I made a mistake in substitution :blush:

:blush:

This post has been edited 2 times. Last edited by ayan_mathematics_king, Jul 23, 2021, 7:12 AM

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#10 h Jul 23, 2021, 3:03 AM

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ayan_mathematics_king wrote:

$J_{m-1}=\int_0^{\frac{\pi}{2}} {\cos(u)}^{2(m-1)} \sin^2 u du=-\frac{1}{2m-1}$

This is obviously incorrect. For $m=2$ , $0<\int_0^{\frac{\pi}{2}} {\cos^2(u)} \sin^2 u du=-\frac{1}{2\times 2-1} < 0$ NOTE

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lifeismathematics

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lifeismathematics

#11 h Apr 21, 2024, 10:56 AM

Y by

$I=\int_{0}^{\infty} (1+x^2)^{-(m+1)} dx$ , $m \in \mathbb{N}$

we substitute $x=\tan{\theta}$ so we get $I=\int_{0}^{\frac{\pi}{2}} \left(\sec^2{\theta}\right)^{-m} d\theta$

$I_{m}:=\int_{0}^{\frac{\pi}{2}} \left(\cos(\theta)\right)^{2m} d\theta$

$I_{m}=\int_{0}^{\frac{\pi}{2}} \cos{\theta}\cdot \cos{\theta}^{2m-1} d\theta$ ,

Integrating by parts we get:
$\displaystyle (\cos^{2m-1}{\theta}\cdot \sin{\theta})|_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} (2m-1)\cos^{2m-2}{\theta}\cdot (-\sin^{2}{\theta})d\theta$ so we get:

$I_{m}=2(m-1)I_{m-1}-(2m-1)I_{m} \implies I_{m}=\frac{2m-1}{2m}\cdot I_{m-1}$ ,

Now solving this recursively we finally get $\boxed{I_{m}=\binom{2m}{m} \cdot \frac{\pi}{2^{2m+1}}}$ . $\square$

This post has been edited 9 times. Last edited by lifeismathematics, Apr 21, 2024, 11:03 AM

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#12 h Mar 21, 2025, 4:23 PM

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storage, same as aboves

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#13 h Mar 21, 2025, 5:17 PM • 1 Y

Y by Hairy_Ball_Theorem

nicer alt

This post has been edited 1 time. Last edited by quasar_lord, Yesterday at 1:30 PM

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