Wali's beta funtion

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Source: CMI 2019 #3

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#1 h Feb 22, 2020, 4:40 AM • 2 Y

Y by Gerninza, Adventure10

Evaluate the following integral

$\int_0^{\infty} (1+x^2)^{-(m+1)} dx$ .

Where $m\in \mathbb{N}$ .

This post has been edited 1 time. Last edited by ftheftics, Feb 22, 2020, 4:41 AM
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#2 h Feb 22, 2020, 4:54 AM • 1 Y

Y by Adventure10

Method $1$ : Substitute $x=\tan\theta$ and use Beta function (involves memorizing formulae and lots of stuff), approach for a high-schooler.

Method $2$ : Note that the integrand is even function, so $\int_0^{\infty} (1+x^2)^{-(m+1)} dx=\frac{1}{2}\int_{-\infty}^{\infty} (1+x^2)^{-(m+1)} dx$ . So we consider an semicircular contour on the upper half plane and using Cauchy residue theorem we have $\int_{-\infty}^{\infty} (1+x^2)^{-(m+1)} dx=\frac{2\pi i}{m!}\lim_{z\to i}\frac{d^m}{dz^m}\bigg(\frac{1}{(z+i)^{m+1}}\bigg)=\frac{(2m)!}{2^{2m}(m!)^2}\pi$ Finally dividing both sides by $2$ we have that $\int_0^{\infty} (1+x^2)^{-(m+1)} dx=\frac{(2m)!}{2^{2m}(m!)^2}.\frac{\pi}{2}$ which is the most simplified form possible.

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#3 h Feb 22, 2020, 4:58 AM • 2 Y

Y by Gerninza, Adventure10

am_11235... wrote:

Method $1$ : Substitute $x=\tan\theta$ and use Beta function (involves memorizing formulae and lots of stuff), approach for a high-schooler.

Method $2$ : Note that the integrand is even function, so $\int_0^{\infty} (1+x^2)^{-(m+1)} dx=\frac{1}{2}\int_{-\infty}^{\infty} (1+x^2)^{-(m+1)} dx$ . So we consider an semicircular contour on the upper half plane and using Cauchy residue theorem we have $\int_{-\infty}^{\infty} (1+x^2)^{-(m+1)} dx=\frac{2\pi i}{m!}\lim_{z\to i}\frac{d^m}{dz^m}\bigg(\frac{1}{(z+i)^{m+1}}\bigg)=\frac{(2m)!}{2^{2m}(m!)^2}\pi$ Finally dividing both sides by $2$ we have that $\int_0^{\infty} (1+x^2)^{-(m+1)} dx=\frac{(2m)!}{2^{2m}(m!)^2}.\frac{\pi}{2}$ which is the most simplified form possible.

Actually cauchy counter Integral is not applicable for high school competition.but ur method 1 is as similar as mine .

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#4 h Feb 22, 2020, 5:25 AM • 2 Y

Y by Gerninza, Adventure10

Consider, $I_m = \int_0^{\infty} (1+x^2)^{-(m+1)} dx$
Let's substitute $x=\tan a$ .

So our integral become $I_m =\int_0 ^{\frac{\pi}{2}} \cos ^{2m} a da$ .

DEFINITION ,

$\int_0 ^{\frac{\pi}{2}} \cos ^n x dx$ Can be define as below,( $n\in \mathbb{N}$ ).

$= \frac{(n-1)!!}{n!!}$ for $n$ is odd .
$=\frac{(n-1)!!}{n!!} \frac{\pi}{2}$ for $n$ is even.

We get $I_m = \frac{(2m-1)!!}{(2m)!!} .\frac{\pi}{2}$ .
So,

$\boxed{l_m =\frac{(2m)! }{2^{2m } (m!)^2} }$ .

This post has been edited 5 times. Last edited by ftheftics, Mar 2, 2020, 3:58 AM
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#5 h Feb 22, 2020, 5:26 AM • 2 Y

Y by Gerninza, Adventure10

See Beta funtion .

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3ch03s

#6 h Feb 22, 2020, 5:57 AM • 1 Y

Y by Adventure10

ftheftics wrote:

Consider, $I_m = \int_0^{\infty} (1+x^2)^{-(m+1)} dx$
Let's substitute $x=\tan a$ .

So our integral become $I_m =\int_0 ^{\frac{pi}{2}} \cos ^{2m} a da$ .

You can find a recurrence relation using integration by parts and then deduce the same result. this is usually most calculus-friendly approach and we dont need to use beta or gamma functions since $m$ is natural.

Regards
Claudio.

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#7 h Feb 22, 2020, 6:32 AM • 2 Y

Y by Gerninza, Adventure10

3ch03s wrote:

ftheftics wrote:

Consider, $I_m = \int_0^{\infty} (1+x^2)^{-(m+1)} dx$
Let's substitute $x=\tan a$ .

So our integral become $I_m =\int_0 ^{\frac{pi}{2}} \cos ^{2m} a da$ .

You can find a recurrence relation using integration by parts and then deduce the same result. this is usually most calculus-friendly approach and we dont need to use beta or gamma functions since $m$ is natural.

Regards
Claudio.

Yes of course indeed one have to write the recurring series in such exam.

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#8 h Feb 22, 2020, 3:09 PM • 3 Y

Y by Gerninza, Adventure10, Mango247

$I_m =\int_0 ^{\frac{\pi}{2}} \cos ^{2m} a da$ .

$I_m =\int_0^{\frac{\pi}{2}} \cos ^{2m-2} a (1-\sin ^2 a) .da$ .

$I_m =I_{m-1} -\int_0^{\frac{\pi}{2}} \cos ^{2m-2} a .\sin ^2 a .da$ .

$I_m =I_{m-1} -\frac{I_m}{2m-1}$ .

$\implies I_m \frac{2m}{2m-1} =I_{m-1}$ .

Get, $\frac{I_m}{I_{m-1}}=\frac {2m-1}{2m}$ .

$\implies \frac{I_m}{I_1} = \frac{(2m-1)....3}{2m.(2m-2).....4}$ .

Clearly,
$I_m = \frac{(2m-1)!!}{(2m)!!} .\frac{\pi}{2}$ .

This post has been edited 1 time. Last edited by ftheftics, Feb 22, 2020, 3:10 PM
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