# Evaluate $\int_{0}^{∞}(1+x^2)^{−(m+1)}dx$ where m is a natural number.

QUESTION: Evaluate $$\int_{0}^{∞}(1+x^2)^{−(m+1)}dx$$ where m is a natural number.

MY ANSWER: Observing the question, the first substitution I made is changing $$x$$ to $$tan\theta$$, since $$1+tan^2\theta=sec^2\theta$$, this should lead me somewhere.

Now, changing the limits of the integration and after some easy calculation, I arrive at- $$\int_{0}^{\frac{π}2}(cos\theta)^{2m}d\theta$$ Since, we do not know that value of $$m$$ how do we solve such an integration?

I am stuck here. Any answers or alternative methods are much appreciated. Thank you so much.

• Integration by parts of the original integral (I mean indefinite integral, without limits) leads to a reccurent formula. $\cos^m x$ could be linearized using $\cos x=\frac{e^{ix}+e^{-ix}}{2}$ and the binomial. Don't know which way is esaier.) May 24 '20 at 8:29
• This page on Wallis Integrals should be useful: en.wikipedia.org/wiki/Wallis%27_integrals. In particular, see the formula for $W_{2p}$ on that page. May 24 '20 at 8:31
• @AlexeyBurdin AKA the reduction formula for $\cos^n(x)$ yea it's very useful May 24 '20 at 8:31

Using the fact that $$\cos (x) = (e^{ix} + e^{-ix})/2$$, expand your integrand to obtain $$\frac{2^{-2m}}{4}\sum_{k=0}^{2m} {2m \choose k}\int_0^{2\pi} e^{i2kx} e^{-i2mx}\,dx.$$ But the orthogonality of the complex exponentials on the interval $$[0, 2\pi]$$ immediately tells us that the integral is proportional to the kronecker delta as $$2\pi\delta_{m, k}$$. This your integral reduces to $$\pi\,2^{-2m+1}{2m \choose m}.$$

• What do you mean by the kronecker delta?.. can you provide some reference?.. and one more, what is orthogonality of complex exponentials..? May 24 '20 at 10:19
• I read about the kronecker delta.. it's a brilliant process to solve this sum.. just tell me one thing, the summation running from $k=0$ to $k=2m$ for $\binom{2m}k$ simply results in $2^{2m}$.. how did you write it as $\binom{2m}m$ ? May 24 '20 at 10:49
• I wish to accept your answer.. can you please resolve the last doubt?.. I will be much obliged... May 24 '20 at 16:15
• The kronecker delta $\delta_{m,k}$ is $0$ if $k \neq m$, and $1$ if $k = m$. Hence, the only term that survives in the sum $\sum_{k=0}^{2m} {2m \choose k} 2\pi \delta_{m,k}$ is when $k=m$, giving $2\pi {2m \choose m}$ for the summation. May 24 '20 at 22:25
• Thank you so much. You helped me a lot.. May 25 '20 at 8:27

In terms of the Beta function,$$\int_0^{\pi/2}\cos^{2m}\theta d\theta=\frac12\operatorname{B}\left(m+\frac12,\,\frac12\right)=\frac{\Gamma\left(m+\frac12\right)\Gamma\left(\frac12\right)}{2\Gamma(m+1)}=\frac{(2m)!}{m!^22^{2m+1}}\pi.$$

Denote

$$I_m= \int_{0}^{\infty}\frac{1}{(1+x^2)^{m+1}}\ dx$$

You have $$I_0= \frac{\pi}{2}$$ and with integration by parts

\begin{aligned}I_m &= \int_{0}^{\infty}\frac{1+x^2}{(1+x^2)^{m+2}}\ dx = I_{m+1} + \int_{0}^{\infty}\frac{x^2}{(1+x^2)^{m+2}}\ dx\\ &=I_{m+1}+\left[x \left(\frac{-1}{2m+2}\frac{1}{(1+x^2)^{m+1}}\right)\right]_0^\infty + \frac{1}{2m+2} I_m \end{aligned}

Therefore the relation

$$I_{m+1} = \frac{2m+1}{2m+2} I_m$$

From there you can compute $$I_m$$.