# Can $\int_{-\infty}^{\infty} C(1+x^2)^{-m} \,\mathrm dx$ be evaluated without trig substitution?

The question is to find the parameter $C$ such that $f(x) = C(1+x^2)^{-m}$
is a probability density function of a continuous random variable.
So I need to show that: $$\int_{-\infty}^{\infty} C(1+x^2)^{-m}\,\mathrm dx = 1$$ Is there another way to integrate this besides the painful trigonometric substitution?
I am wondering if I should just find the values of $m$ such that the integral converges
and say that $C$ is the multiplicative inverse of the integral.
Maybe I am overlooking something. A clue is sufficient. Thanks ....

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First note that, we can evaluate the integral using the beta function

$$\int_{-\infty}^{\infty}\frac{dx}{(1+x^2)^m}= 2\int_{0}^{\infty}\frac{dx}{(1+x^2)^m}={\frac {\sqrt {\pi }\,\Gamma \left( m-\frac{1}{2} \right) }{\Gamma \left( m \right) }},$$

where $\Gamma(n)$ is the gamma function. Now, you can see that

$$C = {\frac {\,\Gamma \left( m \right) }{\sqrt {\pi }\,\Gamma \left( m-\frac{1}{2} \right)}}.$$

See here for how to evaluate the integral. You can use the change of variables

$$t=\frac{1}{1+x^2}$$

to related to the beta function.

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TY TY TY! Very timely review for me. – Andy Tam Jun 19 '13 at 6:28
@AndyTam: It is a good exercise. – Mhenni Benghorbal Jun 19 '13 at 6:30

Here is an elementary solution based on partial integration.

Put \begin{eqnarray}I_m &=&\int_{-\infty}^\infty\frac{dx}{(1+x^2)^m}=\left[\frac{x}{(1+x^2)^m}\right]_{-\infty}^\infty-(-2m)\int_{-\infty}^\infty\frac{x^2dx}{(1+x^2)^m}\\ &=& 0+2m\int_{-\infty}^\infty\frac{(x^2+1)dx}{(1+x^2)^m} -2m\int_{-\infty}^\infty\frac{1\cdot dx}{(1+x^2)^m}=2mI_m-2mI_{m+1} \end{eqnarray} In other words $$I_{m+1}=\frac{2m-1}{2m}I_{m}$$ and since $I_1=\pi$ (by direct evaluation of the arctan derivative) we get $$I_2=\frac{1}{2}\pi,\,I_3=\frac{1}{2}\frac{3}{4}\pi,\,\ldots,\,I_m=\frac{1}{2}\frac{3}{4}\frac{5}{6}\cdots\frac{2m-1}{2m}\pi$$

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To add another approach: For $t>0$ $$\int_{-\infty}^\infty {1\over t+x^2}\,dx = {1\over \sqrt{t}}\int_{-\infty}^\infty {1\over 1+x^2}\,dx = {\pi\over \sqrt{t}}. \tag{1}$$ Differentiating both sides $m-1$ times with respect to $t$ gives $$(-1)^m (m-1)!\int_{-\infty}^\infty{1\over (t+x^2)^m}\,dx = \left(-{1\over2}\right)\left(-{3\over2}\right)\cdots\left(-{2m-1\over2}\right){\pi\over t^{1/2+m}},$$ and setting $t = 1$ now allows you to evaluate the integral.

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Differentiation is a powerful tool. +1 – AD. Jun 22 '13 at 11:14
@AD. It is! I think yours would be the most natural method for me though (+1 to you as well). – Nick Strehlke Jun 22 '13 at 17:08