MHB Evaluate the integral $\displaystyle \int_0^{\infty}\frac{dx}{(1+x^2)^{\alpha/2}}$ for $\alpha>1$

[Ackbach]

Here is this week's POTW:

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Evaluate the integral $\displaystyle \int_0^{\infty}\frac{dx}{(1+x^2)^{\alpha/2}}$ for $\alpha>1$. Express your answer using Gamma functions, where
$$\Gamma(x) :=\int_{0}^{\infty}t^{x-1}e^{-t} \, dt.$$

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[Euge]

Re: Problem Of The Week # 282 - Sep 25, 2017

No one answered this week’s problem. You can read my solution below.

Spoiler

Let $I(\alpha)$ represent the integral. By the trig substitution $x = \tan \theta$, $I(\alpha)$ becomes $\int_0^{\pi/2} \cos^{\alpha - 2}(\theta)\, d\theta$. Multiplying $I(\alpha)$ by $\Gamma(\alpha/2)$ yields the double integral

$$\int_0^\infty \int_0^{\frac{\pi}{2}} r^{\frac{\alpha}{2}-1}e^{-r}\cos^{\alpha-2}(\theta)\, d\theta\, dr$$

which can also be written

$$\int_0^\infty \int_0^{\frac{\pi}{2}} e^{-r\cos^2(\theta)} e^{-r\sin^2(\theta)} (r\cos^2(\theta))^{\frac{\alpha-3}{2}} (r\sin^2(\theta))^{-\frac{1}{2}} r\cos(\theta)\sin(\theta)\, d\theta\, dr$$

Letting $x = r\cos^2(\theta)$ and $y = r\sin^2(\theta)$, the double integral is transformed to

$$0.5 \int_0^\infty \int_0^\infty e^{-x}e^{-y} x^{\frac{\alpha-3}{2}} y^{-\frac{1}{2}}\, dx\, dy$$

which is the same as

$$0.5 \int_0^\infty e^{-x} x^{\frac{\alpha-3}{2}}\, dx \int_0^\infty e^{-y} y^{-\frac{1}{2}}\, dy$$

The integral with respect to $x$ is $\Gamma((\alpha-1)/2)$, and using the $u$-substitution $u = \sqrt{y}$, the integral with respect to $y$ is the same as

$$2\int_0^\infty e^{-u^2}\, du = \int_{-\infty}^\infty e^{-u^2}\, du = \sqrt{\pi}$$

In summary,

$$\Gamma\left(\frac{\alpha}{2}\right)I(\alpha) = 0.5\sqrt{\pi}\,\Gamma\left(\frac{\alpha-1}{2}\right)$$

Dividing both sides by $\Gamma(\alpha/2)$ results in

$$I(\alpha) = \frac{0.5\sqrt{\pi}\,\Gamma\left(\frac{\alpha-1}{2}\right)}{\Gamma\left(\frac{\alpha}{2}\right)}$$