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

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In summary, the value of the integral when α=1 is $\frac{\pi}{2}$. There is a general formula for the value of the integral when α>1, which is $\frac{\pi}{2^{α/2-1}\Gamma(α/2)}$. The integral can be evaluated using the substitution x=tan(u). As α increases, the value of the integral decreases. The integral can also be evaluated using numerical methods such as the trapezoidal rule or Simpson's rule.
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Ackbach
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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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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Re: Problem Of The Week # 282 - Sep 25, 2017

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

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

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

What is the value of the integral when α=1?

When α=1, the integral becomes $\displaystyle \int_0^{\infty}\frac{dx}{1+x^2}$, which is a well-known integral with the value of $\frac{\pi}{2}$.

Is there a general formula for the value of the integral?

Yes, there is a general formula for the value of the integral when α>1, which is $\frac{\pi}{2^{α/2-1}\Gamma(α/2)}$, where $\Gamma$ denotes the gamma function.

Can the integral be evaluated using a substitution?

Yes, the integral can be evaluated using the substitution x=tan(u).

How does the value of the integral change as α increases?

As α increases, the value of the integral decreases. This can be seen by the fact that the denominator $(1+x^2)^{\alpha/2}$ becomes larger, making the fraction smaller and decreasing the overall value of the integral.

Can the integral be evaluated using numerical methods?

Yes, the integral can be evaluated using numerical methods such as the trapezoidal rule or Simpson's rule. These methods involve breaking the integral into smaller intervals and approximating the area under the curve using a series of straight lines or parabolas.

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