integration by substitution with an absolute value

**Nappy** · October 9th 2012, 06:22 AM

Hi Guys,

I need help with a problem I cannot get past. The question below gives me the hint to make the substitution to make the integration easier and I understand that I the area under the curve must be one which will help me solve for c. But I can't see how the original equation simplifies to the integral the question provides. Could someone please show me how the original equation simplifies to the given integral once the substitution has been made. Once I understand that I can solve for c.
$integration by substitution with an absolute value-math.jpg$
Thanks!

Nappy

**TheEmptySet** · October 9th 2012, 10:31 AM

The integral

$\frac{1}{c_n}\int_{-\infty}^{\infty}|\xi|^{2n+1}e^{-\frac{\xi^2}{2}}d\xi$

Can be rewritten as

$\frac{1}{c_n}\int_{-\infty}^{0}-\xi^{2n+1}e^{-\frac{\xi^2}{2}}d\xi + \frac{1}{c_n}\int_{0}^{\infty}\xi^{2n+1}e^{-\frac{\xi^2}{2}}d\xi$

If you replace $-\xi=\zeta$

$\frac{1}{c_n}\int_{0}^{\infty}\zeta^{2n+1}e^{-\frac{\zeta^2}{2}}d\zeta + \frac{1}{c_n}\int_{0}^{\infty}\xi^{2n+1}e^{-\frac{\xi^2}{2}}d\xi$

So both integrals are the same this gives

$\frac{2}{c_n}\int_{0}^{\infty}\xi^{2n+1}e^{-\frac{\xi^2}{2}}d\xi$

let

$u=\frac{\xi^2}{2} \implies du = \xi d\xi$

and we have that $\sqrt{2u}=\xi$

$\frac{2}{c_n}\int_{0}^{\infty}(\sqrt{2u})^{2n}e^{-u}du$

$\frac{2^{n+1}}{c_n}\int_{0}^{\infty}u^ne^{-u}du$

This can be evaluated by integration by parts, but it is also the the gamma function evaluated at n+1

So this gives

$\frac{2^{n+1}}{c_n}n!$

**Nappy** · October 11th 2012, 05:41 AM

How would I go about calculating the first moment of the above equation? I tried both ways by first integrating with the multiplication of x to the integral and the other way by tring to calculate the moment generating function and setting t=0 for the first moment. I keep getting to a point where my answer is (n+1/2)! Is it possible to have a (n+1/2)! factorial. The calculator reads error. I'd really appreciate if someone could show me how to calculate the first moment. I need to find all moments. Thanks for the help!

**HallsofIvy** · October 11th 2012, 05:55 AM

Your original function, because of the absolute value, is an even function. Multiplying by [itex]\xi[/itex], makes the integrand odd so the integral is 0.

**Nappy** · October 11th 2012, 06:17 AM

[itex]\xi[/itex]

**TheEmptySet** · October 11th 2012, 07:55 AM

Originally Posted by Nappy

How would I go about calculating the first moment of the above equation? I tried both ways by first integrating with the multiplication of x to the integral and the other way by tring to calculate the moment generating function and setting t=0 for the first moment. I keep getting to a point where my answer is (n+1/2)! Is it possible to have a (n+1/2)! factorial. The calculator reads error. I'd really appreciate if someone could show me how to calculate the first moment. I need to find all moments. Thanks for the help!

To find the kth moment we multiply the integrand by $\xi^k$

Using the integral from post #2

$\frac{2}{c_n}\int_{0}^{\infty}\xi^k \xi^{2n+1}e^{-\frac{\xi^2}{2}}d\xi$

Using the same substitution we get

$\frac{2}{c_n}\int_{0}^{\infty}(\sqrt{2u})^k (\sqrt{2u})^{2n}e^{-u}du$

This gives

$\frac{2^{\frac{k}{2}+n+1}}{c_n}\int_{0}^{\infty}u^ {n+\frac{k}{2}}e^{-u}du$

This integral is the gamma function. This generalizes the factorial function.

Gamma function - Wikipedia, the free encyclopedia

This gives

$\frac{2^{\frac{k}{2}+n+1}}{c_n}\Gamma\left(n+\frac {k}{2}+1\right)$

This values can be calculated for any value of n and k

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