Intergration from zero to infiniti of (X^-0.5)(e^-x)dx the answer is square root of pi
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Intergration from zero to infiniti of (X^-0.5)(e^-x)dx the answer is square root of pi

[From: ] [author: ] [Date: 11-10-10] [Hit: ]
then2 du = x^{-1/2} dx,I = integral (X^-0.This is a famous integral.You can calculate this with the Gamma function, valued at 1/2, i.......
I need the solution step by step so i can understand it correctly please.. thank you guys

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Do you know how to calculate Gaussian integrals?

let u = x^{1/2}, then 2 du = x^{-1/2} dx, so your integral is translated to

I = integral (X^-0.5)(e^-x)dx = 2 integral { e^(-u^2) du }

This is a famous integral. You can calculate this with the Gamma function, valued at 1/2, i.e. Gamma(1/2) = sqrt{pi} (look up this function if you do not know what it is).

You can also calculate it analytically by pursuing a solution to the square of the integral and rewriting it in terms of an iterated integral with a dummy variable v

I^2 = ( integral { e^(-u^2) du } )^2 = integral { e^(-u^2) du } integral { e^(-v^2) dv }

= integral { e^(-(u^2 +v^2) du dv}

This should look really friendly to you. By writing it this way we can shift to polar coordinates r^2 = u^2 + v^2, dudv = r dr d(theta), and we integrate over all space. The integral then becomes integrable directly. To retrieve the result for semi-infinite bounds, realize the integrand is an even function so that we may append a factor of (1/2) to recover the result I^2 / 2 = pi,

thus I = sqrt{pi}

You should be able to understand this given you are at the point where you have encountered a Gaussian integral. If you require more steps, then google it. As I said, it is a famous integral so you can find solutions all over. For example: http://www.umich.edu/~chem461/Gaussian%2…
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