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…
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…