Reply to thread

Message

<blockquote data-quote="BentSnowman" data-source="post: 2615290" data-attributes="member: 715805">Do you know how to calculate Gaussian integrals?let u = x^{1/2}, then&nbsp; 2 du = x^{-1/2} dx, so your integral is translated toI = integral (X^-0.5)(e^-x)dx = 2 integral { e^(-u^2) du }This is a famous integral.&nbsp; 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 vI^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.&nbsp; 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.&nbsp; The integral then becomes integrable directly.&nbsp; 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.&nbsp; If you require more steps, then google it.&nbsp; As I said, it is a famous integral so you can find solutions all over.&nbsp; For example: http://www.umich.edu/~chem461/Gaussian%20Integrals.pdf</blockquote>

[QUOTE="BentSnowman, post: 2615290, member: 715805"] 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%20Integrals.pdf [/QUOTE]

Name

Verification

Loading…