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Take a look at the integral below. I need to determine $\alpha$. $A$ is actually a function but can be considered as a constant. How can I perform this integration?

$\int A e^{-\alpha^2 s^2}\, ds = -1$

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    I think it should be over (0,infinity), the constant is N (number of electrons)2012-02-15

1 Answers 1

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Write it as

$\int\limits_0^\infty {A{e^{ - {{\left( {\alpha s} \right)}^2}}}ds = - 1} $

Put $\alpha s = u$

$\int\limits_0^\infty {A{e^{ - {u^2}}}\frac{{du}}{\alpha } = - 1} $

$\frac{A}{\alpha }\frac{{\sqrt \pi }}{2} = - 1$

So this must always hold:

$A\sqrt \pi = - 2\alpha $

If you're asking about the primitive,

$\int {A{e^{ - {\alpha ^2}{s^2}}}ds} $

you will find no elementary solution, but there are a lot of series and asympotic expansions that will help.