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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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Note that $\int e^{-s^2/(2\sigma^2)} \, ds = \sigma \sqrt{2\pi}$. So, if we suppose your "constant" $A$ is $-1/(\sigma\sqrt{2})$, then you have $\alpha = 1/(\sigma\sqrt{2})$. – William DeMeo Feb 15 '12 at 5:27
Is this over $(-\infty,\infty)$? If so, use the substitution $v=\alpha s$ and refer to the Wikipedia article on the Gaussian integral. – anon Feb 15 '12 at 6:39
I think it should be over (0,infinity), the constant is N (number of electrons) – heavenly7 Feb 15 '12 at 21:42

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.

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