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