How to integrate $\frac{\int _{-w}^w\:e^{-\frac{x^2}{w^a}}dx}{\int _{-\infty \:}^{\infty \:}\:e^{-\frac{x^2}{w^a}}dx}$

Question

I have found $\frac{\frac{\sqrt{\pi }}{2}\text{erf}\left(\sqrt{w}\right)-\frac{\sqrt{\pi }}{2}\text{erf}\left(-\sqrt{w}\right)}{\pi ^{\frac{1}{2}}}$ for $\frac{\int _{-w}^w\:e^{-\frac{x^2}{w}}dx}{\int _{-\infty }^{\infty }\:e^{-\frac{x^2}{w}}dx}$. However, I need to find how raising $w$ to the power of $a$ will change the solution.

Answer 1

$$ \frac{ \displaystyle \int_{-w}^w e^{-x^2/w^a} \, dx}{ \displaystyle \int_{-\infty}^\infty e^{-x^2/w^a} \, dx}$$ $$ \begin{align} & u = x/w^{a/2} \\ {} \\ & u^2 = x^2/w^a \\ {} \\ & w^{a/2} \, du = dx \end{align} $$ As $x$ goes from $-w$ to $w,\,\,\,$ $u$ goes from $-w^{1-a/2}$ to $w^{1-a/2}.$ Thus we have $$ \frac{\displaystyle\int_{-w^{1-a/2}}^{w^{1-a/2}} e^{-u^2}\, du}{ \displaystyle\int_{-\infty}^{+\infty} e^{-u^2} \, du }. $$ The factor $w^{a/2}$ in the numerator and denominator has canceled. $\qquad$