Integral $\displaystyle{\int_{-\infty}^\infty \frac{e^{-w^2/(4\pi)}}{\sqrt{2w-3}}} \, dw$

Question

How do I find $\displaystyle{\int_{-\infty}^\infty \frac{e^{-w^2/(4\pi)}}{\sqrt{2w-3}}} \, dw$? Integrand doesn't have elementary antiderivative.

Answer 1

Using appropriate changes of variable, the integral over $(3/2, \infty)$ is $$ e^{-9/(16 \pi)} \frac{\pi^{1/4}}{2} \int_0^\infty e^{-s -3\sqrt{s}/(2\sqrt{\pi})} \dfrac{ds}{s^{3/4}} $$ and the integral over $(-\infty, 3/2)$ is $$ - i e^{-9/(16 \pi)} \frac{\pi^{1/4}}{2} \int_0^\infty e^{-s + 3 \sqrt{s}/(2\sqrt{\pi})} \dfrac{ds}{s^{3/4}} $$

Now expand in power series: $$\eqalign{\int_0^\infty e^{-s + b \sqrt{s}} \dfrac{ds}{s^{3/4}} &= \sum_{n=0}^\infty \frac{b^n}{n!} \int_0^\infty e^{-s} s^{n/2-3/4}\;ds\cr &= \sum_{n=0}^\infty \frac{b^n \Gamma\left(\frac{n}{2} + \frac{1}{4}\right)}{n!}\cr &= \frac{\pi}{\sqrt{2}} e^{b^2/8} \left( (b^2)^{1/4} I_{-1/4}(b^2/8) + b (b^2)^{-1/4} I_{1/4}(b^2/8)\right)}$$ so the end result becomes $$e^{-9/(32 \pi)} \frac{\sqrt{3} \pi}{4} \left( (1-i) I_{-1/4}\left(\frac{9}{32\pi}\right) - (1+i) I_{1/4}\left(\frac{9}{32\pi}\right)\right) $$

Answer 2

Assuming that $\sqrt{2w-3}$ is defined as $i\sqrt{3-2w}$ when $2w-3<0$, the given integral equals

$$ I=\frac{1}{\sqrt{2}}\int_{-\infty}^{+\infty}\exp\left[-\frac{\left(w+\frac{3}{2}\right)^2}{4\pi}\right]\frac{dw}{\sqrt{w}} \tag{1}$$ where $$ \text{Re }I = \frac{1}{\sqrt{2}}\int_{0}^{+\infty}\exp\left[-\frac{\left(w+\frac{3}{2}\right)^2}{4\pi}\right]\frac{dw}{\sqrt{w}}\stackrel{w\mapsto u^2}{=}\sqrt{2}\int_{0}^{+\infty}\exp\left[-\frac{\left(u^2+\frac{3}{2}\right)^2}{4\pi}\right]\,du\tag{2}$$ leads to $$ \text{Re }I = \sqrt{\frac{3}{8}} e^{-\frac{9}{32 \pi }} K_{\frac{1}{4}}\left(\frac{9}{32 \pi }\right)\tag{3} $$ with $K$ being a modified Bessel function of the second kind.
The imaginary part of $I$ can be computed in a similar way.