# How to evaluate $\int_{0}^{\infty }\frac{e^{-x^{2}}}{\sqrt{t^{2}+x}}\mathrm{d}x$

How to evaluate the integral below $$\int_{0}^{\infty }\frac{e^{-x^{2}}}{\sqrt{t^{2}+x}}\mathrm{d}x~~~~~~(t＞0)$$ The WolframAlpha gave me a horrible answer $$\frac{t}{2}e^{-\frac{t^{4}}{2}}\left \{ \pi \left [ \mathrm{I}_{-\frac{1}{4}}\left ( \frac{t^{4}}{2} \right )+\mathrm{I}_{\frac{1}{4}}\left ( \frac{t^{4}}{2} \right ) \right ]-4e^{\frac{t^{4}}{2}}\, _{2}F_{2}\left ( \frac{1}{2},1;\frac{3}{4},\frac{5}{4};-t^{4} \right ) \right \}$$ where $\mathrm{I}_{n}(z)$ is the modified Bessel function of the first kind and $_{2}F_{2}$ is the generalized hypergeometric function.

But I have no idea how to prove it.Is there a more simple solution for this integral.

• See Gradshteyn and Ryzhik, 3.478, for a similar integral.
– Nemo
Apr 20, 2016 at 7:21

## 1 Answer

Hint:

$\int_0^\infty\dfrac{e^{-x^2}}{\sqrt{t^2+x}}~dx$

$=2\int_0^\infty e^{-x^2}~d\left(\sqrt{t^2+x}\right)$

$=2\int_t^\infty e^{-(x^2-t^2)^2}~dx$

$=2\int_t^\infty e^{-x^4+2t^2x^2-t^4}~dx$