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.

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

1 Answer 1




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

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

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

Similar to Evaluating $\int_{1}^{\infty}\exp(-(x(2n-x)/b)^2)\,\mathrm dx$


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