Closed form for this integral $\int_{0}^{\infty}\frac{dx}{\sqrt{x}}\, e^{-x^{2}-\frac{b^{2}}{x}}$

How would you evaluate this integral? \begin{equation}\int_{0}^{\infty}\frac{dx}{\sqrt{x}}\, e^{-x^{2}-\frac{b^{2}}{x}}\end{equation} It reminds me of the form of a modified Bessel function of the second kind, but is slightly different because of the $x^{2}$ term. I can series expand it in terms of a sum of Gamma functions, but I don't know what that sum converges to.

Solution 1:

I rewrite the integral in the form $$ I=\tfrac{1}{2}\int^{+\infty}_{-\infty}e^{-t^2}\frac{e^{-b^2/|t|}}{\sqrt{|t|}}dt. $$ With the help of Fourier transform $$ \widehat{f}(x)=\int^{+\infty}_{-\infty}f(t)e^{-itx}dt $$ the Parseval indentity $$ \int^{+\infty}_{-\infty}f(t)\overline{g(t)}dt=\tfrac{1}{2\pi}\int^{+\infty}_{-\infty}\widehat{f}(x)\overline{\widehat{g}(x)}dx $$ and the pairs $$ e^{-t^2}\leftrightarrow \sqrt{\pi}e^{-x^2/4} $$ $$ \frac{e^{-b^2/|t|}}{\sqrt{|t|}}\leftrightarrow \sqrt{\frac{2\pi}{|x|}}e^{-b\sqrt{2|x|}}\left(\cos\left(b\sqrt{2|x|}\right)-\sin\left(b\sqrt{2|x|}\right)\right)\textrm{, }b>0 $$ we get $$ I=\frac{1}{\sqrt{2}}\int^{+\infty}_{0}\frac{\cos\left(b\sqrt{2x}\right)-\sin\left(b\sqrt{2x}\right)}{\sqrt{x}}e^{-x^2/4-b\sqrt{2x}}dx $$ Setting $x=w^2/2$ and then $w=s/b$, we arive to $$ I=\int^{+\infty}_{0}\exp\left(-w^4/16-bw\right)(\cos(bw)-\sin(bw))dw= $$ $$ =b^{-1}\int^{+\infty}_{0}\exp\left(-\frac{s^4}{16b^4}\right)e^{-s}(\cos s-\sin s)ds\tag 1 $$ But $$ \int^{+\infty}_{0}\exp\left(-\frac{s^4}{16b^4}\right)s^kds=2^{k-1}b^{k+1}\Gamma\left(\frac{k+1}{4}\right)\textrm{, for }k=0,1,2,\ldots $$ Hence using the Taylor expansions of $e^{-s}\cos s$ and $e^{-s}\sin s$: $$ I=\tfrac{1}{2}\sum^{\infty}_{k=0}(-1)^k2^{3k/2}\left(\cos\left(\frac{\pi k}{4}\right)+\sin\left(\frac{\pi k}{4}\right)\right)\Gamma\left(\frac{k+1}{4}\right)\frac{b^{k}}{k!}= $$ $$ =\tfrac{1}{\sqrt{2}}\sum^{\infty}_{k=0}(-1)^k2^{3k/2}\sin\left(\frac{\pi (k+1)}{4}\right)\Gamma\left(\frac{k+1}{4}\right)\frac{b^{k}}{k!}=\ldots $$ $$ =2\Gamma(\tfrac{5}{4})\cdot {}_0F_2\left(\tfrac{1}{2},\tfrac{3}{4};-\tfrac{b^4}{4}\right)-2b\sqrt{\pi}\cdot {}_0F_2\left(\tfrac{3}{4},\tfrac{5}{4};-\tfrac{b^4}{4}\right)+2b^2\Gamma(\tfrac{3}{4})\cdot {}_0F_2\left(\tfrac{3}{2},\tfrac{5}{4};-\tfrac{b^4}{4}\right) $$ Another way is to write (1) as $$ I=\left(1-i\right)I_0(2(1+i)b)+(1+i)I_0\left(2(1-i)b\right),\tag 2 $$ where $$ I_0(z)=\int^{+\infty}_{0}\exp\left(-s^4\right)e^{-zs}ds\tag 3 $$