Find $\int\limits^{\infty}_{0}\int\limits^{\infty}_{0}{\frac{1}{(x+y)^{3/2}}\exp\left\{-\frac{a^2}{2(x+y)}\right\}}\,dy\,dx$. In my posterior probability computation, I got the following integration and I could not figure it out.

$$\int\limits^{\infty}_{0}\int\limits^{\infty}_{0}{\frac{1}{(x+y)^{3/2}}\exp\left\{-\frac{a^2}{2(x+y)}\right\}}\,dy\,dx$$     where $a$ is a positive constant

Any help is appreciated.

I also think that this integral diverges.
$$\int\frac{e^{-\frac{a^2}{2 (x+y)}}}{(x+y)^{3/2}}\,dy=-\frac{\sqrt{2 \pi } }{a}\text{erf}\left(\frac{a}{\sqrt{2} \sqrt{x+y}}\right)$$ $$\int_0^\infty\frac{e^{-\frac{a^2}{2 (x+y)}}}{(x+y)^{3/2}}\,dy=\frac{\sqrt{2 \pi } }{a}\text{erf}\left(\frac{a}{\sqrt{2x} }\right)$$ provided $\Re(x)>0\lor x\notin \mathbb{R}$ Now, consider $$\int\text{erf}\left(\frac{1}{\sqrt{t} }\right)\,dt=(t+2) \text{erf}\left(\frac{1}{\sqrt{t}}\right)+\frac{2 e^{-1/t} \sqrt{t}}{\sqrt{\pi
   }}$$ $$I=\int_0^b\text{erf}\left(\frac{1}{\sqrt{t} }\right)\,dt=(b+2) \text{erf}\left(\frac{1}{\sqrt{b}}\right)+\frac{2 \sqrt{b} e^{-1/b}}{\sqrt{\pi
   }}-2$$ Now, for large values of $b$, $$I=-2+\frac{4 \sqrt{b}}{\sqrt{\pi }}+\frac{4 }{3 \sqrt{\pi b
   }}+O\left(\frac{1}{b^{3/2}}\right)$$ which would then make $$\frac{\sqrt{2 \pi } }{a}\int_0^b\text{erf}\left(\frac{a}{\sqrt{2x} }\right)=-\sqrt{2 \pi } a+4 \sqrt{b}+\frac{2}{3} a^2
   \sqrt{\frac{1}{b}}+O\left(\frac{1}{b^{3/2}}\right)$$ 
