How do I evaluate $ \int_{0}^{\infty} \frac{1}{\sqrt{2 \pi s}} e^{-z^{2}/2s} \cdot \frac{1}{2}e^{-s/2} \, ds$? How do you evaluate:

$$\displaystyle \int_{0}^{\infty}  \frac{1}{\sqrt{2 \pi s}} e^{-z^{2}/2s} \cdot \frac{1}{2}e^{-s/2} \, ds=?$$

 A: Here is the outline of a way forward.
Let $I(z)$ be the integral defined as
$$I(z)=\int_0^\infty e^{-\frac z2\left(\frac{s}{z}+\frac{z}{s}\right)}\frac{1}{\sqrt{s}}\,ds \tag 1$$
Now, writing $\left(\frac{s}{z}+\frac{z}{s}\right)=\left(\sqrt{\frac sz}-\sqrt{\frac zs}\right)^2+2$, $(1)$ becomes
$$I(z)=e^{-z}\int_0^\infty e^{-\frac z2\left(\sqrt{\frac sz}-\sqrt{\frac zs}\right)^2}\frac{1}{\sqrt{s}}\,ds \tag 2$$
Next, substituting $\sqrt{s/z}\to s$ in $(2)$reveals
$$I(z)=2\sqrt{z}\,e^{-z}\int_0^\infty e^{-\frac z2\left(s-\frac 1s\right)^2}\,ds \tag 3$$
Then, substituting $s\to 1/s$ in $(3)$ yields
$$I(z)=2\sqrt{z}\,e^{-z}\int_0^\infty e^{-\frac z2\left(s-\frac 1s\right)^2}\frac{1}{s^2}\,ds \tag 4$$
Adding $(3)$ and $(4)$, we obtain
$$I(z)=\sqrt{z}\,e^{-z}\int_0^\infty e^{-\frac z2\left(s-\frac 1s\right)^2}\left(1+\frac{1}{s^2}\right)\,ds \tag 5$$
One final substitution, $s-\frac1s\to s$ gives
$$I(z)=\sqrt{z}\,e^{-z}\int_0^\infty e^{-\frac z2\left(s\right)^2}\,ds \tag 6$$
Now evaluate the Gaussian integral in $(6)$.  Can you finish this now?
