Assume $X|v$~ $N(0,v)$ and $v$ ~ $\Gamma(a,b)$, derive the marginal distribution of $X$. Specifically, when $a = 1$, what is the distribution of $X$?
Here is my attempt:
$f(x) = \int f(x|v)f(v)dv$
$\Rightarrow \int \frac{1}{\sqrt{2\pi}} \frac{b^a}{\Gamma(x)} v^{a-\frac{1}{2}-1} \exp(-(\frac{x^{2}}{2v}+bv))dv $
Then I have a difficulty in solving the integral. My intuition is that this integral can be reduced to some constant independent of $v$ times an integral of the kernel of some simple statistical distribution of $v$. However, I don't find any distribution having $\exp(-(\frac{x^{2}}{2v}+bv))$.
Can anybody help, please?