Integrate $\frac{\lambda y^{2}}{\sqrt{2\pi}} e^{-(\frac{1}{2}+ \lambda x)y^{2}}$ with respect to $y$ $$F(x,y)= \frac{\lambda y^{2}}{\sqrt{2\pi}} e^{-(\frac{1}{2}+ \lambda x)y^{2}}$$
Please show that the function when integrated with respect to $y$ is 
$F_X(x)= \frac{\lambda}{(2\sqrt{2}(\frac{1}{2}+ \lambda x)^\frac{3}{2})}$ using $u =(\frac{1}{2}+ \lambda x)y^{2}$ and the gamma function.
 A: So, the two main passages are the change of variable and the gamma function. I'm supposing you want to integrate it between $0$ and $+\infty$.Therefore:
\begin{equation}
F_x(x) = \int_0^{+\infty} \frac{\lambda y^2}{\sqrt{2\pi}} e^{-(\frac{1}{2} + \lambda x)y^2} dy= \int_0^{+\infty} \frac{\lambda}{\sqrt{2\pi}} \frac{1}{(\frac{1}{2} + \lambda x)} (\frac{1}{2} + \lambda x) y^2 e^{-(\frac{1}{2} + \lambda x)y^2} dy =
\end{equation}
Let's change the variable by imposing $u=(\frac{1}{2} + \lambda x)y^2$ so that we have:
\begin{align}
& du= 2(\frac{1}{2} + \lambda x)y dy\\[5pt]
& y = \sqrt{\frac{u}{(\frac{1}{2} + \lambda x)}}
\end{align}
Applying the variable change leads to:
\begin{align}
Fx(x) &= \frac{\lambda}{\sqrt{2\pi}} \frac{1}{(\frac{1}{2} + \lambda x)} \int_0^{+\infty} u\,\, e^{-u} \frac{1}{2(\frac{1}{2} + \lambda x)\,y(u)} du \\[10pt]
&= \frac{\lambda}{\sqrt{2\pi}} \frac{1}{(\frac{1}{2} + \lambda x)} \int_0^{+\infty} u\,\, e^{-u}\frac{\sqrt{\frac{1}{2} + \lambda x}}{2(\frac{1}{2} + \lambda x)\,\sqrt{u}} du \\[5pt]
&= \frac{\lambda}{2\sqrt{2\pi} (\frac{1}{2} + \lambda x)^{3/2}} \int_0^{+\infty} u^{\frac{1}{2}}\,\, e^{-u} \,\,du
\end{align}
The latter integral is the gamma function of $\frac{3}{2}$ as the gamma function is defined as:
\begin{equation}
\Gamma(z) = \int_0^{+\infty} t^{z-1} e^{-t} dt
\end{equation}
Since that $\Gamma(\frac{3}{2})= \frac{\pi}{2}$ (the reason is not easy to prove, but you can retrieve those particular values from the wikipedia page) you have that:
\begin{equation}
Fx(x) = \frac{\lambda}{2\sqrt{2\pi} (\frac{1}{2} + \lambda x)^{3/2}} \frac{\sqrt{\pi}}{2} = \frac{\lambda}{4\sqrt{2} (\frac{1}{2} + \lambda x)^{3/2}}
\end{equation}
I still have a $\frac{1}{2}$ difference with your solution and honestly at the moment I can't figure out exactly where the mistake is, but I hope it can be still a useful reply.
