Obtaining the pdf of the sum of two iid random variables with pdf $f_{X}(x)= \sqrt{\frac{1}{2\pi x}e^{-x/2}}$ X1, and X2 are independent with the pdf: $f_{X}(x)= \sqrt{\frac{1}{2\pi x}e^{-x/2}}$ defined for x>0
Y=X1+X2
What is the pdf of Y?
This is what I did so far:
$$f_{Y}(y) = \int_{0}^{\infty } f_{X_{1}}(y-x_{2})f_{X_{2}}(x_{2})dx_{2}$$
$$ =  \int_{0}^{\infty }  \sqrt{\frac{1}{2\pi (y-x_{2})}e^{-(y-x_{2})/2}}   \sqrt{\frac{1}{2\pi x_{2}}e^{-x_{2}/2}}  dx_{2}$$
$$=\int_{0}^{\infty }  \sqrt{\frac{1}{2\pi (y-x_{2})}e^{(-y+x_{2})/2}}   \sqrt{\frac{1}{2\pi x_{2}}e^{-x_{2}/2}}  dx_{2}$$
$$=\int_{0}^{\infty }  \sqrt{\frac{1}{2\pi (y-x_{2})}e^{(-y+x_{2})/2} \frac{1}{2\pi x_{2}}e^{-x_{2}/2}}  dx_{2}$$
$$=\int_{0}^{\infty }  \sqrt{\frac{1}{2\pi (y-x_{2})} \frac{1}{2\pi x_{2}}e^{-y/2}}  dx_{2}$$
$$=\frac{1}{2\pi} \sqrt{e^{-y/2}}  \int_{0}^{\infty }  \sqrt{\frac{1}{ (y-x_{2})x_{2}} }  dx_{2}$$
$$=\frac{1}{2\pi} \sqrt{e^{-y/2}}  \int_{0}^{\infty }  \sqrt{\frac{1}{ yx_{2}-(x_{2})^{2}} }  dx_{2}$$
At this point I wasn't sure how to proceed with the integral, so I went to wolfram alpha. According to it, this integral does not converge, so not sure what to do now.

Effectively $x_2$ only goes to $y$, not to $\infty$. (The density is $0$ at negatives.)
For the work that remains, complete the square. After making the right substitution, you should end up needing to find $\int\frac{du}{\sqrt{1-u^2}}$, which is $\arcsin u +C$.  
