It's a well known result shown with the equality of the moment generating functions that if random variables $X_1$ and $X_2$ are independent and identically distributed $\chi^2(1)$ random variables, then $Y = X_1 + X_2$ is distributed according to a $\chi^2(2)$.
When I tried to show this result with convolution, I did not arrive at the same answer:
With the given distributions, the probability density functions (pdfs) of $X_1$ and $X_2$ are $f_{x_{1}}(x) = \frac{e^{-x/2}}{\sqrt{2\, \pi \,x}}$ and $f_{x_{2}}(x) = \frac{e^{-x/2}}{\sqrt{2\, \pi \,x}}$, respectfully, with $0 <x< \infty$.
Then the pdf of $Y = X_1 + X_2$ is the convolution of the pdfs of $X_1$ and $X_2$:
$$ \begin{align*} f_y(t) =& \int_{-\infty}^{\infty} f_{x_1}(x) \; f_{x_2}(t - x) \; dx \\ =& - \frac{e^{-t/2}}{\pi} \cdot \arctan\left( \frac{\sqrt{x}}{\sqrt{t - x}} \right) \bigg|_{0}^{\infty} \end{align*} $$
But $\lim_{x \to \infty} \, \frac{\sqrt{x}}{\sqrt{t - x}} = \sqrt{-1} = i$
What am I missing, or where am I going wrong?