probability density function of sum of variables

Question

Let $X, Y$ be independent each having probability density function: $f(x)=e^{-x}x^{-\frac{1}{2}}$. Find the probability density function of $X+Y$.

My solution: Define the new random variables $U=X$ and $V=X+Y$. Then the determinant of the Jacobian is 1.

Since $X, Y$ are independent then their joint distribution is $f_{X, Y}(x, y)=e^{-(x+y)}(xy)^{-\frac{1}{2}}$ so $f_{U, V}(u, v)=|J| \cdot f_{X, Y}=e^{-(u+(v-u))}(u(v-u))^{-\frac{1}{2}}=e^{-v}(u(v-u))^{-\frac{1}{2}}$

So $f_U(u)=\int_0^{\infty} e^{-v}(u(v-u))^{-\frac{1}{2}} dv$. The problem is that the integral doesn't converge. Where is my mistake?

Answer 1

I assume both X and Y are non-negative. Using the definition of convolution. $f_V(v)=e^{−v}∫_0^v\frac{du}{v^\frac{1}{2}(v−u)^\frac{1}{2}}$.