I have arrived at the following integral: $$\int_y^\infty \frac{e^{-x/2}}{2x^2}$$ The limits I'm not so certain, but the function is correct. I have tried integration by parts but I arrived at a more complicated function. Any help is appreciated.
edit: I have $f_X(x)$ with $exp(\frac{1}{2})$ distribution and $f_{Y|X}(y|x)$ with $Uc[0, x^2]$ distribution. I found the joint distribution by multiplying both and I wanted to find $f_Y(y)$ so I could calculate $E(X)$, $E(Y)$, $Var(X)$, $Var(Y)$ and $Cov(X,Y)$. But, by looking at the answer to the integral I think I took a wrong turn somewhere...
$$I=\int_y^\infty\frac{e^{-x/2}}{2x^2}dx=\frac 14\int_{y/2}^\infty u^{-2}e^{-u}du$$ Now note that the incomplete gamma function is defined as: $$\Gamma(s,x)=\int_x^\infty t^{s-1}e^{-t}dt$$ So we can say: $$I=\frac 14\Gamma\left(-1,\frac y2\right)$$