Computing a Gaussian-like integral

Question

I am given the following joint probability function:

\begin{align} f(x) = \begin{cases} \frac{1}{2}ye^{-y-\frac{x}{y}} & \text{if} \ x\geq 0 \ \text{and} \ y\geq 0 \\ 0 & \text{else} \end{cases} \end{align} and I'm asked to compute $\Pr(X \leq Y^3)$. So far I have

\begin{align} \Pr(X \leq Y^3) & = \frac{1}{2} \int_{y=0}^\infty \int_{x=0}^{y^3}ye^{-y-\frac{x}{y}} dx \ dy \\ & = \frac{1}{2} \int_0^\infty y^2e^{-y}(1-e^{-y^2}) \ dy \\ & = \frac{1}{2} \left(\int_0^\infty y^2e^{-y} \ dy - \int_0^\infty y^2e^{-y-y^2} \ dy\right) \end{align}

I know that

$$ \int_0^\infty y^2e^{-y} \ dy = 2 $$

However, I don't know how to compute the second term. I tried many things such as substitution, multiplying the term by itself and trying to express the integral in polar coordinates, but to no avail. A hint is that I should be able to express the integral in terms of the cdf of the standard normal distribution. Any help would be highly appreciated.

Thanks

Answer 1

Hint

Considering $$I=\int y^2e^{-y-y^2} \, dy$$ first, complete the square $$y^2+y=\left(y+\frac{1}{2}\right)^2-\frac 14$$ Now, change variable $t=y+\frac 12$ to make $$I=\frac{\sqrt[4]{e}}{4}\int e^{-t^2} (1-2 t)^2\,dt$$

Answer 2

Hint

$$e^{-(y^2+y)}=e^{-(y^2+y+1/4)}e^{1/4}=e^{-(y+1/2)^2}e^{1/4}$$

Substitute $z=y+\frac 12$

$$e^{-(y^2+y)}=e^{-z^2}e^{1/4}$$

$$\int y^2e^{-(y^2+y)} \ dy=e^{1/4}\int (z-\frac 12)^2e^{-z^2} \ dz $$