Happy Birthday Mehul

Let I = $\large \int_{0}^{\infty}{e^2y^3e^{-ey^2}dy}$

= $\int_{0}^{\infty} e^{2-ey^2} y^3 dy$

Now let $2 - ey^2 = -u$

$2ey dy= du$

$y^2 = \frac{u+2}{e}$

So the Inegral is equal to : $= \int_{0}^{\infty}{e^{2-ey^2}y^2 \frac{2e}{2e}ydy}$ $= \int_{0}^{\infty}{e^{-u}\frac{u+2}{2e^2}du}$ [ Since $y^2 = \frac{u+2}{e}$ ]

$= \int_{0}^{\infty}{e^{-(u+2)}\frac{u+2}{2}du}$

$= \frac{1}{2} \int_{0}^{\infty}{e^{-(u+2)}(u+2)^{2-1}d(u+2)}$

$= \frac{1}{2}\Gamma(2)$

$= \frac{1}{2}$ [Since $\Gamma(2) = 1$ ]

$=\large 0.5$

$\begin{aligned} I & = \int_0^\infty e^2y^3e^{-ey^2} dy \quad \quad \small \color{#3D99F6}{\text{Let }x=ey^2 \space \Rightarrow dx = 2ey\space dy} \\ & = \int_0^\infty \frac{e^2y^3e^{-x}}{2ey} dx \\ & = \frac{1}{2} \int_0^\infty ey^2e^{-x} dx \\ & = \frac{1}{2} \int_0^\infty xe^{-x} dx \\ & = \frac{\Gamma(2)}{2} = \frac{1!}{2} = \boxed{0.5} \end{aligned}$