How to solve infinite repeating integration. I am having trouble solving the following integration
$$\frac{1}{y!}\int^\infty_0\!x^ye^{-2x}\,\mathrm{dx}$$
We see that in order to solve this, we need integration by parts.
$u = x^y$
$du = yx^{y-1}$
Even without continuing any further, we can see that this will create an infinite loop since we have no way of discovering when y, the exponent of x, will reach $0$. Thus, there will be an infinite repetition of using integration by parts.
 A: Hint: $y!$ is only defined for non-negative integer values of $y$.
The integral is a function of $y$, so let us name it $F(y)$
$$F(y)=\frac{1}{y!}\int^\infty_0 x^y\,\mathsf e^{-2x}\operatorname dx$$
Now clearly, $F(0) = \tfrac 12$
Also for $y\in\Bbb N^+$, as you said, we use $\int u\operatorname d v =uv-\int v\operatorname d u$
When  $u = x^y, \operatorname dv=\mathsf e^{-2x}\operatorname d x$
Then $\operatorname du = y\,x^{y-1}\operatorname d x, v = {-\tfrac{1}{2}}\mathsf e^{-2x}$
Thus obtaining an recursive expression of $\;F(y)$.
Put it together to find a closed form.

 For advanced points, show this holds for the extension $\frac 1{\Gamma (y+1)}\int^\infty_0 x^y\,\mathsf e^{-2x}\operatorname dx$

A: Evaluating this integral we find that
$$\frac{1}{y!}\int x^ye^{-2x}dx=-\frac{\Gamma(y+1,2x)}{2^{y+1}y!}$$
Where $\Gamma(a, z)$ and $\Gamma(z)$ satisfy
$$\Gamma(a,z)=\int_{z}^{\infty}t^{a-1}e^{-t}dt,$$
$$\Gamma(z)=\int_{0}^{\infty}t^{z-1}e^{-t}dt$$
This means that
$$\frac{1}{y!}\int_{0}^{\infty} x^ye^{-2x}dx=\lim_{n\to\infty}\frac{\Gamma(y+1, 2n)}{2^{y+1}y!}-\frac{\Gamma(y+1)}{2^{y+1}y!}$$
$$=-\frac{\Gamma(y+1)}{2^{y+1}y!}=-\frac{1}{2^{y+1}},$$
in your case this is for $y\ge0$ since the original expression involves $y!$.
