How to solve infinite repeating integration.

Question

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.

Answer 1

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 $\frac 1{\Gamma (y+1)}\int^\infty_0 x^y\,\mathsf e^{-2x}\operatorname dx$

Answer 2

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 $y!$.