Expand the incomplete gamma function $\int_0^x e^{-t}t^n dt$

Expand the incomplete gamma function $\int_0^x e^{-t}t^n dt$ in a series of powers of x.

What I have tried:

I know the expansions: $$t^n = 1 + t + t^2+\cdots+t^n \\ e^{-t} = 1-t+\frac{t^2}{2!} + \cdots+\frac{(-1)^nt^n}{n!}$$ Combining this into the integral $$\int_0^x \left(1-t+\frac{t^2}{2!} + \cdots+\frac{(-1)^nt^n}{n!} \right) \left(1 + t + t^2+\cdots+t^n \right)dt$$

I'm not sure if this step is allowed, but I integrated as is: $$\int_0^x e^{-t}t^n dt = \left(x-\frac{x^2}{2}+\frac{x^3}{3!} + \cdots+\frac{(-1)^nx^{n+1}}{(n+1)!} \right) \left(x + \frac{x^2}{2} + \frac{x^3}{3}+\cdots+\frac{t^{n+1}}{n+1} \right)$$

How do I simplify this even further?

Solution 1:

Sorry, but what you wrote is wrong under many respects.

1. You cannot truncate $e^{-t}$
2. $t^n\ne 1+t+\dots+t^n$

You can instead expand $$ t^ne^{-t}=t^n\sum_{k=0}^\infty \frac{(-1)^kt^k}{k!}= \sum_{k=0}^\infty \frac{(-1)^kt^{k+n}}{k!} $$ Integrating term by term from $0$ to $x$, we get $$ \int_0^x t^ne^{-t}\,dt=\sum_{k=0}^\infty \frac{(-1)^kx^{k+n+1}}{(n+k+1)k!} $$