Formula for evaluating $\int x^n e^x dx$.

Question

In this link How to integrate $ \int x^n e^x dx$?

Has the following formula:

$$\int {x^n e^x dx} = \bigg[\sum\limits_{k = 0}^n {( - 1)^{n - k} \frac{{n!}}{{k!}}x^k } \bigg]e^x + C$$

I tried to evaluate $$\int_0^a {x^n e^x dx},$$ but there is a singularity at $x=0$. How we can fix this?

Answer 1

In this answer, an alternative form of the equation at the end is

$$ \int {x^n e^x dx} = \left[(-1)^n n! + \sum\limits_{k=1}^n {(-1)^{n - k} \frac{n!}{k!} x^k} \right]e^x + C. $$

Now you do not get $0^0$ when $x = 0.$

The "$x^0$" term in the original formula was always meant to be a constant, not actually the zero-th power of a variable. Look at the several examples for $n = 1,2,3,4,5$ before they are summarized in a general formula, and it should be quite obvious that this is how the formula is meant to be read.

Answer 2

We usually define $0^0=1$, especially in contexts like power series, combinatorics, and anywhere else where exponents are mainly integers. Some of the reasons we do this is that it is consistent with other instances of the empty product, like $0!=1$, it leads to no problems (that I'm aware of), and it is very convenient. So there is no singularity.