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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?

gt6989b
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Pinteco
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    What do you mean by the term "singularity"? There is no problem at all to evaluate a polynomial (if $n$ is not a nonnegative integer then your sum from $k = 0$ to $k=n$ doesn't make sense) multiplied by $e^x$ at $x = 0$. – KCd May 10 '18 at 23:12
  • In the formula at $x=0$ there is $0^0$. – Pinteco May 10 '18 at 23:13
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    Stop thinking symbolically and look at actual examples (try $n = 0, 1, 2, 3$) to see why your objection has no merit. – KCd May 10 '18 at 23:16
  • In slight contrast to KCd, I wouldn't say your thinking has no merit. It makes a lot of sense actually, and is an important question to ask. The thing is, we are letting $x$ be arbitrary here and letting $n$ vary, so we are appealing to the fact $\lim_{x\to0} x^0 = 1$ even though $0^0$ is undefined. In this sense, it is a removable singularity and we define it in such a way that as a function of $n$, $x^n$ is analytic for any constant $n$. – Brevan Ellefsen May 11 '18 at 00:13

2 Answers2

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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.

David K
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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.

Arthur
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