# Reduced formula: $\int x^m \cdot e^x dx$

Please, I need find a reduced formula to $$\int x^m \cdot e^x dx$$ by parts integration method, I've found that every time I compute the m is reduced so: $m-1$, $m-2$, $m-3$, but I couldn't found the general formula.

Thanks so much.

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$$I_m = \int x^m e^x dx = \int x^m d(e^x) = x^m e^x - \int m x^{m-1} e^x dx= x^m e^x - mI_{m-1} + c_{m-1}$$ \begin{align} I_m & = x^m e^x - m \left(x^{m-1} e^x - (m-1) I_{m-2} \right) + c_{m-2}\\ & = x^m e^x - m x^{m-1} e^x + m(m-1) I_{m-2} + c_{m-2}\\ & = x^m e^x - m x^{m-1} e^x + m(m-1) x^{m-2} e^x - m(m-1)(m-2)I_{m-3} + c_{m-3}\\ & = e^x \left(x^m - m x^{m-1} + m(m-1) x^{m-2} - \cdots + (-1)^m m!\right) + c_{0}\\ & = e^x \left(\sum_{k=0}^m P(m,k)(-1)^k x^{m-k}\right) + c_0\\ & = e^x \left(\sum_{k=0}^m P(m,k)(-1)^k x^{m-k}\right) + c_0\\ & = (-1)^m \Gamma(m+1,-x) + c_0 \end{align} where $P(m,k) = \dfrac{m!}{(m-k)!}$.

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Very good, but, what do you mean with $P(m,k)$? – calbertts Oct 28 '12 at 0:26
He appears to mean $P(m,k)=\underbrace{m(m-1)(m-2)\cdots(m-k+1)}_{k\text{ factors}}$, which is the number of permutations of size $k$ of members of a set of size $m$. – Michael Hardy Oct 28 '12 at 0:30
Thanks! now understand. – calbertts Oct 28 '12 at 0:31