How to integrate $\int_{0}^{a}x^{n-1}e^{-x}dx$ We know that $$\int_{0}^{\infty }x^{n-1}e^{-x}dx = \Gamma (n)$$
But how do we integrate this?
$$\int_{0}^{a}x^{n-1}e^{-x}dx$$
 A: This integral can be viewed as a recurrance
$$\begin{align*}
I_n &= \int_0^a x^{n-1}e^{-x}dx\\
&= -\int_0^a x^{n-1}de^{-x}\\
&= -\left[x^{n-1}e^{-x}\right]_0^a + \int_0^a e^{-x}dx^{n-1}\\
&= -a^{n-1}e^{-a} + (n-1)\int_0^a x^{n-2}e^{-x}dx\\
&= -a^{n-1}e^{-a} + (n-1)I_{n-1}
\end{align*}$$
With $$I_1 = \int_0^a e^{-x}dx = -\left[e^{-x}\right]_0^a = -e^{-a}+1$$
A: Notice that if $P$ is a polynomial of degree $n$, then the derivative of $x \mapsto P(x)e^{-x}$ is of the form $x \mapsto Q(x) e^{-x}$ where $Q$ is also a polynomial of degree (exactly) $n$ given by $Q = P'-P$. Now the map $P \mapsto Q = P'-P$ is a bijection from $\mathbb{R}[X]$ to itself because it preserves degree. In particular, there exists a unique polynomial $P_n$ such that $x \mapsto P_n(x) e^{-x}$ is a primitive of $x \mapsto x^n e^{-x}$. We know $\deg P_n = n$ and $P_n' - P_n = X^n$. So you can find $P_n$ (and thus a primitive of $x \mapsto x^n e^{-x}$) by solving a simple linear system. If I'm not mistaken, you can check that $P_n$ is given by
$$P_n = -n! \sum_{k = 0}^n \frac{X^k}{k!}$$
To do the computation easily, define recursively $Q_0 = 1$, $Q_{n+1} = Q_n + \frac{X^{n+1}}{(n+1)!}$. And check (also recursively) that $Q_{n+1}' = Q_n$ so $Q_n-Q_n' = \frac{X^{n}}{n!}$, and finally $P_n = -n! Q_n$ solves your problem.
A: If $n$ is taken to be integer you can manipulate the integral representation of the Gamma function to obtain a finite polynomial in $a$ multiplied by $e^{-a}$ as follows:
$$\begin{aligned}
\int _{0}^{a}\!{x}^{n-1}{{\rm e}^{-x}}{dx}&=\int _{0}^{\infty }\!{x}^{n
-1}{{\rm e}^{-x}}{dx}-\int _{a}^{\infty }\!{x}^{n-1}{{\rm e}^{-x}}{dx}\\
&=\Gamma  \left( n \right) -{
{\rm e}^{-a}}\int _{0}^{\infty }\! \left( y+a \right) ^{n-1}{{\rm e}^{
-y}}{dy}\quad:\quad y=x+a\\
&=\Gamma  \left( n \right) -{
{\rm e}^{-a}}\sum _{k=0}^{n-1} \left( {n-1\choose k}{a}^{n-k-1}\int _{0
}^{\infty }\!{y}^{k}{{\rm e}^{-y}}{dy} \right)\\
&=\Gamma  \left( n \right) -{
{\rm e}^{-a}}\sum _{k=0}^{n-1}{n-1\choose k}{a}^{n-k-1}\Gamma  \left( 
k+1 \right) \\
&=\left( n-1 \right) !\left(1 -{
{\rm e}^{-a}}\sum _{k=0}^{n-1}{\frac { \,{a}^{n-k-
1}}{ \left( n-k-1 \right) !}}\right)\\
&=\left( n-1 \right) !\left(1 -{
{\rm e}^{-a}}\sum _{j=0}^{n-1}{\frac { \,{a}^{j}}{j!}}\right)
\end{aligned}$$
or alternatively you can give the answer in terms of the incomplete gamma function 
A: there is a trick
we know that $$ \int _{0}^{b}dxe^{-ax}=\frac{e^{-ab}-1}{-a} $$
also if we take the n-th derivaitve with resept to 'b' we have
$$ D^{n-1}e^{-bx}=(-1)^{n-1}x^{n-1}e^{-ax} $$ at the poitn $ a=1 $
so $$ \int_{0}^{a}dxx^{n-1}e^{-x}=\left.(-1)^{n-1}D^{n-1} \frac{e^{-ab}-1}{-b}\right\vert_{b=1} $$
with $ D= \frac{d}{db}$ derivation with respect to 'b'
A: You can write it as $\gamma(n,a)$ or get the series form solution as follows:
$\int_0^ax^{n-1}e^{-x}dx$
$=\int_0^ax^{n-1}\sum\limits_{k=0}^\infty\dfrac{(-1)^kx^k}{k!}dx$
$=\int_0^a\sum\limits_{k=0}^\infty\dfrac{(-1)^kx^{k+n-1}}{k!}dx$
$=\left[\sum\limits_{k=0}^\infty\dfrac{(-1)^kx^{k+n}}{k!(k+n)}\right]_0^a$
$=\sum\limits_{k=0}^\infty\dfrac{(-1)^ka^{k+n}}{k!(k+n)}$
A: If $P(x)$ is a polynomial and $c$ is a nonzero constant, the formula is$$\int e^{cx}P(x)dx=e^{cx}[\frac{P(x)}c-\frac{P'(x)}{c^2}+\frac{P''(x)}{c^3}-\frac{P'''(x)}{c^4}+\cdots]$$which you can check by differentiating. For $n\in\mathbb N$, setting $P(x)=x^{n-1}$ and $c=-1$, we have$$\int e^{-x}x^{n-1}dx=-e^{-x}[x^{n-1}+(n-1)x^{n-2}+(n-1)(n-2)x^{n-3}+\cdots].$$
