How to evaluate $\int_{0}^{1}x^{m-1}\exp(-x)dx$?

Question

How to evaluate $\int_{0}^{1}x^{m-1}\exp(-x)dx$?

I tried expanding in taylor series as follows:

$\int_{0}^{1}x^{m-1}\exp(-x)dx = \int_{0}^{1}\sum_{i=0}^{+\infty}\frac{(-1)^ix^{m-1+i}}{i!} dx = \sum_{i=0}^{+\infty}\frac{(-1)^i}{i!(m+i)} = \sum_{i=0}^{+\infty}\Bigg[ \frac{1}{(2i)!(m+2i)}- \frac{1}{(2i+1)!(m+2i+1)} \Bigg] = \sum_{i=0}^{+\infty} \frac{2i(m+2i+1)+1}{(2i)!(m+2i)(m+2i+1)}$

I was trying to get to this result: $\int _{0}^{1}x^{n}e^{-x}\,dx=n!\left[1-e^{-1}\sum _{i=0}^{n}{\frac {1}{i!}}\right]$

Thanks!

Answer 1

As @LordSharktheUnknown commented, you will have to use integration by parts. If we use $m$ as a parameter then we can calculate the integral $I_m$ recursively by using the following formula.

$$I_m=\int_0^1x^me^{-x}dx=\left.-x^me^{-x}\right|_{x=0}^{x=1}+m\int_{0}^1x^{m-1}e^{-x}dx=-e^{-1}+mI_{m-1}$$

Answer 2

As mentioned by other answers/comments, repeated integration by parts is the direct way to compute your integral.

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For fun, here is an overkill approach for those who are familiar with Poisson processes. [Of course, proving the referenced properties of the Poisson process would boil down to the direct approach mentioned above.]

The integrand resembles the density $f(x) = \frac{1}{(m-1)!}x^{m-1} e^{-x}$ of the $\text{Gamma}(m, 1)$ distribution.

The time of the $m$th arrival in a Poisson process with rate $1$ follows this distribution, so $\frac{1}{(m-1)!} \int_0^1 x^{m-1} e^{-x}$ is the probability that the $m$th arrival occurs before time $1$.

Since the number of arrivals in the time interval $[0,1]$ follows the $\text{Poisson}(1)$ distribution, this probability is $1 - e^{-1} \sum_{i=0}^{m-1} \frac{1}{i!}$.