How to integrate $\int_{0}^{T} t^{m-1}\cdot \mathrm{e}^{-ut}\,\mathrm{d}t$?

Question

Task: Compute the following integral.

$$\int_{0}^{T} t^{m-1}\cdot \mathrm{e}^{-ut}\,\mathrm{d}t $$

Answer 1

Use Reduction formulae or gamma functions...

Integrate by parts treating polynomial as first function and exponential part as second function.then use recurrence relation.

set $ut=z, t=\frac{z}{u}, dt=\frac{dz}{u}$ $\int t^{m-1} \textrm e^{-ut} \textrm dt= \frac{\int z^{m-1} \textrm e^{-z} \textrm dz}{u^m}=\frac{I_m}{u^m}$ where $I_m=\int z^{m-1} \textrm e^{-z} \textrm dz$ Integrate $I_m$ by part

$I_m=z^{m-1}e^{-z}+(m-1)\int z^{m-2} \textrm e^{-z} \textrm dz=z^{m-1}e^{-z}+(m-1)I_{m-1}$ Now replace $m$ by $m-1$

$I_{m-1}=z^{m-2}e^{-z}+(m-2)I_{m-2}$ $I_1=0$ and $I_2=ze^{-z}$ replacing these values, we get $I_m=e^{-z}[z^{m-1} +(m-1) z^{m-2}+(m-1)(m-2) z^{m-3}+···+(m-1)!]$... Put this value in original integral you are done and finally put limits.

Answer 2

For $m=0$, $$ a_{0}=\int_{0}^{T}e^{-ut}dt=-\frac{e^{-ut}}{u}\mid_{0}^{T}=\frac{1}{u}\left(1-e^{-uT}\right) $$ For integer $m\geq1$, \begin{align*} a_{m} & =\int_{0}^{T}t^{m}e^{-ut}dt\\ & =-t^{m}\frac{e^{-ut}}{u}\mid_{0}^{T}+\frac{m}{u}\int_{0}^{T}t^{m-1}e^{-ut}dt\\ & =-T^{m}\frac{e^{-uT}}{u}+\frac{m}{u}a_{m-1}. \end{align*} This gives you a recurrence relation for $a_{m}$, which is probably most amenable to computation (please double check my math; IBP is prone to errors).

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Also, Wolfram computes this as $$ a_{m}=\frac{\Gamma(m+1)-\Gamma(m+1,uT)}{u^{m+1}} $$ where $\Gamma(\cdot,\cdot)$ is the incomplete gamma function. I believe this holds for noninteger $m$ too.