How to integrate $t^{m-1} \textrm e^{-ut}$?

Question

Could you please help me to compute $$\int t^{m-1} \textrm e^{-ut} \textrm dt$$?

Answer 1

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..

Answer 2

Hint

Substitute $x=ut$ and have a look at the gamma-function.

Answer 3

Change of variable

$$ut = z ~~~~~~~ \text{d}z = u\text{d}t ~~~~~~~ t = \frac{z}{u}$$

Hence

$$\frac{1}{u^m}\int z^{m-1}\ e^{-z}\ \text{d}z$$

TO evaluate this, if you have the indefinite integral, you can either integrate by parts $m-1$ times or try with a Taylor series of the exponential. In se second case:

$$\frac{1}{u^m}\int z^{m-1} \sum_{k = 0}^{+\infty} \frac{(-z)^k}{k!}\ \text{d}z$$

$$\sum_{k = 0}^{+\infty} \frac{(-1)^k}{k!} \frac{1}{u^m}\int z^{m-1} z^k\ \text{d}z$$

$$\frac{1}{u^m}\sum_{k = 0}^{+\infty} \frac{(-1)^k}{k!}\frac{z^{m+k}}{m+k}$$

This is a well known series, which gives you:

$$\sum_{k = 0}^{+\infty} \frac{(-1)^k}{k!}\frac{z^{m+k}}{m+k} = \Gamma(m) - \Gamma(m, x)$$

So finally

$$\frac{1}{u^m}\int z^{m-1}\ e^{-z}\ \text{d}z = \frac{\Gamma(m) - \Gamma(m, x)}{u^m}$$

Special Function Involved

• Gamma Function
• Incomplete Gamma Function

THE DEFINITE INTEGRAL

If instead you're searching for something like

$$\int_0^{+\infty} t^{m-1} e^{-ut}\ \text{d}t$$

Then substitute in the same way to get

$$\frac{1}{u^m}\int_0^{+\infty}z^{m-1}e^{-z}\ \text{d}z$$

This integral is very famous and it's nothing but the Gamma Function, so:

$$\int_0^{+\infty}z^{m-1}e^{-z}\ \text{d}z = \Gamma(m)$$

Hence

$$\frac{1}{u^m}\int_0^{+\infty}z^{m-1}e^{-z}\ \text{d}z = \frac{\Gamma(m)}{u^m}$$