How to evaluate the definite integral $\int_0^\infty t^{n-1}e^{-at}dt$

Question

How to evaluate the following definite integral $\int_0^\infty t^{n-1}e^{-at}dt$.

Answer 1

The Gamma function is defined as $$ \Gamma(x)=\int_0^\infty t^{x-1}e^{-t}\,\mathrm{d}t $$ so substituting $t\mapsto t/a$ gives $$ \begin{align} \int_0^\infty t^{n-1}e^{-at}\,\mathrm{d}t &=\frac1{a^n}\int_0^\infty t^{n-1}e^{-t}\,\mathrm{d}t\\ &=\frac1{a^n}\Gamma(n)\\ &=\frac{(n-1)!}{a^n}\qquad\text{if $n$ is a positive integer} \end{align} $$

Answer 2

Apply integration by parts repeatedly. Each time you do, the power of $t$ will drop by one. Assuming $n$ is an integer, this ends after $n-1$ repetitions .

Answer 3

Hint: Check out Gamma functions. And substitute $at=u$. I am assuming $a>0$ here.

Answer 4

By integration by parts we get,

$$\int_0^{\infty}t^{n-1}e^{-at}dt=[-\frac{1}{a}t^{n-1}e^{-at}]_0^{\infty}+\frac{1}{a}\int_0^{\infty}(n-1)t^{(n-2)}e^{-at}dt$$ $$=[-\frac{1}{a}t^{n-1}e^{-at}]_0^{\infty}+[-\frac{1}{a^2}t^{n-1}e^{-at}]_0^{\infty}+\int_0^{\infty}(n-2)t^{(n-3)}e^{-at}dt$$ $$=...=\frac{(n-1)!}{a^{n-1}}\int_0^{\infty}e^{-at}dt=\frac{(n-1)!}{a^{n}}$$

Answer 5

Or directly$$\partial _{a}^{k}\int_{0}^{\infty }dt\exp [-at]=(-1)^{k}\int_{0}^{\infty }dtt^{k}\exp [-at],$$where the left hand side is easily calculated.