How to evaluate the definite integral $\int_0^\infty t^{n-1}e^{-at}dt$ How to evaluate the following definite integral $\int_0^\infty t^{n-1}e^{-at}dt$.
 A: 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
$$
\begin{align}
\int_0^\infty t^{n-1}e^{-at}\,\mathrm{d}t
&=\frac1{a^n}\int_0^\infty t^{n-1}e^{-t}\,\mathrm{d}t\\
&=\frac1{a^n}\Gamma(n)\\
&=\frac{(n-1)!}{a^n}\qquad\text{if $n$ is a positive integer}
\end{align}
$$
A: 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 .
A: Hint: Check out Gamma functions. And substitute $at=u$. I am assuming $a>0$ here.
A: 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.
A: By integration by parts we get,
$$\int_0^{\infty}t^{n-1}e^{-at}dt=[-\frac{1}{a}t^{n-1}e^{-at}]_0^{\infty}+\frac{1}{a}\int_0^{\infty}(n-1)t^{(n-2)}e^{-at}dt$$ 
$$=[-\frac{1}{a}t^{n-1}e^{-at}]_0^{\infty}+[-\frac{1}{a^2}t^{n-1}e^{-at}]_0^{\infty}+\int_0^{\infty}(n-2)t^{(n-3)}e^{-at}dt$$
$$=...=\frac{(n-1)!}{a^{n-1}}\int_0^{\infty}e^{-at}dt=\frac{(n-1)!}{a^{n}}$$
