General formula for evaluating integrals of the form $\displaystyle\int_0^\infty x^ae^{-bx}dx$ 
Find a general formula for evaluating integrals of the form $\displaystyle\int_0^\infty x^a e^{-bx}dx.$

I've been pondering about this question for some time. Obviously integration by parts is way too inefficient. I know there's recursion involved and maybe there's a way to find a general formula through intuition.
 A: Hint: Use the change of variables $u=bx$ and then use the gamma function. 
A: You can do this nicely using the Feynman trick: let
$$I(y) = \int_0^\infty e^{yx}e^{-bx} dx$$
Compute this integral (for $y<b$ and $b\ge1$). Then
$$\int_0^\infty x^ae^{-bx}dx=\left.\frac{d^a}{dy^a}\right|_{y=0}I(y) = -a!b^{-a-1}$$
I leave it to you to fill in the details (and to show that it is allowed to exchange the integral with the $a$ derivatives in $y$).
A: $$\int_0^\infty x^a e^{-bx}\,dx=\frac1{b^{a+1}}\int_0^\infty y^ae^{-y}\,dy
=\frac{\Gamma(a+1)}{b^{a+1}}$$
where $\Gamma$ denotes the gamma function.
The gamma function satisfies
$$\Gamma(a+1)=a\Gamma(a)$$
and so for integers $n\ge0$,
$$\Gamma(n+1)=n!.$$
A: Example: $\displaystyle \int_0^\infty x^3 e^{-2x}dx$ using the shortcut to multiple "integration-by-parts":
$$\begin{align}
x^3 \quad &\searrow^+ \quad e^{-2x}\\
3x^{2} \quad &\searrow^- \quad -e^{-2x}/2\\
6x \quad &\searrow^+ \quad e^{-2x}/4\\
6 \quad &\searrow^- \quad -e^{-2x}/8\\
0 \quad &\;\;\;\;\;\; \quad e^{-2x}/16\\
\end{align}
\\
$$
$\text{such that}\;\displaystyle \int_0^\infty x^3 e^{-2x}dx=\left[-x^3\frac{e^{-2x}}{2}-3x^2\frac{e^{-2x}}{4}-6x\frac{e^{-2x}}{8}-6\frac{e^{-2x}}{16} \right]_0^\infty$
$\mathbf{Note:}$ The above process is tedious for large a and/or b. It is then preferrable to switch to the gamma function instead.
A: Notice that for $a\gt0$,
$$
\begin{align}
F(a,b)
&=\int_0^\infty x^ae^{-bx}\,\mathrm{d}x\\
&=-\frac1b\int_0^\infty x^a\,\mathrm{d}e^{-bx}\\
&=\left[-\frac1bx^ae^{-bx}\right]_0^\infty+\frac ab\int_0^\infty x^{a-1}e^{-bx}\,\mathrm{d}x\\
&=\frac{a}{b}F(a-1,b)\tag{1}
\end{align}
$$
and when $a=0$,
$$
\begin{align}
F(0,b)
&=\int_0^\infty e^{-bx}\,\mathrm{d}x\\
&=\frac1b\tag{2}
\end{align}
$$
Now, check the first few values of $a$ to see if a pattern emerges:
$$
F(0,b)=\frac1b\\
F(1,b)=\frac1b\frac1b\\
F(2,b)=\frac{2\cdot1}{b^2}\frac1b\\
F(3,b)=\frac{3\cdot2\cdot1}{b^3}\frac1b
$$
When you see a pattern, e.g. $\frac{n!}{b^n}\frac1b$, prove it using induction (and $(1)$ and $(2)$).
A: Hint: Look at the definition of the Gamma function and consider what substitution might be appropriate.
A: This may be a bit of a simplistic shortcut compared to previous answers, but hey, I guess the more the merrier: you can use the facts that 


*

*the Laplace transform of $f(t)=t^n$ is $F(s)=\frac{n!}{s^{n+1}}$

*in general the Laplace transform is defined as $\int_0^\infty{e^{-st}f(t)dt}$


Obviously sub $t=x$, $s=b$ and $n=a$ into the first bullet point and you are home and dry.
