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.
$$\int_0^\infty x^a e^{-bx}\,dx=\frac1{b^{a+1}}\int_0^\infty y^ae^{-y}\,dy =\frac{\Gamma(a+1)}{b^{a+1}}$$ where $\Gamma$ denotes the gamma function. The gamma function satisfies $$\Gamma(a+1)=a\Gamma(a)$$ and so for integers $n\ge0$, $$\Gamma(n+1)=n!.$$
Hint: Use the change of variables $u=bx$ and then use the gamma function.
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 $y$).
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.
Hint: Look at the definition of the Gamma function and consider what substitution might be appropriate.
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 $(2)$).
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
Obviously sub $t=x$, $s=b$ and $n=a$ into the first bullet point and you are home and dry.
Generating functions also work.
With a slight change of notation define, $$I(m,n)=\int_0^{+\infty} x^n e^{-mx}\ dx,\quad G(t)=\sum_{n=0}^\infty\frac{I(n,m)}{n!}t^n,$$ we have, $$\begin{align*} G(t)&=\sum_{n=0}^\infty\frac{I(m,n)}{n!}t^n=\int_0^{+\infty} e^{-mx}\sum_{n=0}^\infty \frac{(xt)^n}{n!}\ dx\\&=\int_0^{+\infty} e^{-mx}e^{xt}\ dx =\int_0^{+\infty} e^{(t-m)x}\ dx=\frac{1}{m-t} \end{align*}$$ for $t<m$. Expanding the result as a geometric series in the same region, $$\frac{1}{m-t}=\frac{1}{m}\frac{1}{1-t/m}=\frac{1}{m}\sum_{n=0}^\infty\frac{t^n}{m^n}=\sum_{n=0}^\infty\frac{t^n}{m^{n+1}}=\sum_{n=0}^\infty\frac{n!}{m^{n+1}}\frac{t^n}{n!}$$ comparing coefficients, $$I(n,m)=\left[t^n/n!\right]G(t)=\frac{n!}{m^{n+1}}.$$
I think we need to assume that $b$ is a positive real number. Let us make the substitution $u=bx$. Then $\small\displaystyle\int_0^\infty x^a e^{-bx}dx=\frac{1}{b^{a+1}}\int_0^\infty u^ae^{-u}du$
Here, the integral $\int_0^\infty u^ae^{-u}du$ is convergent iff $\Re a>-1$, $a\in\Bbb C$ and by definition of Gamma function, it is equal to $\Gamma(a+1).$
