\begin {align*} \int _{0}^{\infty }x^{n}e^{-x}dx & =n!\\ \int _{0}^{\infty }x^{n}e^{-ax}dx & =n!\frac {1}{a^{n+1}}\qquad \text {use }y=ax\\ \int _{0}^{\infty }x^{3}e^{-x}dx & =3!\\ \int _{0}^{\infty }\frac {x^{3}}{e^{x}-1}dx & =3!\xi \relax (4) \end {align*}
Start by multiplying numerator and denominator by \(e^{-x}\) using \(\frac {1}{1-y}=1+y+y^{2}+\cdots \) which becomes \(\int _{0}^{\infty }x^{3}\sum _{n=1}^{\infty }e^{-nx}dx\) or \(\sum _{n=1}^{\infty }\int _{0}^{\infty }x^{3}e^{-nx}dx\), then use \(z=nx\), this gives \(\sum _{n=1}^{\infty }\frac {1}{n^{4}}\int _{0}^{\infty }z^{3}e^{-z}dx\) or \(\left (3!\right ) \sum _{n=1}^{\infty }\frac {1}{n^{4}}\) or \(3!\xi \relax (4) \)\[ \int _{0}^{\infty }e^{-x^{4}}dx=\frac {1}{4}\Gamma \left (\frac {1}{4}\right ) \] Start by using \(x^{4}=y\) or \(x=y^{\frac {1}{4}}\). then \(\frac {dy}{dx}=\frac {1}{4}y^{\left (\frac {1}{4}-1\right ) }\), now the integral becomes \(\frac {1}{4}\int _{0}^{\infty }y^{\left (\frac {1}{4}-1\right ) }e^{-y}dy\) and compare this to \(\int _{0}^{\infty }y^{\left (s-1\right ) }e^{-x}dx=\Gamma \left ( s\right ) \)\[ \int _{0}^{\infty }e^{-\sqrt {x}}dx=\int _{0}^{\infty }e^{-x^{\frac {1}{2}}}dx \] Use same method as above. Will get \(2\Gamma \relax (2) =2\)\begin {align*} \zeta \relax (s) \Gamma \relax (s) & =\int _{0}^{\infty }\frac {x^{\left (s-1\right ) }}{e^{x}-1}dx\qquad s>1\\ \zeta \left (n+1\right ) \left (n!\right ) & =\int _{0}^{\infty }\frac {x^{n}}{e^{x}-1}dx\qquad n>0\\ \zeta \relax (2) & =\frac {\pi ^{2}}{6}\\ \zeta \relax (4) & =\frac {\pi ^{4}}{90}\\ \zeta \relax (s) & =\sum _{n=1}^{\infty }\frac {1}{n^{s}}\qquad s>1\\ \zeta \relax (4) & =\frac {1}{1^{4}}+\frac {1}{2^{4}}+\frac {1}{3^{4}}+\cdots \end {align*}
So given \(\int _{0}^{\infty }\frac {x^{3}}{e^{x}-1}dx\), write as \(\int _{0}^{\infty }\frac {x^{\left (4-1\right ) }}{e^{x}-1}dx=\zeta \relax (4) \Gamma \relax (4) \) or \(\left (3!\right ) \zeta \relax (4) \)\begin {align*} \int _{0}^{\infty }x^{n}e^{-x}dx & =n!\\ \int _{0}^{\infty }x^{1-n}e^{-x}dx & =\Gamma \relax (n) =\left ( n-1\right ) !\qquad \end {align*}
\begin {align*} I & =\int \frac {dx}{\sqrt {a^{2}-x^{2}}}\qquad \text {use }x=a\sin \theta \\ I & =\int \frac {dx}{x^{2}+a^{2}}\qquad \text {use }x=a\tan \theta \\ I & =\int _{0}^{\infty }xe^{-ax^{2}}dx\qquad \text {use }u=x^{2}\\ I & =\int _{0}^{\infty }e^{-ax^{2}}dx\qquad \text {use }I=\frac {1}{2}\int _{-\infty }^{\infty }e^{-ax^{2}}dx=\frac {1}{2}\sqrt {\frac {\pi }{a}} \end {align*}
For \(I=\int _{0}^{\infty }x^{n}e^{-ax^{2}}dx\) or \(I=\int _{-\infty }^{\infty }x^{n}e^{-ax^{2}}dx\). If \(n\) is even, use the trick of \(I\relax (a) =\int _{-\infty }^{\infty }e^{-ax^{2}}dx\) and repeated \(I^{\prime }\left ( a\right ) \). if \(n\) is odd, use \(I\relax (a) =\int _{-\infty }^{\infty }xe^{-ax^{2}}dx=\frac {1}{2a}\) (integration by parts) and then repeated \(I^{\prime }\relax (a) \).
GAMMA:
\begin {align*} \Gamma \relax (n) & =\int _{0}^{\infty }x^{n-1}e^{-x}dx\\ \Gamma \left (\frac {1}{2}\right ) & =\int _{0}^{\infty }x^{\frac {-1}{2}}e^{-x}dx \end {align*}
use \(u=x^{\frac {1}{2}}\), then \(\frac {du}{dx}=\frac {1}{2}x^{-\frac {1}{2}}\) and the integral becomes \(\int _{0}^{\infty }x^{\frac {-1}{2}}e^{-x}dx=\int _{0}^{\infty }\frac {1}{u}e^{-u^{2}}\left (2udu\right ) =2\int _{0}^{\infty }e^{-u^{2}}du=\sqrt {\pi }\)\begin {align*} I & =\int _{0}^{\infty }xe^{-ax}\sin kx\ dx\\ I & =\int _{0}^{\infty }xe^{-ax}\cos kx\ dx \end {align*}
For these, we will be given \(I=\int _{0}^{\infty }e^{-ax}\sin kx\ dx\) and then use \(I\relax (a) =\int _{0}^{\infty }e^{-ax}\sin kx\ dx\) and then do the \(I^{\prime }\relax (a) \) method.
\begin {align*} \int _{-\infty }^{\infty }e^{-x^{2}}dx & =\sqrt {\pi }\\ \int _{-\infty }^{\infty }e^{-ax^{2}}dx & =\sqrt {\frac {\pi }{a}}\qquad a>0\\ \int _{-\infty }^{\infty }e^{-a\left (x+b\right ) ^{2}}dx & =\sqrt {\frac {\pi }{a}}\qquad a>0\\ \int _{-\infty }^{\infty }x^{n}e^{-ax^{2}}dx & =I\qquad \text {for }n\text { even, use the }I^{\prime }\relax (a) \text { method} \end {align*}