How can I calculate this integral by hand? In solving this problem, I come up with the following integral:
$$\int_{-1/(4\pi)}^0\frac{(s\log{(-4\pi s)})^{(2+n)/2}}{s^2}ds$$
where $n=1,2,3...$
By using Mathematica, I could get that the integral is equal to $$\frac{8\pi^{-n/2}\Gamma(2+n/2)}{2^{1+n/2}n^{(4+n)/2}}$$
My question is: How do we calculate it by hand? Is there any clever way to do it?
 A: Rewrite your integral as $$\frac{(2n)^{1+n/2}}{2+n}\int_{-1/(4\pi)}^{0}s^{-1+n/2}[\log(-4\pi s)]^{1+n/2}ds$$
Then make the substitution $u=\log(-4\pi s),$ giving $du=1/s$ and $s^{n/2}=(-\frac{1}{4\pi} e^{u})^{n/2}$. Your limits become $0$ (upper) and $-\infty$ (lower) - switch them, using up a minus sign. The gamma function should become very apparent.  
Note: You should be much more careful than I have been with minus signs in logarithms and square roots. This is purely a sketch to give you some ideas.
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\expo}[1]{{\rm e}^{#1}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
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\begin{align}
&\overbrace{\int_{-1/\pars{4\pi}}^{{0}}{%
\bracks{s\log{\pars{-4\pi s}}}^{\pars{2 + n}/2}
 \over
 s^{2}}\,\dd s}^{\ds{s\ =\ -\,{x \over 4\pi}\,,\quad x = -4\pi s}}\
=\
\pars{-1}^{\pars{n/2} -1}\,\,{\pi^{-n/2} \over 2^{n}}\quad
\overbrace{\int_{0}^{1}x^{\pars{n/2} - 1}\ln^{\pars{n/2} + 1}\pars{x}\,\dd x}
^{\ds{x\ =\ \expo{-z}\,,\quad z = -\ln\pars{x}}}
\\[3mm]&=
\pars{-1}^{\pars{n/2} - 1}\,\,
{\pi^{-n/2} \over 2^{n}}\,\pars{-1}^{\pars{n/2} + 1}
\int_{0}^{\infty}\expo{-nz/2}\,z^{\pars{n/2} + 1}\,\dd z
\\[3mm]&=
\pars{-1}^{n}\,\,{\pi^{-n/2} \over 2^{n}}\,{1 \over \pars{n/2}^{\pars{n/2} + 2}}
\quad
\overbrace{\int_{0}^{\infty}\expo{-z}\,z^{\bracks{\pars{n/2} + 2} - 1}\,\dd z}
^{\ds{\Gamma\pars{{n \over 2} + 2}}}
\\[3mm]&=
\pars{-1}^{n}\,\,{\pi^{-n/2} \over 2^{-2 + \pars{n/2}}}\,
{1 \over n^{\pars{n/2} + 2}}\Gamma\pars{2 + {n \over 2}}
=
\pars{-1}^{n}\,\,{8\pi^{-n/2} \over 2^{1 + \pars{n/2}}}\,
{1 \over n^{\pars{4 + n}/2}}\Gamma\pars{2 + {n \over 2}}
\\[1cm]&
\end{align}
$$
{\large\int_{-1/\pars{4\pi}}^{{0}}{%
\bracks{s\log{\pars{-4\pi s}}}^{\pars{2 + n}/2}
 \over
 s^{2}}\,\dd s
=
\color{#ff0000}{\large%
\pars{-1}^{n}\,
{8\pi^{-n/2}\,\Gamma\pars{2 + n/2}
 \over
 2^{1 + n/2\,}n^{\pars{4 + n}/2}}}}
$$
Notice that we got a pre factor $\pars{-1}^{n}$ besides the result the OP reports from the Mathematica package. We check explicitly ( we started from the initial integral with $n = 1$ ) our result for $n = 1$ and we found it's negative !!!.
We would like the OP recheck it with Mathematica ( Sorry, I don't have Mathematica ). 
