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?
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.
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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 ).
$$\int_{-1/(4\pi)}^0\frac{(s\log{(-4\pi s)})^{(2+n)/2}}{s^2},ds$$
$$=\int_0^1\frac{\left((-x/4\pi)\ln x\right)^{n/2+1}}{(-x/4\pi)^2}(-1/(4\pi),dx$$
$$=\int_0^\infty \frac{((-\exp({-z})/4\pi)(-z))^{n/2+1}}{(-\exp({-z})/4\pi)^2}(-1/(4\pi)(-\exp({-z})),dz$$
$$=\frac{8\pi^{-n/2}\Gamma(2+n/2)}{2^{1+n/2}n^{(4+n)/2}}$$
– Pedro Oct 18 '13 at 08:46
