Evaluation of $\frac{1}{2\pi i }\int_{c-i\infty}^{c+i\infty}\frac{\log(s)^{n}\log(1-s)^{m}}{s(1-s)}ds$

Question

I am having trouble trying to evaluate the integral : $$\frac{1}{2\pi i }\int_{c-i\infty}^{c+i\infty}\frac{\log(s)^{n}\log(1-s)^{m}}{s(1-s)}ds$$ Where $0<c<1$ and $n,m$ are positive integers. The integrand has two branch cuts : $(-\infty,0]$ and $[1,\infty)$, so shifting the line of integration doesn't work. I tried bending the line of integration so it lies just above and just below one of the branch cuts, but that didn't work either, or i have made a mistake. any insight is appreciated.

EDIT :

After a couple of transformations, i obtained : $$\frac{1}{2\pi i }\int_{c-i\infty}^{c+i\infty}\frac{\log(s)^{n}\log(1-s)^{m}}{s(1-s)}ds=$$ $$\frac{1}{2\pi i }\int_{-\infty}^{\infty}\frac{\log(is)^{n}\log(1-is)^{m}}{s(1-is)}ds=-\Re\left[\frac{1}{\pi i }\int_{0}^{\infty}\frac{y ^{n}\log(1-e^{y})^{m}}{1-e^{y}}dy \right ]$$ Still, i have no idea on how to evaluate the integral !!

Answer 1

Your integral can be expressed in simple combinations of zeta values. Let $$I_{n,m} = \frac{1}{2\pi i }\int_{c-i\infty}^{c+i\infty}\frac{\ln(s)^{n}\ln(1-s)^{m}}{s(1-s)}ds$$ then bending contour gives $$\begin{aligned} I_{n,m} &= \frac{{ - 1}}{{2\pi i}}\int_{ - \infty }^0 {\frac{{{{\ln }^m}(1 - s)}}{{s(1 - s)}}\left[ {{{(\ln \left| s \right| + \pi i)}^n} - {{(\ln \left| s \right| - \pi i)}^n}} \right]ds} \\ &= \frac{1}{{2\pi i}}\int_0^\infty {\frac{{{{\ln }^m}(1 + s)}}{{s(1 + s)}}\left[ {{{(\ln s + \pi i)}^n} - {{(\ln s - \pi i)}^n}} \right]ds} \end{aligned}$$

Thus it suffices to evaluate $$J_{n,m} = \int_0^\infty {\frac{{{{\ln }^n}s{{\ln }^m}(1 + s)}}{{s(1 + s)}}ds} $$ this is easy, simply note $$\int_0^\infty {{s^{a - 1}}{{(1 + s)}^{b - 1}}ds} = B(a,1 - a - b) = \frac{{\Gamma (a)\Gamma (1 - a - b)}}{{\Gamma (1 - b)}}$$ so $J_{n,m}$ can be calculated via differentiation of $a,b$.

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The following Mathematica code calculates $I_{n,m}$, with same notations as above

int[n0_, m0_] := Module[{n = n0, m = m0}, f[a_, b_] := Gamma[a]*Gamma[1 - a - b]/Gamma[1 - b]; 
    j[n_, m_] := D[Normal[Series[f[a, b], {a, 0, n}, {b, 0, m}]], {a, n}, {b, m}] 
    /. {a -> 0, b -> 0} // FullSimplify; 
    coefflist[n_] := CoefficientList[(x + Pi*I)^n - (x - Pi*I)^n, x]; 
    For[i = 0; sum = 0, i < n, i++,
    sum = If[coefflist[n][[i + 1]] == 0, sum, sum + j[i, m]*coefflist[n][[i + 1]]]];
FullSimplify[sum/(2 Pi*I)]]

For example, $$I_{2,3} = 2 \pi ^2 \zeta (3)+24 \zeta (5) \qquad I_{1,4}=I_{4,1}=24\zeta(5)$$$$I_{4,2} = 24 \zeta (3)^2+\frac{68 \pi ^6}{315} \qquad I_{3,3} = 36 \zeta (3)^2+\frac{107 \pi ^6}{420}$$ $$I_{4,4} = 96 \pi ^2 \zeta (3)^2+2304 \zeta (5) \zeta (3)+\frac{3701 \pi ^8}{3150}$$