Let $\alpha >0$ and $2 n > \alpha$ and $n \in {\mathbb N}$. We consider a following integral:
\begin{equation} {\mathcal A}_{2 n}(\alpha) := \int\limits_0^\infty \left( \frac{1-e^{-x}}{1+e^{-x}} \right)^{2 n} \frac{dx}{x^{1+\alpha}} \end{equation}
We have computed this integral in a little bit dodgy way and then verified numerically that the result is correct. Below we want to present this derivation and then ask the reader to help us fix the details to satisfy the mathematical rigor. Here we go:
We start from the following identity: \begin{eqnarray} \left( \frac{1-e^{-x}}{1+e^{-x}} \right)^{2 n} &=& 1+ \sum\limits_{k=1}^\infty e^{-k x} \cdot \sum\limits_{p=0}^{2 n} \binom{2n}{p} (-2)^p (-1)^k \binom{k+p-1}{p-1} \quad (i) \end{eqnarray} which follows from decomposing the quantity in the parentheses into partial fractions and then expanding the result using the binomial expansion. We now have:
\begin{eqnarray} {\mathcal A}_{2 n}(\alpha) &:=& \sum\limits_{p=0}^{2 n} \binom{2 n}{p} (-2)^p \frac{1}{(p-1)!} \cdot \sum\limits_{k=1}^\infty (-1)^k \cdot (k+1)^{(p-1)} \cdot k^\alpha \cdot \Gamma(-\alpha) \\ &=& \Gamma(-\alpha) \cdot \sum\limits_{p=0}^{2 n} \binom{2 n}{p+1} \frac{(-2)^p}{p!} \sum\limits_{q=0}^{p+1} \left[{p+1\atop q}\right] \cdot \underbrace{\sum\limits_{k=1}^\infty \frac{(-1)^k}{k^{1-\alpha-q}}}_{-\eta(1-\alpha-q)} \\ &=& \Gamma(-\alpha) \cdot \sum\limits_{p=0}^{2 n} \binom{2 n}{p+1} \frac{(-2)^p}{p!} \sum\limits_{q=0}^{p+1} \left[{p+1\atop q}\right] \cdot (2^{\alpha+q}-1) \cdot \zeta(1-\alpha-q) \\ &=&\sum\limits_{p=0}^{2n-1} \sum\limits_{q=0}^{p+1} \frac{(-1)^p}{p!} \binom{2 n}{p+1} \left[{p+1\atop q}\right] \cdot \left(2^{\alpha +q}-1\right) 2^{-\alpha +p-q+2} \pi ^{-\alpha -q} \Gamma (-\alpha ) \Gamma (\alpha +q) \zeta (\alpha +q) \sin \left(\frac{1}{2} \pi (\alpha +q-1)\right) \end{eqnarray} In the first line above we divided both sides of $(i)$ by $x^{1+\alpha}$ and integrated over the half line. Here is the dodgy part, firstly we completely neglected the unity in front of the sum in the right hand side and secondly we also completely neglected the fact that the actual integrals under the infinite sum are divergent and we simply replaced them by the analytic continuation of the Gamma function. We also completely neglect the fact that the last sum over $k$ is divergent (not even asymptotically convergent). In the second line we substituted $p \leftarrow p+1$ then we used the fact that $(k+1)^{(p)} \cdot k^\alpha = k^{(p+1)} \cdot k^{\alpha-1} $ and then we expressed the Pochhammer symbol through powers of $k$ through unsigned Stirling numbers of the first kind and then we recognized the last sum over $k$ as the Dirichlet eta function. In the third line we expressed the eta function through the Riemann zeta function and in the fourth line we used the reflection formula for the zeta function and we simplified the result.
We verify the result numerically using the code below:
In[1596]:= alpha = RandomReal[{1, 2}, WorkingPrecision -> 50];
n = RandomInteger[{1, 5}];
l1 = Table[
NIntegrate[
If[x < 1/1000,
x^(2 n)/2^(2 n) - (x^(2 n + 2) n)/(
2^(2 n) 6 ), ((Exp[x] - 1)/(Exp[x] + 1))^(2 n )] 1/
x^(1 + alpha), {x, 0, Infinity}, WorkingPrecision -> 20], {n, 1,
5}]
l2 = Table[-Gamma[-alpha] Sum[
Binomial[2 n, p + 1] (-2)^(p + 1)/p! Abs[
StirlingS1[p + 1, q]] (1 - 2^(alpha + q)) Zeta[
1 - alpha - q], {p, 0, 2 n - 1}, {q, 0, p + 1}], {n, 1, 5}];
l3 = Table[
Sum[((-1)^p 2^(
2 - alpha + p - q) (-1 + 2^(alpha + q)) \[Pi]^(-alpha - q)
Abs[StirlingS1[1 + p, q]] Binomial[2 n,
1 + p] Gamma[-alpha] Gamma[alpha + q] Sin[
1/2 \[Pi] (-1 + alpha + q)] Zeta[alpha + q])/
p!, {p, 0, 2 n - 1}, {q, 0, p + 1}], {n, 1, 5}];
Abs[l1/l2 - 1]
Abs[l1/l3 - 1]
Out[1598]= {0.77764847559837969768, 0.28278293320917398700, \
0.20653325308677948682, 0.17307652728606858385, 0.15356015000807632272}
Out[1601]= {1.91*10^-17, 3.38*10^-17, 1.31816*10^-15, 1.1*10^-19,
0.*10^-20}
Out[1602]= {1.91*10^-17, 3.38*10^-17, 1.31816*10^-15, 1.1*10^-19,
0.*10^-20}
As I said it already before how to we make this derivation mathematically rigorous, especially how do we fix the first step in this derivation.