Solution containing Riemann Zeta function for an integral involving the EGF of the Bernoulli/Euler polynomials In this post, the first of the following integrals is questioned. I added the second one.
$$
\begin{align}
&2\int_{0}^{\infty}\left(\sum_{k=0}^{n}\frac{\left(-1\right)^{k}B_{k}(1)}{k!}x^{k-n-1}-\frac{1}{x^{n}\left(e^{x}-1\right)}\right)dx\\\\
=&\ \frac{1}{1-2^{n}}\int_{0}^{\infty}\left(\sum_{k=0}^{n-1}\frac{\left(-1\right)^{k}E_k(1)}{k!}x^{k-n}-\frac{2}{x^{n}\left(e^{x}+1\right)}\right)dx\\\\
=&\ \frac{\zeta (n)}{(2\pi)^{n-1}},\quad\text{for odd } n.
\end{align}
$$
The result is conjectured. How can we prove it?
This problem is derived from integrating the EGF of the Bernoulli & Euler polynomials after dividing it by a power of $x$. See my previous post, which outlines the far more generalized problem (and a generalized conjecture).
 A: The first integral can be simplified using the property of the Bernoulli polynomials
\begin{equation}
 B_{n}\left(0\right)=(-1)^{n}B_{n}\left(1\right)=B_{n}
\end{equation}
where $B_n$ are the Bernoulli numbers. Moreover, for $n=1,2,\cdots$
\begin{equation}
  B_{2n+1}=0
\end{equation}
and $B_1=-1/2$. Then, for $n=2p+1$, an odd integer,
\begin{align}
 I_{2p+1}&=2\int_{0}^{\infty}\left(\sum_{k=0}^{2p+1}\frac{\left(-1\right)^{k}B_{k}(1)}{k!}x^{k-2p-2}-\frac{1}{x^{2p+1}\left(e^{x}-1\right)}\right)\,dx\\
 &=2\int_{0}^{\infty}\left(\frac1x\sum_{k=0}^{2p+1}\frac{B_{k}}{k!}x^{k}-\frac{1}{e^{x}-1}\right)\,\frac{dx}{x^{2p+1}}\\
 &=2\int_{0}^{\infty}\left(-\frac12+\frac1x\sum_{r=0}^{p}\frac{B_{2r}}{(2r)!}x^{2r}-\frac{1}{e^{x}-1}\right)\,\frac{dx}{x^{2p+1}}
\end{align}
By changing $x=-y$ in the integral, we have also
\begin{align}
 I_{2p+1}&=-2\int_{-\infty}^0\left(-\frac12-\frac1y\sum_{r=0}^{p}\frac{B_{2r}}{(2r)!}y^{2r}-\frac{1}{e^{-y}-1}\right)\,\frac{dy}{y^{2p+1}}\\
 &=-2\int_{-\infty}^0\left(\frac12-\frac1y\sum_{r=0}^{p}\frac{B_{2r}}{(2r)!}y^{2r}+\frac{1}{e^{y}-1}\right)\,\frac{dy}{y^{2p+1}}
\end{align}
as
\begin{equation}
 \frac{1}{e^{-y}-1}=-1-\frac{1}{e^{y}-1}
\end{equation}
Thus, by taking the half-sum of both representations
\begin{equation}
 I_{2p+1}=\int_{-\infty}^{\infty}\left(-\frac12+\frac1x\sum_{r=0}^{p}\frac{B_{2r}}{(2r)!}x^{2r}-\frac{1}{e^{x}-1}\right)\,\frac{dx}{x^{2p+1}}
\end{equation}
To evaluate this integral with the residue method, we close the contour by the upper half circle. At $x=0$, the integrand
\begin{equation}
 f(x)=x^{-2p-1}\left(-\frac12+\frac1x\sum_{r=0}^{p}\frac{B_{2r}}{(2r)!}x^{2r}-\frac{1}{e^{x}-1}\right)
\end{equation}
has a removable singularity. Indeed, from the EGF of the Bernoulli numbers, for $x\to0$,
\begin{equation}
 f(x)=O\left( x \right)
\end{equation}
The contribution of the half-circle vanishes as for $x\to\infty$
\begin{equation}
 f(x)=O\left( x^{-2} \right)
\end{equation}
The poles in the contour are at $x=2ik\pi$, with $k=1,2,\cdots$. Their residues are $-\frac{1}{(2ik\pi)^{2p+1}}$. Finally,
\begin{align}
 I_{2p+1}&=2i\pi\sum_{k=1}^\infty\frac{-1}{(2ik\pi)^{2p+1}}\\
 &=\frac{1}{(2\pi)^{2p}}\zeta(2p+1)
\end{align}
as expected.
The same method applies for the second integral by remarking that (DLMF)
\begin{equation}
 E_{n}\left(1\right)=\frac{2}{n+1}(2^{n+1}-1)B_{n+1}
\end{equation}
one can extend the integral over $(-\infty,\infty)$. Here again, the singularity at the origin is removable due to the EGF of the Euler polynomials. The poles are at $x=(2k+1)i\pi$ and considering that
\begin{equation}
 \sum_{k=0}^\infty\frac{1}{(2k+1)^{2p+1}}=\left( 1-2^{-2p-1} \right)\zeta(2p+1)
\end{equation}
the identity follows.
