TITLE: Proof of a few equations involving $\int_{\alpha}^{\infty}\frac{1}{t\left(e^{t}\pm1\right)}dt$
QUESTION [10 upvotes]: I derived these formulas with the Laurent series and Euler-Maclaurin summation formula. I can demonstrate this later if anyone's curious.  I'm wondering if there is another way.  I'm also interested in finding generalized formulas.
$$\lim\limits_{\alpha\to 0}\left[\frac{\ln\left(\alpha\right)}{2}+\int_{\alpha}^{\infty}\frac{1}{t\left(e^{t}+1\right)}dt\right]=\frac{1}{2}\left(\ln\left(\pi\right)-\ln\left(2\right)-\gamma\right)$$
$$\lim\limits_{\alpha\to 0}\left[\frac1\alpha+\frac{\ln\left(\alpha\right)}{2}-\int_{\alpha}^{\infty}\frac{1}{t\left(e^{t}-1\right)}dt\right]=\frac{1}{2}\left(\ln\left(\pi\right)+\ln\left(2\right)-\gamma\right)$$
A manipulation of these equations yields 
$$\lim\limits_{s\to -1}\left[\frac{1}{\ln|s|}+\left(-\frac{1}{s+1}+\frac{1}{2}\right)\ln|\ln|s||+\int_{s}^{\infty}\frac{\ln\left(\ln\left(u\right)\right)}{\left(u+1\right)^{2}}du\right]=\frac{1}{2}\left(\ln\left(\pi\right)-3\ln\left(2\right)-\gamma\right)$$
$$\lim\limits_{s\to 1}\left[\frac{1}{\ln\left(s\right)}+\left(\frac{1}{s-1}+\frac{1}{2}\right)\ln\left(\ln\left(s\right)\right)-\int_{s}^{\infty}\frac{\ln\left(\ln\left(u\right)\right)}{\left(u-1\right)^{2}}du\right]=\frac{1}{2}\left(\ln\left(\pi\right)+\ln\left(2\right)-\gamma\right)$$

Here is my elementary method of deriving these:
Begin with the Euler-Maclaurin summation formula:
$$
\begin{align}
\frac{1}{h}\int_a^b f(t)dt
&=\sum_{k=0}^n f(kh+a)-\left(\frac{f(a)+f(b)}{2}\right)
\\&-\sum_{k=1}^n \frac{h^{2k-1}B_{2k}}{(2k)!} \left(f^{(2k-1)}(b)-f^{(2k-1)}(a)\right)
\\&-R
\end{align}
$$
where $h=\frac{b-a}{n}$ and $R$ is the remainder term. Letting $n=\frac{b-a}{x}$ and rearranging we get
$$
\begin{align}
\sum_{k=1}^{(b-a)/x} \frac{x^{2k-1}B_{2k}}{(2k)!} \left(f^{(2k-1)}(b)-f^{(2k-1)}(a)\right)
&=\sum_{k=0}^{(b-a)/x} f(kx+a)-\left(\frac{f(a)+f(b)}{2}\right)
\\&-\frac{1}{x}\int_a^b f(t)dt
\\&-R
\end{align}
$$ 
Limiting $b\to 0$ and $a\to -\infty$, we have
$$
\begin{align}
\lim\limits_{\substack{%
     a \to -\infty\\
     b \to 0}}
\sum_{k=1}^{-a/x} \frac{x^{2k-1}B_{2k}}{(2k)!} \left(f^{(2k-1)}(b)-f^{(2k-1)}(a)\right)
&=\lim\limits_{\substack{%
     a \to -\infty\\
     b \to 0}}\left(\sum_{k=0}^{-a/x} f(kx+a)-\left(\frac{f(a)+f(b)}{2}\right)-\frac{1}{x}\int_a^b f(t)dt\right)
\end{align}
$$
The remainder disappears as $n\to\infty$. Now make a variable substitution in the limit $a\to -ax$.
$$
\begin{align}
\lim\limits_{\substack{%
     a \to \infty\\
     b \to 0}}\sum_{k=1}^{a} \frac{x^{2k-1}B_{2k}}{(2k)!} \left(f^{(2k-1)}(b)-f^{(2k-1)}(-ax)\right)
&=\lim\limits_{\substack{%
     a \to \infty\\
     b \to 0}}\left(\sum_{k=0}^{a} f((k-a)x)-\left(\frac{f(-ax)+f(b)}{2}\right)-\frac{1}{x}\int_{-ax}^b f(t)dt\right)
\\&=\lim\limits_{\substack{%
     a \to \infty\\
     b \to 0}}\left(\sum_{k=0}^{a} f(-kx)-\left(\frac{f(-ax)+f(b)}{2}\right)-\frac{1}{x}\int_{-ax}^b f(t)dt\right)
\end{align}
$$
Now use the following hint.
$$\frac{1}{z(e^z-1)}=\frac{1}{z^2}-\frac{1}{2z}+\sum_{k=1}^\infty\frac{B_{2k}}{(2k)!}z^{2k-2}$$
which when we integrate we get
$$\begin{align}
\int_x^\infty\frac{1}{z(e^z-1)}dz
&=K-\left(-\frac{1}{x}-\frac{\ln{|x|}}{2}+\sum_{k=1}^\infty\frac{B_{2k}}{(2k)!(2k-1)}x^{2k-1}\right)
\end{align}$$
Where $K$ stands for the integral evaluated at $\infty$. 
Let $f(x)=\text{Ei}\left(x\right)-\ln\left|x\right|-\gamma$. Note that $\lim\limits_{t\to 0}f^{(m)}(t)=\frac1m$ and $\lim\limits_{t\to -\infty}f^{(m)}(t)=0$ for $m\ge1$. Further note that $\lim\limits_{t\to 0} f(t) = 0$ so we may substitute the sum: $\sum_{k=0}^{a} f(-kx)=\sum_{k=1}^{a} f(-kx)$. Now we have
$$
\begin{align}
\sum_{k=1}^\infty\frac{B_{2k}}{(2k)!(2k-1)}x^{2k-1}
&=\lim\limits_{\substack{%
     a \to \infty\\
     b \to 0}}\left(\sum_{k=1}^{a} f(-kx)-\left(\frac{f(-ax)+f(b)}{2}\right)-\frac{1}{x}\int_{-ax}^b f(t)dt\right)
\\&=\lim\limits_{\substack{%
     a \to \infty\\
     b \to 0}}\left(\sum_{k=1}^{a}\text{Ei}(-kx)-\sum_{k=1}^{a}\ln\left|-kx\right|-\sum_{k=1}^{a}\gamma-\left(\frac{f(-ax)+f(b)}{2}\right)-\frac{1}{x}\int_{-ax}^b f(t)dt\right)
\\&=-\int_x^\infty\frac{1}{z(e^z-1)}dz+\lim\limits_{\substack{%
     a \to \infty\\
     b \to 0}}\left(-a\ln|x|-\ln|a!|-a\gamma-\left(\frac{f(-ax)+f(b)}{2}\right)-\frac{1}{x}\int_{-ax}^b f(t)dt\right)
\\&=-\int_x^\infty\frac{1}{z(e^z-1)}dz+\frac{1}{x}+\frac{\ln\left|x\right|}{2}+\frac{1}{2}\left(\gamma-\ln\left(2\pi\right)\right)
\end{align}
$$
(The limit is tricky which is why I left out some steps). Therefore
$$\begin{align}
\int_x^\infty\frac{1}{z(e^z-1)}dz=\frac12 (\gamma-\ln(2\pi))+\frac{1}{x}+\frac{\ln{|x|}}{2}-\sum_{k=1}^\infty\frac{B_{2k}}{(2k)!(2k-1)}x^{2k-1}
\end{align}$$
We can derive the other equation with $f(x)=-\text{Ei}(x)+2\text{Ei}(2x)-\ln|4x|-\gamma$.

REPLY [3 votes]: The result can also be found using a Mellin transform technique:
\begin{align}
I^+&=\int_{\alpha}^{\infty}\frac{1}{t\left(e^{t}+1\right)}\,dt\\
&=\frac{1}{2}\int_1^\infty \frac{e^{-\alpha u/2}}{\cosh (\alpha u/2)} \,du\\
&=\frac{1}{2}\int_0^\infty f(u)h\left( \frac{\alpha u}{2} \right)\,du
\end{align} 
where
\begin{align}
f(z)&=\begin{cases}
     0\text{ if } z\le 1\\
     z^{-1}\text{ if } z> 1
     \end{cases}\\
h(z)&=\frac{e^{-z}}{\cosh z}
\end{align}
Mellin transforms of these functions are (see Ederlyi TI 6.6.6, for example)
\begin{align}
\mathcal{M}\left[f(z)\right]&=\frac{1}{1-s} \text{ for }\Re s<1\\
\mathcal{M}\left[h(z)\right]&=2^{1-s}\left( 1-2^{1-s} \right)\Gamma(s)\zeta(s) \text{ for }\Re s>0
\end{align} 
with the Mellin convolution transform (DLMF),
we can choose $c>0$ to express
\begin{align}
I^+&=\frac{1}{4i\pi}\int_{c-i\infty}^{c+i\infty} 2^{1-s}\left( 1-2^{1-s} \right)\Gamma(s)\zeta(s)\left( \frac{\alpha}{2} \right)^{-s}\,\frac{ds}{s}
\end{align}
The oles of the function lie at $s =0,-1,-3,-5,\cdots$, with $s=0$ being double. Corresponding residues are $\ln\pi -2\ln2-\gamma-\ln\frac{\alpha}{2},\frac{\alpha}{2},-\frac{\alpha^3}{72},\cdots$. By closing the integral with the left half-circle, we obtain the expansion 
\begin{equation}
I^+\sim \frac{1}{2}\left( \ln\frac{\pi}{2\alpha} -\gamma\right)+\frac{\alpha}{4}-\frac{\alpha^3}{144}+O\left( \alpha^5 \right)
\end{equation} 
The same method applies for the second integral, with the tabulated transform
\begin{equation}
\mathcal{M}\left[\frac{e^{-z}}{\sinh z}\right]=2^{1-s}\Gamma(s)\zeta(s) \text{ for }\Re s>1
\end{equation} 
As $\zeta(-2n)=0$, the poles lie  at $s=1,0,-1,-3,\cdots$ ($s=0$ being double) with the corresponding residues $\frac{2}{\alpha},\ln\left( \frac{\alpha}{2\pi} \right)+\gamma,\frac{2}{\alpha},-\frac{\alpha}{6},\cdots$. This leads to the result
\begin{equation}
I^{-}\sim \frac{1}{2}\left( \ln\left( \frac{\alpha}{2\pi} \right)+\gamma \right)+\frac{1}{\alpha}-\frac{\alpha}{12}+\cdots
\end{equation} 
In both cases, the proposed expression in terms of the Bernoulli numbers can be retrieved as $\zeta(-2n-1)=-\frac{B_{2n+2}}{2n+2}$.