Trying to evaluate $\int_{0}^{\infty}\frac{\ln^{s}(t)\ln(1+t)}{t(1+t^2)}\mathrm dt$ $$G(s)=\int_{0}^{\infty}\frac{\ln^{s}(t)\ln(1+t)}{t(1+t^2)}\mathrm dt$$
Trying a substitution of $t=e^x$
$$G(s)=\int_{-\infty}^{\infty}\frac{x^s\ln(1+e^x)}{1+e^{2x}}\mathrm dx$$
I trying to evaluate $G(s)$, but unable to. Can anyone help? Thank you.
 A: Assuming $s\in\mathbb{N}$, you may consider that
$$ \int_{0}^{+\infty}\frac{t^{a-1}\log(1+t)}{1+t^2}\,\mathrm dt=\frac{\pi}{4\sin(\pi a)}\left[H_{-1/2-a/4}-H_{-a/4}+2\log(2)\cos\frac{\pi a}{2}+\pi\sin\frac{\pi a}{2}\right] $$
for any $a$ in a neighbourhood of the origin. In order to compute the wanted integral you may simply differentiate (wrt to $a$) the RHS multiple times, then consider the limit as $a\to 0$.
This should work also by considering fractional derivatives, if defined in terms of the Laplace (inverse) transform. Ultimately we are just dealing with $D^{\alpha}\log\Gamma(z)$.
A: Partial answer.
Assuming $s$ is an even positive integer,
$\begin{align}G(s)&=\int_{0}^{\infty}\frac{\ln^{s}(t)\ln(1+t)}{t(1+t^2)}\mathrm dt\\
&=\int_{0}^{1}\frac{\ln^{s}(t)\ln(1+t)}{t(1+t^2)}\mathrm dt+\int_{1}^{\infty}\frac{\ln^{s}(t)\ln(1+t)}{t(1+t^2)}\mathrm dt\end{align}$
In the latter integral perform the change of variable $y=\dfrac{1}{t}$,
$\begin{align}G(s)&=\int_{0}^{1}\frac{\ln^{s}(t)\ln(1+t)}{t(1+t^2)}\mathrm dt+\int_{0}^{1}\frac{(-1)^st\ln^{s}(t)\ln\left(\frac{1+t}{t}\right)}{1+t^2}\mathrm dt\\
&=\int_{0}^{1}\frac{\ln^{s}(t)\ln(1+t)}{t}\mathrm dt-\int_{0}^{1}\frac{t\ln^{s}(t)\ln(1+t)}{1+t^2}\mathrm dt+\int_{0}^{1}\frac{(-1)^st\ln^{s}(t)\ln\left(1+t\right)}{1+t^2}\mathrm dt+\\
&\int_{0}^{1}\frac{(-1)^{s+1}t\ln^{s+1}(t)}{1+t^2}\mathrm dt\\
&=\int_{0}^{1}\frac{\ln^{s}(t)\ln(1+t)}{t}\mathrm dt+\int_{0}^{1}\frac{(-1)^{s+1}t\ln^{s+1}(t)}{1+t^2}\mathrm dt
\end{align}$
In the latter integral perform the change of variable $\displaystyle y=t^2$,
$\begin{align}G(s)&=\int_{0}^{1}\frac{\ln^{s}(t)\ln(1+t)}{t}\mathrm dt+\frac{(-1)^{s+1}}{2^{s+2}}\int_{0}^{1}\frac{\ln^{s+1}(t)}{1+t}\mathrm dt\\
&=\int_{0}^{1}\frac{\ln^{s}(t)\ln(1+t)}{t}\mathrm dt-\frac{1}{2^{s+2}}\int_{0}^{1}\frac{\ln^{s+1}(t)}{1+t}\mathrm dt\\
\end{align}$
Perform integration by parts,
$\begin{align}H(s)&=\int_0^1\frac{\ln(1+t)\ln^{s} t}{t}\mathrm dt\\
&=\Big[\ln t \ln^s t\ln(1+t)\Big]_0^1-\int_0^1 \frac{\ln^{s+1} t}{1+t}\mathrm dt-s\int_0^1 \frac{\ln(1+t)\ln^{s} t}{t}\mathrm dt\\
&=-\int_0^1 \frac{\ln^{s+1} t}{1+t}\mathrm dt-sH(s)
\end{align}$
Therefore,
$\begin{align}H(s)=-\frac{1}{s+1}\int_0^1 \frac{\ln^{s+1} t}{1+t}\mathrm dt\end{align}$
$\begin{align}G(s)&=-\left(\frac{1}{2^{s+2}}+\frac{1}{s+1}\right)\int_0^1 \frac{\ln^{s+1} t}{1+t}\mathrm dt\end{align}$
On the other hand,
$\begin{align}\int_0^1 \frac{\ln^{s+1} t}{1+t}\mathrm dt&=\int_0^1 \frac{\ln^{s+1} t}{1-t}\mathrm dt-\int_0^1 \frac{2t\ln^{s+1} t}{1-t^2}\mathrm dt\end{align}$
In the latter integral perform the change of variable $y=x^2$,
$\begin{align}\int_0^1 \frac{\ln^{s+1} t}{1+t}\mathrm dt&=\int_0^1 \frac{\ln^{s+1} t}{1-t}\mathrm dt-\frac{1}{2^{s+1}}\int_0^1 \frac{\ln^{s+1} t}{1-t}\mathrm dt\\
&=\left(1-\frac{1}{2^{s+1}}\right)\int_0^1 \frac{\ln^{s+1} t}{1-t}\mathrm dt\\
&=(-1)^{s+1}\left(1-\frac{1}{2^{s+1}}\right)(s+1)!\zeta(s+2)\\
&=-\left(1-\frac{1}{2^{s+1}}\right)(s+1)!\zeta(s+2)
\end{align}$
Therefore,
If $s$ is an even positive integer,
$\begin{align}\boxed{G(s)=\left(\frac{1}{2^{s+2}}+\frac{1}{s+1}\right)\left(1-\frac{1}{2^{s+1}}\right)(s+1)!\zeta(s+2)}\end{align}$
A: For $n \in \mathbb{N}_0$ we can use the substitution $x \to -x$ in the third integral and a well-known integral representation of the Dirichlet $\eta$-function to find
\begin{align}
G(n) &= \int \limits_0^\infty \frac{x^{n+1}}{\mathrm{e}^{2x} +1} \, \mathrm{d} x + \int \limits_0^\infty \frac{x^n \ln(1+\mathrm{e}^{-x})}{\mathrm{e}^{2x}+1} \, \mathrm{d} x + \int \limits_{-\infty}^0 \frac{x^n \ln(1+\mathrm{e}^{x})}{\mathrm{e}^{2x}+1} \, \mathrm{d} x \\
&= \frac{(n+1)!}{2^{n+2}} \eta(n+2) + \int \limits_0^\infty x^n \ln(1+\mathrm{e}^{-x}) \frac{(-1)^n \mathrm{e}^{2x}+1}{\mathrm{e}^{2x}+1} \, \mathrm{d} x \, .
\end{align}
For even $n$ the remaining integral reduces to the $\eta$-function again after integrating by parts and we obtain
$$ G(n) = \left[\frac{1}{2^{n+2}} + \frac{1}{n+1}\right] (n+1)! \, \eta(n+2) \, , \, n \in 2 \mathbb{N}_0 \, . $$
For odd $n$ I do not know how to simplify the remaining integral:
$$ G(n) = \frac{(n+1)!}{2^{n+2}} \eta(n+2) - \int \limits_0^\infty x^n \ln(1+\mathrm{e}^{-x}) \tanh(x) \, \mathrm{d} x \, . $$
If we use series expansions, we end up with
$$ G(n) = \frac{(n+1)!}{2^{n+2}} \eta(n+2) - n! \sum \limits_{k,l=0}^\infty \frac{(-1)^{k+l}}{k+1} \left[\frac{1}{(k+2l+1)^{n+1}} - \frac{1}{(k+2l+3)^{n+1}}\right] \, , $$
but this does not look too pleasant either.
