TITLE: How to write a limit in terms of finite summation
QUESTION [5 upvotes]: I managed to find$$\int\limits_0^\infty \frac{\ln^{2a}(x)\ln(1+x)}{\sqrt{x}(1+x)}\mathrm{d}x=-\pi\lim_{m\to \frac12 }\frac{d^{2a}}{d m^{2a}} \frac{\psi(1-m) + \gamma}{\sin(m\pi)}.$$
To show this relation, we follow the same approach here:
Reduce $n$ by $m$ in the beta function:
$$\int_0^\infty\frac{x^{m-1}}{(1+x)^{m+n}}\mathrm{d}x=\operatorname{B}(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)},$$
we have
$$\int_0^\infty\frac{x^{m-1}}{(1+x)^{n}}\mathrm{d}x=\frac{\Gamma(m)\Gamma(n-m)}{\Gamma(n)}.$$
Differentiate both sides $2a$ times with respect to $m$ and once with respect to $n$,
\begin{gather*}
\frac{\partial^{2a}}{\partial m^{2a}} \frac{\partial}{\partial n}  \frac{\Gamma(m)\Gamma(n-m)}{\Gamma(n)}=\frac{\partial^{2a}}{\partial m^{2a}} \frac{\partial}{\partial n}\int\limits_0^\infty \frac{x^{m-1}}{(1+x)^n}\mathrm{d}x\\
\{\text{use differentiation under the integral sign theorem}\}\\
=\int\limits_0^\infty \frac{\partial^{2a}}{\partial m^{2a}} \frac{\partial}{\partial n}\frac{x^{m-1}}{(1+x)^n}\mathrm{d}x\\
=-\int\limits_0^\infty \frac{\ln^{2a}(x)\ln(1+x)x^{m-1}}{(1+x)^n}\mathrm{d}x.
\end{gather*}
Now take the limit on both sides letting $m\to 1/2$ and $n\to1$,
\begin{gather*}
-\int\limits_0^\infty \frac{\ln^{2a}(x)\ln(1+x)}{\sqrt{x}(1+x)}\mathrm{d}x=\lim_{\substack{m\to 1/2 \\ n \to 1}}\frac{\partial^{2a}}{\partial m^{2a}} \frac{\partial}{\partial n} \frac{\Gamma(m)\Gamma(n-m)}{\Gamma(n)}\\
=\lim_{\substack{m\to 1/2 \\ n \to 1}}\frac{\partial^{2a}}{\partial m^{2a}}\Gamma(m)\left( \frac{\partial}{\partial n} \frac{\Gamma(n-m)}{\Gamma(n)}\right)\\
=\lim_{\substack{m\to 1/2 \\ n \to 1}}\frac{\partial^{2a}}{\partial m^{2a}} \Gamma(m)\left(\frac{\Gamma(n-m)[{\psi}(n-m) -\psi(n)]}{\Gamma(n)}\right)\\
\{\text{evaluate the limit when $n\to 1$ and use $\psi(1)=-\gamma$}\}\\
=\lim_{m\to 1/2 }\frac{\partial^{2a}}{\partial m^{2a}} \Gamma(m)\Gamma(1-m)[\psi(1-m) + \gamma]\\
\left\{\text{use  $\Gamma(m)\Gamma(1-m)=\frac{\pi}{\sin(m\pi )}$}\right\}\\
\left\{\text{and write $\frac{\partial}{\partial m}$ as $\frac{d}{dm}$, since we have one variable left}\right\}\\
=\pi\lim_{m\to \frac12 }\frac{d^{2a}}{d m^{2a}} \frac{\psi(1-m) + \gamma}{\sin(m\pi)}.
\end{gather*}

The question here is can we write the limit in terms of finite summation?

REPLY [2 votes]: Thanks to @user178256 for the reference for the Euler number $( E_k)$ which is the key to solve the problem.

In the question body, we showed that
$$\int\limits_0^\infty \frac{\ln^{2a}(x)\ln(1+x)}{\sqrt{x}(1+x)}\mathrm{d}x=-\pi\lim_{m\to \frac12 }\frac{d^{2a}}{d m^{2a}} \frac{\psi(1-m) + \gamma}{\sin(m\pi)}$$
using
$$\frac{d^a}{dm^a}(f*g)=\sum_{k=0}^a \binom{a}{k} \frac{d^k}{dm^k}f*\frac{d^{a-k}}{dm^{a-k}} g$$
we have
$$I=-\pi\sum_{k=0}^{2a} \binom{2a}{k} \lim_{m\to \frac12}\frac{d^{k}}{dm^k} (\psi(1-m)+\gamma)*\lim_{m\to \frac12}\frac{d^{2a-k}}{dm^{2a-k}} \csc(m\pi)$$
since we have
$$\lim_{m\to \frac12}\frac{d^{0}}{dm^0} \csc(m\pi)=1$$
$$\lim_{m\to \frac12}\frac{d^{1}}{dm^1} \csc(m\pi)=0$$
$$\lim_{m\to \frac12}\frac{d^{2}}{dm^2} \csc(m\pi)=\pi^2$$
$$\lim_{m\to \frac12}\frac{d^{3}}{dm^3} \csc(m\pi)=0$$
$$\lim_{m\to \frac12}\frac{d^{4}}{dm^4} \csc(m\pi)=5\pi^4$$
or
$$\lim_{m\to \frac12}\frac{d^{r}}{dm^r} \csc(m\pi)=|E_{r}|\pi^r,$$
we consider only the even-$k$ terms and so
$$I=-\pi\sum_{k=0}^{a} \binom{2a}{2k} \lim_{m\to \frac12}\frac{d^{2k}}{dm^{2k}} (\psi(1-m)+\gamma)*\lim_{m\to \frac12}\frac{d^{2a-2k}}{dm^{2a-2k}} \csc(m\pi)$$
$$=-\pi^{2a+1}\sum_{k=0}^{a} \binom{2a}{2k} \lim_{m\to \frac12}\frac{d^{2k}}{dm^{2k}} (\psi(1-m)+\gamma)*|E_{2a-2k}|\pi^{-2k}$$
separate the first term using $\psi(1/2)+\gamma=-2\ln(2)$, we have
$$I=2\ln(2)|E_{2a}|\pi^{2a+1}-\pi^{2a+1}\sum_{k=1}^{a} \binom{2a}{2k} \lim_{m\to \frac12}\frac{d^{2k}}{dm^{2k}} (\psi(1-m)+\gamma)*|E_{2a-2k}|\pi^{-2k}$$
using
$$\lim_{m\to \frac12}\frac{d^{2k}}{dm^{2k}} (\psi(1-m)+\gamma)=\psi^{(2k)}\left(\frac12\right)=-(2k)!(2^{2k+1}-1)\zeta(2k+1)$$
we finally get
$$I=2\ln(2)|E_{2a}|\pi^{2a+1}+(2a)!{\pi^{2a+1}}\sum_{k=1}^{a} \frac{|E_{2a-2k}|}{(2a-2k)!}{\pi^{-2k}}(2^{2k+1}-1)\zeta(2k+1).$$

Bonus: By using
$$\frac{1}{(2n+1)^{2a+1}}=\frac{1}{(2a)!}\int_0^1 x^{2n}\ln^{2a}(x)\mathrm{d}x,$$
we have
\begin{gather*}
\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{(2n+1)^{2a+1}}=\frac{1}{(2a)!}\int_0^1\ln^{2a}(x)\left(\sum_{n=1}^\infty H_n(-x^2)^n\right)\mathrm{d}x\\
=\frac{1}{(2a)!}\int_0^1\ln^{2a}(x)\left(-\frac{\ln(1+x^2)}{1+x^2}\right)\mathrm{d}x\\
=-\frac{1}{(2a)!}\int_0^1\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x
\end{gather*}
where
\begin{gather*}
\int_0^1\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x
=\left(\int_0^\infty-\int_1^\infty\right)\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x\\
=\int_0^\infty\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x-\underbrace{\int_1^\infty\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x}_{x\to 1/x}\\
=\int_0^\infty\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x-\int_0^1\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x\\
+2\int_0^1\frac{\ln^{2a}(x)\ln(x)}{1+x^2}\mathrm{d}x\\
\left\{\text{add $\int_0^1\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x$ to both sides then divide by 2}\right\}\\
=\frac12\int_0^\infty\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\ln^{2a+1}(x)}{1+x^2}\mathrm{d}x\\
\overset{x^2\to x}{=}4^{-a-1}\int_0^\infty\frac{\ln^{2a}(x)\ln(1+x)}{\sqrt{x}(1+x)}\mathrm{d}x-(2a+1)!\beta(2a+2).
\end{gather*}
Therefore
$$\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{(2n+1)^{2a+1}}=(2a+1)\beta(2a+2)-\frac{\ln(2)|E_{2a}|}{(2a)!}\left(\frac{\pi}{2}\right)^{2a+1}$$
$$-\frac12\left(\frac{\pi}{2}\right)^{2a+1}\sum_{k=1}^{a} \frac{|E_{2a-2k}|}{(2a-2k)!}{\pi^{-2k}}(2^{2k+1}-1)\zeta(2k+1).$$

In case the reader is currious about the Mathematica command of the Euler number $(E_r)$, it is EulerE[r]