How to show that
$$\lim_{m\to 0}\frac{\partial^{2a-1}}{\partial m^{2a-1}}\pi \csc(m\pi) \left[\operatorname{\psi}^{(0)} (1-m) + \gamma\right] =2(2a-1)!\sum_{k=0}^{a-1}\frac{\eta(2k)\psi^{(2a-2k)}(1)}{(2a-2k)!}\tag{1}$$
where $\psi^{(a)}(x)$ is the polygamma function and $\eta(x)$ is the dirichlet eta function.
I obtained this formula by observing the pattern (using Mathematica):
$$\lim_{m\to 0}\frac{\partial}{\partial m}\pi \csc(m\pi) \left[\operatorname{\psi}^{(0)} (1-m) + \gamma\right] =\frac12\psi^{(2)}(1)$$
$$\lim_{m\to 0}\frac{\partial^3}{\partial m^3}\pi \csc(m\pi) \left[\operatorname{\psi}^{(0)} (1-m) + \gamma\right] =\frac14\left[2\pi^2\psi^{(2)}(1)+\psi^{(4)}(1)\right]$$
$$\lim_{m\to 0}\frac{\partial^5}{\partial m^5}\pi \csc(m\pi) \left[\operatorname{\psi}^{(0)} (1-m) + \gamma\right] =\frac16\left[7\pi^4\psi^{(2)}(1)+5\pi^2\psi^{(4)}(1)+\psi^{(6)}(1)\right]$$
$$\lim_{m\to 0}\frac{\partial^7}{\partial m^7}\pi \csc(m\pi) \left[\operatorname{\psi}^{(0)} (1-m) + \gamma\right] =\frac1{24}\left[124\pi^6\psi^{(2)}(1)+98\pi^4\psi^{(4)}(1)+28\pi^2\psi^{(6)}(1)+3\psi^{(8)}(1)\right]$$
But I could not prove it. Here is what I tried: Using
$$\frac{d^a}{dm^a}(f*g)=\sum_{k=0}^a \binom{a}{k} \frac{d^k}{dm^k}f*\frac{d^{a-k}}{dm^{a-k}} g$$
it follows that
$$\lim_{m\to 0}\frac{\partial^{2a-1}}{\partial m^{2a-1}}\pi \csc(m\pi) \left[\operatorname{\psi}^{(0)} (1-m) + \gamma\right]$$
$$=\pi\sum_{k=0}^{2a-1}\binom{2a-1}{k}\lim_{m\to 0}\frac{\partial^{k}}{\partial m^{k}}\csc(m\pi)\lim_{m\to 0}\frac{\partial^{2a-k-1}}{\partial m^{2a-k-1}}\left[\operatorname{\psi}^{(0)} (1-m) + \gamma\right]$$
We have $$\lim_{m\to 0}\frac{\partial^{2a-k-1}}{\partial m^{2a-k-1}}\left[\operatorname{\psi}^{(0)} (1-m) + \gamma\right]=-\lim_{m\to 0}\psi^{(2a-k-1)}(1-m)=-\psi^{(2a-k-1)}(1)$$
But $\displaystyle\lim_{m\to 0}\frac{\partial^{k}}{\partial m^{k}}\csc(m\pi)$ is either indeterminate for even $k$ or $-\infty$ for odd $k$. So this approach does not work.
The integral representation of the limit in $(1)$ is $\displaystyle-\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx$ as shown here.
Note: I can prove the formula in $(1)$ if I am allowed to use the well known generalization:
$$\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{n^{2a}}=-\frac{2a+1}{2}\eta(2a+1)+\sum_{j=0}^{a-1}\eta(2j)\zeta(2a+1-2j)$$
so I need a proof without using this generalization.
Best Answer
We have $$ \psi (1 - z) + \gamma = \sum\limits_{k = 1}^\infty {\frac{1}{{k!}}\left[ {\frac{{d^k \psi (1 - z)}}{{dz^k }}} \right]_{z = 0} z^k } = \sum\limits_{k = 1}^\infty {( - 1)^k \frac{{\psi ^{(k)} (1)}}{{k!}}z^k } $$ for $|z|<1$. Thus, $$ \frac{{\psi (1 - z) + \gamma }}{z} = \sum\limits_{k = 0}^\infty {( - 1)^{k + 1} \frac{{\psi ^{(k + 1)} (1)}}{{(k + 1)!}}z^k } $$ for $|z|<1$ (the left-hand side is defined as a limit when $z=0$). We also have $$ \pi z\csc (\pi z) = \sum\limits_{k = 0}^\infty {\frac{{( - 1)^{k - 1} 2\pi ^{2k} (2^{2k - 1} - 1)B_{2k} }}{{(2k)!}}z^{2k} } = 2\sum\limits_{k = 0}^\infty {\eta (2k)z^{2k} } $$ for $|z|<1$ (the left-hand side is defined as a limit when $z=0$). Thus, by Taylor's formula \begin{align*} & \mathop {\lim }\limits_{z \to 0} \frac{{d^{2n - 1} }}{{dz^{2n - 1} }}\pi \csc (\pi z)\left[ {\psi (1 - z) + \gamma } \right] \\ & = 2(2n - 1)![z^{2n - 1} ]\left( {\left( {\sum\limits_{k = 0}^\infty {\eta (2k)z^{2k} } } \right)\left( {\sum\limits_{k = 0}^\infty {( - 1)^{k + 1} \frac{{\psi ^{(k + 1)} (1)}}{{(k + 1)!}}z^k } } \right)} \right) \end{align*} where $[z^n]$ is the coefficient extraction operator. Perform the Cauchy product of the series and extract the coefficient of $z^{2n-1}$. The result will be the formula in your question. Note that the values of $\psi^{k}(1)$ can be expressed in terms of the Riemann zeta function.