Integral $\int_0^1 \log x \frac{(1+x^2)x^{c-2}}{1-x^{2c}}dx=-\left(\frac{\pi}{2c}\right)^2\sec ^2 \frac{\pi}{2c}$ Hi I am trying to prove this result $$
I:=\int_0^1 \log x \frac{(1+x^2)x^{c-2}}{1-x^{2c}}dx=-\left(\frac{\pi}{2c}\right)^2\sec ^2 \frac{\pi}{2c},\quad c>1.
$$
Thanks.  Since $x\in[0,1]
$ we can write$$
I=\sum_{n=0}^\infty \int_0^1 \log x (1+x^2) x^{c-2+2cn}dx.
$$
since $\sum_{n=0}^\infty x^n= (1-x)^{-1} , |x| < 1.$  Simplifying
$$
I=\sum_n \int_0^1 \log x\, x^{c-2+2n} dx+\sum_n \int_0^1 \log x \, x^{c+2cn}\, dx.
$$
Now we can write
$$
I=-\frac{1}{4}\psi_1 \left(\frac{c+1}{2}\right)-\frac{1}{4} \psi_1\left( \frac{3+c}{2}\right)
$$
where I summed the results of the integrals using general result of $\int_0^1 \log x \, x^n \, dx=-\frac{1}{(n+1)^2}$.  Howeever this is not the result $\sec^2...$.  Thanks
The function $\psi$ is the polygamma function which is defined in general by
$$
\psi_m(z)=\frac{d^{m+1}}{dz^{m+1}} \log \Gamma(z)
$$
and $\Gamma(z)=(z-1)!$, for this case $m=1$.
 A: Using the identity
\begin{align}
\psi(1-x) - \psi(x) = \pi \cot(\pi x)
\end{align}
then the derivative with respect to $x$ yields
\begin{align}
\psi_{1}(1-x) + \psi_{1}(x) = \pi^{2} \csc^{2}(\pi x).
\end{align}
Now the integral
\begin{align}
I &= \int_{0}^{1} \frac{ x^{c-2} (1+x^{2}) \ln(x) }{ 1 - x^{2c} } \, dx
\end{align}
is evaluate as follows.
\begin{align}
I &= \sum_{n=0}^{\infty} \, \int_{0}^{1} (1+x^{2}) \, x^{2cn+c-2} \ln(x) \, dx \\
&= - \sum_{n=0}^{\infty} \left[ \frac{1}{(2cn+c-1)^{2}} + \frac{1}{(2cn+c+1)^{2}} \right] \\
&= - \frac{1}{4 c^{2}} \left[ \psi_{1}\left(\frac{c-1}{2c}\right) + \psi_{1}\left( \frac{c+1}{2c}\right) \right] \\
&= - \left(\frac{\pi}{2c}\right)^{2} \csc^{2}\left(\frac{\pi}{2} + \frac{\pi}{2c}\right) \\
&= - \left(\frac{\pi}{2c}\right)^{2} \sec^{2}\left(\frac{\pi}{2c}\right).  
\end{align}
It can now be stated that
\begin{align}
\int_{0}^{1} \frac{ x^{c-2} (1+x^{2}) \ln(x) }{ 1 - x^{2c} } \, dx = - \left(\frac{\pi}{2c}\right)^{2} \sec^{2}\left(\frac{\pi}{2c}\right).
\end{align} 
A: Another derivation that uses only very elementary complex analysis and no special functions:
(all integrals are to be regarded as Cauchy PV integrals if necessary)
By substituting $x\mapsto 1/x$, we find that the original integral equals
$$\int_1^{\infty} (-\log x)  \frac{(1+1/x^2)x^{2-c}}{1-x^{-2c}}\frac{dx}{x^2} = \int_1^{\infty} \log x \frac{(1+x^2)x^{c-2}}{1-x^{2c}} \,dx$$
Adding both copies of the integral, the problem is reduced to showing that
$$\int_0^{\infty} \log x \frac{(1+x^2)x^{c-2}}{1-x^{2c}} \,dx = -\frac{\pi^2}{2c^2}\sec^2\left(\frac{\pi}{2c}\right)$$
We will do this by the method that is by now well-known: let $I(\mu) = \int_0^{\infty} \frac{(1+x^2)x^{c-2+\mu}}{1-x^{2c}} \,dx$, then we are looking for $I'(0)$.
We have (remember we are taking PV everywhere)$$\int_0^{\infty} \frac{x^a}{1-x^b} \,dx = \frac{\pi}{b}\cot\left(\pi\frac{a+1}{b}\right)$$
which can be easily derived using residues. (Integrate around a pizza slice contour.)
This immediately gives $$I(\mu) = \frac{\pi}{2c} \left[\cot\left(\pi\frac{c-1+\mu}{2c}\right) + \cot\left(\pi\frac{c+1+\mu}{2c}\right)  \right]$$
and hence 
$$ I'(\mu) = -\frac{\pi^2}{4c^2}\left[\csc^2\left(\frac{\pi}{2c} (c-1+\mu)\right) +\csc^2\left(\frac{\pi}{2c}(c+1+\mu)\right)\right]$$
$$ \implies I'(0) = -\frac{\pi^2}{2c^2}\sec^2\left(\frac{\pi}{2c}\right)$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{1}\ln\pars{x}\,{\pars{1 + x^{2}}x^{c - 2} \over 1 - x^{2c}}\,\dd x
     =-\pars{\pi \over 2c}^{2}\sec^{2}\pars{\pi \over 2c}:\ {\large ?}}$

With $\ds{x \equiv t^{1/\pars{2c}}\quad\imp\quad t = x^{2c}}$:
  \begin{align}
&\color{#c00000}{%
\int_{0}^{1}\ln\pars{x}\,{\pars{1 + x^{2}}x^{c - 2} \over 1 - x^{2c}}\,\dd x}
=\int_{0}^{1}
\ln\pars{t^{1/\bracks{2c}}}\,{\pars{1 + t^{1/c}}t^{1/2 - 1/c} \over 1 - t}
\,{1 \over 2c}\,t^{1/\pars{2c} - 1}\,\dd t
\\[3mm]&={1 \over 4c^{2}}\bracks{%
\int_{0}^{1}{\ln\pars{t}t^{-\pars{1 + 1/c}/2} \over 1 - t}\,\dd t
+\int_{0}^{1}{\ln\pars{t}t^{-\pars{1 - 1/c}/2} \over 1 - t}\,\dd t}
\\[3mm]&=-\,{1 \over 4c^{2}}
\lim_{\mu \to -\pars{1 + 1/c}/2}
\partiald{}{\mu}\int_{0}^{1}{1 - t^{\mu} \over 1 - t}\,\dd t + \pars{~c \to -c~}
\\[3mm]&=\color{#c00000}{-\,{1 \over 4c^{2}}
\lim_{\mu \to -\pars{1 + 1/c}/2}
\partiald{\bracks{\Psi\pars{\mu + 1} + \gamma}}{\mu} + \pars{~c \to -c~}}
\end{align}
  where we used the A&S table identity ${\bf\mbox{6.3.22}}$. $\ds{\Psi\pars{z}}$ is the Digamma Function and $\ds{\gamma}$ is the
  Euler-Mascheroni Constant.

\begin{align}
&\color{#c00000}{%
\int_{0}^{1}\ln\pars{x}\,{\pars{1 + x^{2}}x^{c - 2} \over 1 - x^{2c}}\,\dd x}
=
-\,{1 \over 4c^{2}}\bracks{\Psi'\pars{\half + {1 \over 2c}} + \Psi'\pars{\half - {1 \over 2c}}}
\end{align}
With Euler Reflection Formula
${\bf\mbox{6.4.7}}$:
\begin{align}
&\color{#c00000}{%
\int_{0}^{1}\ln\pars{x}\,{\pars{1 + x^{2}}x^{c - 2} \over 1 - x^{2c}}\,\dd x}
=
-\,{1 \over 4c^{2}}\bracks{-\pi\,\totald{\cot\pars{\pi z}}{z}}
_{z\ =\ 1/2\ -\ 1/\pars{2c}}
\\[3mm]&=-\,{\pi^{2} \over 4c^{2}}\,\csc^{2}\pars{\pi\bracks{\half - {1 \over 2c}}}
=-\,{\pi^{2} \over 4c^{2}}\,\sec^{2}\pars{\pi \over 2c}
\end{align}

$$
\color{#00f}{\large%
\int_{0}^{1}\ln\pars{x}\,{\pars{1 + x^{2}}x^{c - 2} \over 1 - x^{2c}}\,\dd x}
=\color{#00f}{\large-\,\pars{\pi \over 2c}^{2}\sec^{2}\pars{\pi \over 2c}}\,,\qquad
\verts{c} > 1
$$

A: Here is an approach. Follow the steps
1) Make the change of variables $\ln(x)=-t$
2) Follow it with another change of variables $y=2ct$
3) Use the Hurwitz Zeta function. See this problem.  
