TITLE: Integral $\int_0^1 \log x \frac{(1+x^2)x^{c-2}}{1-x^{2c}}dx=-\left(\frac{\pi}{2c}\right)^2\sec ^2 \frac{\pi}{2c}$
QUESTION [6 upvotes]: 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 $m=1$.

REPLY [3 votes]: $\newcommand{\+}{^{\dagger}}
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$\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}

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