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How to prove that

$$\int^1_0 \frac{\log(1+x)}{1+a^2x^2}\, dx= \frac{1}{a} \Im \left( \chi_2(ia) - \operatorname{Li}_2\left( \frac{1+ai}{2}\right)\right)\,\,\, a> 0$$

where $\Im $ is the imaginary part , $\chi$ is the Legendre chi function , $\operatorname{Li}_2$ is the dilogarithm .

For the special case $a=1$ we have

$$\int^1_0 \frac{\log(1+x)}{1+x^2}dx = \frac{\pi}{8}\log(2)$$

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  • $\begingroup$ Your problem is a special case of this one. $\endgroup$ – Mhenni Benghorbal Aug 21 '13 at 18:21
  • $\begingroup$ When $a = 2$, $\mbox{LHS} \approx 0.1617$, $\mbox{RHS} \approx 0.2220$. However, they agree when $a = 1$. $\endgroup$ – Felix Marin Aug 13 '14 at 18:09
  • $\begingroup$ @MhenniBenghorbal Please, check my above comment. I guess the proposed OP answer is wrong. $\endgroup$ – Felix Marin Aug 13 '14 at 18:22
  • $\begingroup$ @FelixMarin: Let me check this later. $\endgroup$ – Mhenni Benghorbal Aug 13 '14 at 19:05
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{1}{\ln\pars{1 + x} \over 1 + a^{2}x^{2}}\,\dd x:\ {\large ?}.\qquad a \in {\mathbb R}}$.

I believe the proposed OP solution $\ds{\pars{~{1 \over a}\,\Im\pars{\chi_2\pars{\ic a} -{\rm Li}_{2}\pars{1\ +\ a\ic \over 2}}~}}$ is not correct. Albeit it agrees at $\ds{a = 1}$, we can evaluate it for several values and they yield different values between LHS and RHS.

\begin{align}&\color{#c00000}{\int_{0}^{1}% {\ln\pars{1 + x} \over 1 + a^{2}x^{2}}\,\dd x} =\Im\int_{0}^{1}{\ln\pars{1 + x} \over -\ic + \verts{a}x}\,\dd x =\Im\int_{1}^{2}{\ln\pars{x} \over -\ic - \verts{a} + \verts{a}x}\,\dd x \\[3mm]&=\Im\bracks{-\,{1 \over 1 + \ic\verts{a}} \int_{1}^{2}{\ln\pars{x} \over 1 - \verts{a}x/\pars{\ic + \verts{a}}}\,\dd x} \\[3mm]&=\Im\bracks{-\,{1 \over 1 + \ic\verts{a}} \int_{1}^{2}{\ln\pars{x} \over 1 - \mu x}\,\dd x}\,, \qquad\qquad \mu \equiv {\verts{a} \over \ic + \verts{a}} \end{align}

With $\ds{\mu x\equiv t\ \imp\ x = {t \over \mu}\quad\mbox{and}\quad \dd x={\ic + \verts{a} \over \verts{a}}\,\dd t}$ \begin{align}&\color{#c00000}{\int_{0}^{1}% {\ln\pars{1 + x} \over 1 + a^{2}x^{2}}\,\dd x} ={1 \over \verts{a}}\, \Im\bracks{-\int_{\mu}^{2\mu}% {\ln\pars{t/\mu} \over 1 - t}\,\dd t} \\[3mm]&={1 \over \verts{a}}\,\Im\bracks{% \ln\pars{1 - 2\mu}\ln\pars{2} - \int_{\mu}^{2\mu}{\ln\pars{1 - t} \over t}\,\dd t} \\[3mm]&={1 \over \verts{a}}\,\Im\bracks{% \ln\pars{1 - 2\mu}\ln\pars{2} + {\rm Li}_{2}\pars{2\mu} - {\rm Li}_{2}\pars{\mu}} \end{align}

$$\color{#c00000}{\int_{0}^{1}% {\ln\pars{1 + x} \over 1 + a^{2}x^{2}}\,\dd x} ={1 \over \verts{a}}\,\Im\bracks{% \ln\pars{\ic - \verts{a} \over \ic + \verts{a}}\ln\pars{2} +{\rm Li}_{2}\pars{2\verts{a} \over \ic + \verts{a}} -{\rm Li}_{2}\pars{\verts{a} \over \ic + \verts{a}}} $$

\begin{align}&\color{#66f}{\large\int_{0}^{1}% {\ln\pars{1 + x} \over 1 + a^{2}x^{2}}\,\dd x} \\[3mm]&=\color{#66f}{\large2\ln\pars{2}\,{\arctan\pars{a} \over a} +{1 \over \verts{a}}\,\Im\bracks{% {\rm Li}_{2}\pars{2\verts{a} \over \ic + \verts{a}} -{\rm Li}_{2}\pars{\verts{a} \over \ic + \verts{a}}}} \end{align}

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A related problem. Note that, the Legendre chi function is related to the polylogarithm by the identity

$$ \chi_\nu(z) = \frac{1}{2}\left[\operatorname{Li}_\nu(z) - \operatorname{Li}_\nu(-z)\right]. $$

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