Need help with $\int_0^1 \frac{\ln(1+x^2)}{1+x} dx$ [duplicate]

Solution 1:

I don't want a complete solution, just the parameter would do.

Try \begin{equation} I(a)=\int_0^1\frac{\ln\left(1+a^2x^2\right)}{1+x}\, dx \end{equation}

No upvote yet? Okay...

Differentiating w.r.t. $a$, we have \begin{align} I'(a)&=\int_0^1\left[\frac{2ax^2}{(1+x)\left(1+a^2x^2\right)}\right]\, dx\\ &=\frac{2a}{1+a^2}\int_0^1\frac{1}{1+x}\,dx-\frac{2a}{1+a^2}\int_0^1\frac{1-x}{1+a^2x^2}\,dx\\ &=\frac{2a\ln2}{1+a^2}-\frac{2\arctan a}{1+a^2}+\frac{\ln\left(1+a^2\right)}{a\left(1+a^2\right)}\\ \end{align} then integrating back \begin{align} I(a)&=\ln\left(1+a^2\right)\ln2-\arctan^2(a)+\int\left[\frac{\ln\left(1+a^2\right)}{a}-\underbrace{\frac{a\ln\left(1+a^2\right)}{1+a^2}}_{({\Large\color{red}{\star}})}\right]\,da\\ &=\ln\left(1+a^2\right)\ln2-\arctan^2(a)-\frac{1}{2}{\rm{Li}}\left(-a^2\right)-\frac{1}{4}\ln^2\left(1+a^2\right)+C\\ \end{align} For $({\Large\color{red}{\star}})$ we use integration by parts once by taking $u=\ln\left(1+a^2\right)$.

Now, I'm sure you can you take it from here.

Solution 2:

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} &\color{#66f}{\large\int_{0}^{1}{\ln\pars{1 + x^{2}} \over 1+x}\,\dd x} =\ln^{2}\pars{2} - \int_{0}^{1}\ln\pars{1 + x}\,{2x \over 1 + x^{2}}\,\dd x \\[5mm]&=\ln^{2}\pars{2} -2\,\color{#c00000}{\Im\int_{0}^{1}{\ln\pars{1 + x} \over 1 - \ic x}\,\dd x} =\color{#66f}{\large-\,{\pi^{2} \over 48} + {3 \over 4}\,\ln^{2}\pars{2}} \approx {\tt 0.1547} \end{align}

With $\ds{\mu \equiv {\ic \over 1 + \ic}=\half\,\pars{1 + \ic}}$:

\begin{align} &\color{#c00000}{\Im\int_{0}^{1}{\ln\pars{1 + x} \over 1 - \ic x}\,\dd x} =\Im\int_{1}^{2}{\ln\pars{x} \over 1 + \ic - \ic x}\,\dd x =\Im\bracks{-\ic\int_{1}^{2}{\ln\pars{x} \over 1 - \ic x/\pars{1 + \ic}} \,{\ic\,\dd x \over 1 + \ic}} \\[5mm]&=\Im\bracks{-\ic\int_{1}^{2}{\ln\pars{x} \over 1 - \mu x}\,\mu\,\dd x} =\Im\bracks{-\ic\int_{\mu}^{2\mu}{\ln\pars{x/\mu} \over 1 - x}\,\dd x} \\[5mm]&=\Im\bracks{\left.\ic\ln\pars{1 - x}\ln\pars{x \over \mu} \right\vert_{x\ =\ \mu}^{x\ =\ 2\mu} -\ic\int_{\mu}^{2\mu}\ln\pars{1 - x}\,{1/\mu \over x/\mu}\,\dd x} \\[5mm]&=\Im\braces{\ic\ln\pars{1 - 2\mu}\ln\pars{2} + \ic\int_{\mu}^{2\mu}\ \underbrace{\bracks{-\,{\ln\pars{1 - x} \over x}}} _{\ds{=\ \color{#c00000}{{\rm Li}_{2}'\pars{x}}}}\ \,\dd x} \\[5mm]&=\Re\,{\rm Li}_{2}\pars{1 + \ic} - \Re\,{\rm Li}_{2}\pars{1 + \ic \over 2} \end{align}

Solution 3:

(Edit: Just noticed that the OP specifically requested a hint instead of a complete solution. My apologies.)

Start by integrating by parts:

$$\begin{align} \mathcal{I} &:=\int_{0}^{1}\frac{\ln{\left(1+x^2\right)}}{1+x}\,\mathrm{d}x\\ &=\left[\ln{(1+x)}\ln{(1+x^2)}\right]_{0}^{1}-\int_{0}^{1}\frac{2x\ln{(1+x)}}{1+x^2}\,\mathrm{d}x\\ &=\ln^2{(2)}-2\int_{0}^{1}\frac{x\ln{(1+x)}}{1+x^2}\,\mathrm{d}x\\ &=\ln^2{(2)}-2\mathcal{J}{(1)},\\ \end{align}$$

where we've introduced the parameter $a$ by defining the function $\mathcal{J}{(a)}:=\int_{0}^{1}\frac{x\ln{(1+ax)}}{1+x^2}\,\mathrm{d}x$.