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Evaluation of $$\int_{0}^{\sqrt{2}-1}\frac{\ln(1+x^2)}{1+x}dx$$

$\bf{My\; Try:::}$ Let $$I(a) = \int_{0}^{\sqrt{2}-1}\frac{\ln(1+ax^2)}{1+x}dx$$

Now $$I'(a) = \int_{0}^{\sqrt{2}-1}\frac{x^2}{(1+ax^2)(1+x)}dx = \frac{1}{a}\int_{0}^{\sqrt{2}-1}\frac{(1+ax^2)-1}{(1+ax^2)(1+x)}dx$$

So $$I'(a) = \frac{\ln 2}{2a}-\frac{1}{a^2}\int_{0}^{\sqrt{2}-1}\frac{1}{(x^2+\frac{1}{a})(1+x)}dx$$

So $$I'(a) = \frac{\ln 2}{2a}-\frac{1}{a^2(a+1)}\int_{0}^{\sqrt{2}-1}\left[\frac{1-x}{x^2+\frac{1}{a}}-\frac{1}{1+x}\right]dx$$

Now How can i solve after that, Help required, Thanks

juantheron
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  • IMHO the introduction of a variable worses the issues instead of simplifying it. Going back to the initial integral, Mathematica gives a rather complicated expression involving dilogarithms, even after simplification... – Jean Marie Jun 30 '16 at 20:09
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    Are you sure you didn't make a typo? By contrast $\int_{0}^{\sqrt{2}-1}\frac{\ln(1+x)}{1+x}dx$ has a nice form and is very simple. – John Molokach Jun 30 '16 at 20:27

2 Answers2

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Since the number $\sqrt{2}-1$ is invariant under the transformation $x \mapsto \frac{1-x}{1+x}$, it is natural to make that substitution. Doing so, we find: $$\begin{align} I &= \int_0^{\sqrt{2}-1} \frac{\ln(1+x^2)}{1+x} dx \\&= \int_{\sqrt{2}-1}^1 \cfrac{\ln(1+x^2)+\ln\left(\frac2{(1+x)^2}\right)}{1+x}dx \\&=\frac12 I + \frac12 \int_{\sqrt{2}-1}^1 \frac{\ln(1+x^2)}{1+x} dx +\frac12 \int_{\sqrt{2}-1}^1 \frac{\ln2 - 2 \ln(1+x)}{1+x} dx \\&=\frac12 \int_0^1 \frac{\ln(1+x^2)}{1+x} dx -\frac18 \ln^2 2 \\&= \frac{\ln^2 2}{4}-\frac{\pi^2}{96}.\end{align}$$

You can check the evaluation $$\int_0^1 \frac{\ln(1+x^2)}{1+x} dx =\frac34 \ln^2 2- \frac{\pi^2}{48}$$ here, for example.

Community
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Noam Shalev - nospoon
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The solution will actually involve poylogarithms and basic complex things \begin{align*} \int_{0}^{\sqrt{2}-1} \frac{\ln \left ( 1+x^2 \right )}{1+x} \, {\rm d}x&= \int_{0}^{\sqrt{2}-1} \frac{\ln \left ( 1-i^2 x^2 \right )}{1+x} \, {\rm d}x\\ &= \int_{0}^{\sqrt{2}-1} \frac{\ln (1-ix) + \ln (1+ix)}{1+x} \, {\rm d}x\\ &= \int_{0}^{\sqrt{2}-1} \frac{\ln (1-ix)}{1+x} \, {\rm d}x + \int_{0}^{\sqrt{2}-1} \frac{\ln (1+ix)}{1+x} \, {\rm d}x\\ &=-{\rm Li}_2 \left ( \frac{1-i}{2} \right ) +{\rm Li}_2 \left ( 1 - \frac{1+i}{\sqrt{2}} \right ) + \frac{i}{4} \pi \ln \left [ \left ( 1+i \right ) - i \sqrt{2} \right ] \\ & \quad \; -{\rm Li}_2 \left ( \frac{1+i}{2} \right ) + {\rm Li}_2 \left ( 1 - \frac{1-i}{\sqrt{2}} \right ) - \frac{i}{4} \pi \ln \left [ (1-i) + i \sqrt{2} \right ] \\ &= - \left [ -i \mathcal{G} + \frac{\pi^2}{48} - \frac{1}{2}\left ( -\frac{\ln 2}{2} + \frac{i \pi}{4} \right )^2 \right ] + {\rm Li}_2 \left ( 1 - \frac{1+i}{\sqrt{2}} \right ) \\ &\quad \; -\left [ i \mathcal{G} + \frac{\pi^2}{48} - \frac{1}{2} \left ( -\frac{\ln 2}{2} - \frac{i \pi}{4} \right )^2 \right ] +{\rm Li}_2 \left ( 1 - \frac{1-i}{\sqrt{2}} \right ) + \frac{\pi^2}{16} \\ &= \frac{\ln^2 2}{4} - \frac{5 \pi^2}{48} + \frac{\pi^2}{16} + {\rm Li}_2 \left ( 1 - \frac{1-i}{\sqrt{2}} \right ) +{\rm Li}_2 \left ( 1 - \frac{1+i}{\sqrt{2}} \right ) \\ &= \frac{\ln^2 2}{4} - \frac{\pi^2}{24} + {\rm Li}_2 \left ( 1 - \frac{1-i}{\sqrt{2}} \right ) +{\rm Li}_2 \left ( 1 - \frac{1+i}{\sqrt{2}} \right ) \approx 0.0173049 \end{align*}

Side note: One interesting identity regarding the dilog function is the following: $$\Re{ \left[{\rm Li}_{2}\left(\frac{1}{2}+iq\right) \right]}=\frac{{\pi}^{2}}{12}-\frac{1}{8}{\ln^2{\left(\frac{1+4q^2}{4}\right)}}-\frac{{\arctan^2{(2q)}}}{2}$$ where q∈Q and there is a similar one for the imaginary part which I will post later.

Addendum: In general it holds that:

\begin{align*} \int_{0}^{1} \frac{\log (1+ax)}{1+x} \, {\rm d}x &\overset{u=1+x}{=\! =\! =\!} \int_{1}^{2} \frac{\log \left ( 1-a+au \right )}{u} \, {\rm d}u \\ &=\int_{1}^{2} \frac{\log \left [ \left ( 1-a \right ) \left ( 1+ \frac{a}{1+a} u \right ) \right ]}{u} \, {\rm d}u \\ &= \log \left ( 1-a \right ) \int_{1}^{2} \frac{{\rm d}u}{u} +\int_{1}^{2} \frac{\log \left ( 1+\frac{a}{1-a} u \right )}{u} \, {\rm d}u \\ &= \log 2 \log (1-a) - \sum_{n=1}^{\infty} \frac{1}{n} \left ( -\frac{a}{1-a} \right )^n\int_{1}^{2} u^{n-1} \, {\rm d}u\\ &= \log 2 \log (1-a) - \sum_{n=1}^{\infty} \frac{1}{n^2} \left ( -\frac{a}{1-a} \right )^n \left ( 2^n-1 \right ) \\ &= \log 2 \log (1-a) -{\rm Li}_2 \left ( -\frac{2a}{1-a} \right ) + {\rm Li}_2 \left ( -\frac{a}{1-a} \right ) \end{align*}

How we conclude that $$\bbox[blue, 2pt]{{\color{white}{\int_{0}^{1}\frac{\log (1+ax)}{1+x} \, {\rm d}x = \log 2 \log (1-a) -{\rm Li}_2 \left ( -\frac{2a}{1-a} \right ) + {\rm Li}_2 \left ( - \frac{a}{1-a} \right )}}}$$

Martin.s
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