How to evaluate this integral $$\mathcal{I}=\int_{0}^{\infty }\frac{\ln\left ( 1+x^{3} \right )}{1+x^{2}}\mathrm{d}x$$ Mathematica gave me the answer below $$\mathcal{I}=\frac{\pi }{4}\ln 2+\frac{2}{3}\pi \ln\left ( 2+\sqrt{3} \right )-\frac{\mathbf{G}}{3}$$ where $\mathbf{G}$ is Catalan's constant.

  • $\begingroup$ What is $\;G\;$ ? $\endgroup$ – DonAntonio Feb 24 '16 at 10:23
  • $\begingroup$ @Joanpemo catalan's constant $\endgroup$ – Renascence_5. Feb 24 '16 at 10:25
  • $\begingroup$ Do you know the residue method for integration ? $\endgroup$ – Jean Marie Feb 24 '16 at 10:26
  • $\begingroup$ @EvilNebula Thank you. $\endgroup$ – DonAntonio Feb 24 '16 at 10:27
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    $\begingroup$ @EvilNebula: Just a remark on the wording: To find the value of an integral is not 'to prove' it. One (dis)proves a statement showing why it is (not) true, so if you expect $\int_0^{\infty}f(x)$ to equal $a$ you compute $\int_0^{\infty}f(x)$ to prove the statement "$\int_0^{\infty}f(x)$=a" $\endgroup$ – Octania Feb 24 '16 at 11:42

Lemma 1::$$\int_{0}^{\infty}\dfrac{\ln{(x^2-x+1)}}{x^2+1}=\dfrac{2}{3}\ln{(2+\sqrt{3})}-\dfrac{4}{3}G$$ Use this well known $$\int_{0}^{+\infty}\dfrac{\ln{(x^2+2\sin{a}\cdot x+1)}}{1+x^2}dx=\pi\ln{2\cos{\dfrac{a}{2}}}+a\ln{|\tan{\dfrac{a}{2}}|}+2\sum_{k=0}^{+\infty}\dfrac{\sin{(2k+1)a}}{(2k+1)^2}$$ this indentity proof is very easy consider $\ln{(x^2+2\sin{a}\cdot x+1)}$ Fourier expansions(possion fourier).

then you can take $a=-\dfrac{\pi}{6}$ then we have $$\pi\ln{2\cos{\dfrac{\pi}{12}}}=\dfrac{\pi}{2}\ln{(2+\sqrt{3})}$$ $$-\dfrac{\pi}{6}\ln{\tan{\dfrac{\pi}{12}}}=\dfrac{\pi}{6}\ln{(2+\sqrt{3})}$$ and $$2\sum_{k=0}^{3N}\dfrac{\sin{(2k+1)-\pi/6}}{(2k+1)^2}=-\sum_{k=0}^{3N}\dfrac{(-1)^k}{(2k+1)^2}-3\sum_{k=0}^{N-1}\dfrac{(-1)^k}{(6k+3)^2}\to -G-\dfrac{G}{3}=-\dfrac{4}{3}G$$ so $$\int_{0}^{\infty}\dfrac{\ln{(x^2-x+1)}}{x^2+1}=\dfrac{2}{3}\ln{(2+\sqrt{3})}-\dfrac{4}{3}G$$ By done!

Lemma 2:$$\int_{0}^{+\infty}\dfrac{\ln{(1+x)}}{1+x^2}dx=\dfrac{\pi}{4}\ln{2}+G$$ \begin{align*} \int_{0}^{\infty} \frac{\log (x + 1)}{x^2 + 1} \, dx &= \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx + \int_{1}^{\infty} \frac{\log (x + 1)}{x^2 + 1} \, dx \\ &= \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx + \int_{0}^{1} \frac{\log (x^{-1} + 1)}{x^2 + 1} \, dx \quad (x \mapsto x^{-1}) \\ &= 2 \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx - \int_{0}^{1} \frac{\log x}{x^2 + 1} \, dx\\ &=\dfrac{\pi}{4}\ln{2}+G \end{align*}


$$\int_{0}^{+\infty}\dfrac{\ln{(1+x^3)}}{1+x^2}dx=\int_{0}^{+\infty}\dfrac{\ln{(1+x)}}{1+x^2}dx+\int_{0}^{+\infty}\dfrac{\ln{(x^2-x+1)}}{1+x^2}dx=\frac{\pi }{4}\ln 2+\frac{2}{3}\pi \ln\left ( 2+\sqrt{3} \right )-\frac{\mathbf{G}}{3}$$


We can attack this integral

$$I = \int_0^{\infty} dx \frac{\log{(1+x^3)}}{1+x^2}$$

by considering the complex contour integral

$$\oint_C dz \frac{\log{(1+z^3)} \log{z}}{1+z^2}$$

where $C$ is the following contour


This is a keyhole contour about the positive real axis, but with additional keyholes about the branch points at $z=e^{i \pi/3}$, $z=-1$, and $z=e^{i 5 \pi/3}$. There are simple poles at $z=\pm i$.

I will outline the procedure for evaluation. The integral about the circular arcs, large and small, go to zero as the radii go to $\infty$ and $0$, respectively. Each of the branch points introduces a jump of $i 2 \pi$ due to the logarithm in the integrand. By the residue theorem, we have

$$-i 2 \pi \int_0^{\infty} dx \frac{\log{(1+x^3)}}{1+x^2} - i 2 \pi \int_{e^{i \pi/3}}^{\infty e^{i \pi/3}} dt \frac{\log{t}}{1+t^2} \\ - i 2 \pi \int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{1+t^2} - i 2 \pi \int_{e^{i 5 \pi/3}}^{\infty e^{i 5 \pi/3}} dt \frac{\log{t}}{1+t^2} = \\ i 2 \pi \sum_{\pm} \left[\frac{\log{(1+z^3)} \log{z}}{2 z} \right]_{z=\pm i} $$

Without going into too much detail, I will illustrate how the integrals are done by evaluating one of them. Consider

$$\int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{1+t^2} = -\int_1^{\infty} dy \frac{\log{y}+i \pi}{1+y^2}$$


$$\int_1^{\infty} \frac{dy}{1+y^2} = \int_{\pi/4}^{\pi/2} d\theta = \frac{\pi}{4}$$

$$\begin{align}\int_1^{\infty} dy\frac{\log{y}}{1+y^2} &= G\end{align}$$

so that

$$\int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{1+t^2} = -G - i \frac{\pi^2}{4} $$

Along similar lines,

$$\int_{e^{i \pi/3}}^{\infty e^{i \pi/3}} dt \frac{\log{t}}{1+t^2} = \frac{2}{3} G + \frac{\pi}{6} \log{(2+\sqrt{3})}$$

$$\int_{e^{i 5 \pi/3}}^{\infty e^{i 5 \pi/3}} dt \frac{\log{t}}{1+t^2} = \frac{2}{3} G - \frac{5 \pi}{6} \log{(2+\sqrt{3})} + i \frac{\pi^2}{2}$$

Combining the integrals, I get

$$\frac{G}{3} - \frac{2 \pi}{3} \log{(2+\sqrt{3})} + i \frac{\pi^2}{4}$$

The sum of the residues on the RHS is relatively simple to evaluate; I get

$$\sum_{\pm} \left[\frac{\log{(1+z^3)} \log{z}}{2 z} \right]_{z=\pm i} = \frac{(1/2 \log{2} -i \pi/4)(i \pi/2)}{2 i} + \frac{(1/2 \log{2} + i \pi/4)(i 3 \pi/2)}{-2 i}\\ = -\frac{\pi}{4} \log{2}-i \frac{\pi^2}{4}$$

The integral we seek is then the negative of the sum of the combined integrals and the sum of the residues, which gives us

$$\int_0^{\infty} dx \frac{\log{(1+x^3)}}{1+x^2} = -\frac{G}{3} + \frac{\pi}{4} \log{2} +\frac{2 \pi}{3} \log{(2+\sqrt{3})} $$

which agrees with Mathematica.

  • $\begingroup$ (+1)Nice solution RG!Thank you so much! But I'd like to know can we solve it without using contour integral(^ω^) $\endgroup$ – Renascence_5. Feb 24 '16 at 11:07
  • $\begingroup$ @EvilNebula.: well, perhaps there is a way to do it. However, I do not know of away off the top of my head. For these integrals, contour integration is by far the best way to go. $\endgroup$ – Ron Gordon Feb 24 '16 at 11:09
  • $\begingroup$ alright.thank you again!(◕ω◕) $\endgroup$ – Renascence_5. Feb 24 '16 at 11:11
  • $\begingroup$ Stupid question, where do the minus signs in front of the branch cut integrals come from? $\endgroup$ – tired Feb 24 '16 at 12:52
  • $\begingroup$ @tired: They result from a traversal about the small circles in a clockwise, rather than a counterclockwise sense. $\endgroup$ – Ron Gordon Feb 24 '16 at 13:12

It is possible to calculate the integral via the usual tool of differentiating with respect to a parameter. I don't claim the calculations to be especially nice, but it is nice to have as a comparison to the residue approach which is shorter, and nicer (but needs someone with a very good feeling about what contour to integrate along). We will use the fact that $$ \int_0^1\frac{\log x}{1+x^2}\,dx=-\mathrm G. $$ where $\mathrm G$ denotes Catalan's constant (it must show up somehow).

I will give some details below, but I cannot motivate myself to write everything explicitly.

First, let $$ f(s)=\int_0^{+\infty}\frac{\log(s+x^3)}{1+x^2}\,dx $$ Note that (just split the integral into $\int_0^1+\int_1^{+\infty}$ and do $y=1/x$ in the latter) $$ f(0)=\int_0^{+\infty}\frac{3\log x}{1+x^2}\,dx=0. $$ The integral we want to calculate becomes $$ f(1)=f(0)+\int_0^1 f'(s)\,ds=\int_0^1 f'(s)\,ds. $$ We calculate $f'(s)$ below. Differentiating, making a partial fraction decomposition, and calculating elementary but horrible primitives, we find that $$ \begin{aligned} f'(s)&=\int_0^{+\infty}\frac{1}{(s+x^3)(1+x^2)}\,dx\\ &=\frac{1}{1+s^2}\int_0^{+\infty}\frac{s+x}{1+x^2}+\frac{1-sx-x^2}{s+x^3}\,dx\\ &=\cdots\\ &=\frac{1}{18(1+s^2)}\Bigl(\frac{4\sqrt{3}\pi}{s^{2/3}}-4\sqrt{3}\pi s^{2/3}+9\pi s+6\log s\Bigr) \end{aligned} $$ Next, we calculate more elementary, but horrible, primitives, (let $u=s^{1/3}$) $$ \int \frac{1}{18(1+s^2)}\Bigl(\frac{4\sqrt{3}\pi}{s^{2/3}}-4\sqrt{3}\pi s^{2/3}\Bigr)\,ds=\frac{\pi}{3}\log\Bigl(\frac{1+\sqrt{3}s^{1/3}+s^{2/3}}{1-\sqrt{3}s^{1/3}+s^{2/3}}\Bigr). $$ Hence, $$ \begin{aligned} f(1)&=\int_0^1 f'(s)\,ds\\ &=\biggl[\frac{\pi}{3}\log\Bigl(\frac{1+\sqrt{3}s^{1/3}+s^{2/3}}{1-\sqrt{3}s^{1/3}+s^{2/3}}\Bigr)+\frac{\pi}{4}\log(1+s^2)\biggr]_0^1+\frac{1}{3}\int_0^1\frac{\log s}{1+s^2}\,ds\\ &=\frac{\pi}{4}\log 2+\frac{2\pi}{3}\log(2+\sqrt{3})-\frac{\mathrm G}{3}. \end{aligned} $$

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    $\begingroup$ Good show with the non-complex approach. $\endgroup$ – Ron Gordon Feb 24 '16 at 16:19

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