The challenge here is merely to evaluate the following integral:

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

This integral is a tough one because of the branch points strewn throughout the complex plane, as well as the utter lack of symmetry. It turns out, however, that we may still use the residue theorem to evaluate the integral so long as we choose a suitable contour.

To wit, consider

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

where $C$ is the following contour

(Apologies for the crudeness.) 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 poles of order $2$ 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

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

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

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

Now,

$$\int_1^{\infty} \frac{dy}{(1+y^2)^2} = \int_{\pi/4}^{\pi/2} d\theta \cos^2{\theta} = \frac{\pi}{8}-\frac14$$

$$\begin{align}\int_1^{\infty} dy\frac{\log{y}}{(1+y^2)^2} &= -\int_0^1 du \frac{u^2 \log{u}}{(1+u^2)^2}\\ &= -\sum_{k=0}^{\infty} (-1)^k (k+1) \int_0^1 u^{2 k+2} \log{u} \\ &= \sum_{k=0}^{\infty} (-1)^k \frac{k+1}{(2 k+3)^2} \\ &= \frac{G}{2} – \frac{\pi}{8}\end{align}$$

so that

$$\int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{(1+t^2)^2} = – \left ( \frac{G}{2} – \frac{\pi}{8} \right ) – i \pi \left ( \frac{\pi}{8}-\frac14\right ) $$

Along similar lines,

$$\int_{e^{i \pi/3}}^{\infty e^{i \pi/3}} dt \frac{\log{t}}{(1+t^2)^2} = \frac{G}{3}-\frac{\pi }{8}+\frac{1}{12} \pi \log \left(2+\sqrt{3}\right)+i

\left(\frac{1}{4} \log \left(2+\sqrt{3}\right)-\frac{\pi }{6}\right)$$

$$\int_{e^{i 5 \pi/3}}^{\infty e^{i 5 \pi/3}} dt \frac{\log{t}}{(1+t^2)^2} = \frac{G}{3}-\frac{\pi }{8}-\frac{5}{12} \pi \log \left(2+\sqrt{3}\right)+i

\left(-\frac{5 \pi }{6}+\frac{\pi ^2}{4}-\frac{1}{4} \log

\left(2+\sqrt{3}\right)\right)$$

Combining the integrals, I get

$$\frac{G}{6} -\frac{\pi}{8}-\frac{\pi}{3} \log{(2+\sqrt{3})} + i \left [-\frac{3 \pi}{4} + \frac{\pi^2}{8}\right ] $$

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

$$\sum_{\pm} \frac{d}{dz} \left[\frac{\log{(1+z^3)} \log{z}}{(z\pm i)^2} \right]_{z=\pm i} = \frac{\pi}{2}-\frac{\pi}{8}\log (2)+i \left(\frac{3 \pi }{4}-\frac{\pi ^2}{8}\right)$$

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

$$\int_0^{\infty} dx \frac{\log{(1+x^3)}}{(1+x^2)^2} = -\frac{G}{6} – \frac{3\pi}{8} + \frac{\pi}{8} \log{2} + \frac{\pi}{3} \log{(2+\sqrt{3})} \approx 0.320555$$

which agrees with Mathematica. Note how the imaginary parts fortuitously canceled.

Hi Ron:

We haven’t spoke for a while, so I thought I would drop a line and ask you something.

I like to look over your blog on occasion to see what new stuff you have posted.

Here:

http://math.stackexchange.com/questions/385931/closed-form-for-int-0-infty-frac-arctan-x-ln1x21x2-sqrtx-dx/386090#386090

I have been kind of obsessing over this one for a little while. I am confounded by O.L’s solution. If you feel like it, could you explain how we could evaluate those complex integrals he derived?. He just says, “by properly deforming the contour”. What contour?. A professor of mine suggested using a semicircle whose diameter lies on the line y=x and use an appropriate branch. How can we get to that Catalan in the solution?. I have pieces strung about here and there, but have not put the puzzle together ğŸ™‚ Take care, Cody

Hey Cody,

Sorry for not getting back to you so quickly. I will have a look as soon as I can. Sorry also I haven’t posted in a few weeks. Combination of not having much to say and being very busy at work.

Ron

I certainly understand the ‘busy’ thing. Whenever you get around to it. Thanks for responding and take care.

Cody

Cody – you too. – Ron

hi

how about laplaceinverse(1/sqrt(s^2-1))

Post it as a question in Math.SE and I will answer it.

Hi,

concerning “an integral with difficult branch

points”.I don’t understand the integrals along the branch cuts.The log has a jump od –i2pi.Is it log(z) or log(1+x^3)or both.

Thank yiou for uour help

Gilbert,

The factor log(1+z^3) has jumps where 1+z^3=0 (leaving the factor of log(z)), while log(z) has a jump about z=0 (leaving the factor log(1+z^3)). Once you get that (which really is not too difficult to understand), then the real fun begins.

Thanks for having a look!

Ron

Hi Ron,

Again concerning integral with difficult branch points.

I solve the first integral but I have trouble

with the second.integral between :inf.e^pi*i/3 and e^pi.i/3.I tried again and again but now I’m stuck.

I’m a hobby mathematician and it is the first time I see such an integral.In the

books you have more usual examples.

I hope it’s opportune to ask this question.

Thank you

Gilbert