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I'm looking for a closed form for this integral: $$I=\int_0^\infty\frac{\arctan(x)\,\operatorname{arccot}(x+1)}{x}dx.$$ Mathematica and Maple could not evaluate it symbolically. Numerically, $$I\approx1.3513049368715095284050230093075694014884142059538...$$ WolframAlpha and ISC+ could not find a plausible closed form for this number. Still, I hope that it exists, because the integral looks nice.

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  • $\begingroup$ @OlivierOloa c'est pour toi ça :-) $\endgroup$ – ParaH2 Nov 27 '15 at 23:42
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The integrand has a closed-form antiderivative in terms of elementary functions and polylogarithms. It can be found using Mathematica after expressing inverse trig functions through logarithms of complex arguments, and can be manually checked for correctness using differentiation. After subtracting its limits at $\infty$ and $0$ and simplification, we can get this result: $$\begin{align}I&=\frac1{32}\Big[\!\operatorname{Li}_2\!\left(\tfrac15\right)\cdot\ln2+\operatorname{Li}_3\!\left(\tfrac15\right)+ \operatorname{Li}_3\!\left(\tfrac45\right)\!\Big]+\frac{\pi}{16}\,\left(3\operatorname{Ti}_2(2)-4\,G\right)\\&+\frac{\ln5}{192}\,\left(9\ln2\cdot\ln5-12\ln^22-2\ln^25\right)+\frac{\pi^2}{192}\,\left(\ln5-7\ln2\right)+\frac{17}{16}\zeta(3),\end{align}$$ where $\operatorname{Ti}_2(x)$ is the inverse tangent integral, $G=\operatorname{Ti}_2(1)$ is the Catalan constant.

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  • $\begingroup$ I used the same approach in Mathematica. $\endgroup$ – Ben Longo Nov 28 '15 at 7:32
  • $\begingroup$ Thanks. I learnt something new ( "$\ldots$expressing inverse trig functions$\ldots$" ). $\endgroup$ – Felix Marin Aug 11 '16 at 15:30
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Using Mathematica I have arrived at the result

$$I=\frac{1}{4} (-\text{Li}_3(-1-i)-\text{Li}_3(-1+i)+\text{Li}_3(1-i)+\text{Li}_3(1+i))+\frac{1}{4} \left(-\text{Li}_3\left(-\frac{1}{2}-\frac{i}{2}\right)-\text{Li}_3\left(-\frac{1}{2}+\frac{i}{2}\right)+\text{Li}_3\left(\frac{1}{2}-\frac{i}{2}\right)+\text{Li}_3\left(\frac{1}{2}+\frac{i }{2}\right)+\text{Li}_2\left(-\frac{1}{2}-\frac{i}{2}\right) \log \left(-\frac{1}{2}-\frac{i}{2}\right)+\text{Li}_2\left(-\frac{1}{2}+\frac{i}{2}\right) \log \left(-\frac{1}{2}+\frac{i}{2}\right)-\text{Li}_2\left(\frac{1}{2}-\frac{i}{2}\right) \log \left(\frac{1}{2}-\frac{i}{2}\right)-\text{Li}_2\left(\frac{1}{2}+\frac{i}{2}\right) \log \left(\frac{1}{2}+\frac{i}{2}\right)\right)$$

Which matches your numerical estimates.

$$I\approx1.3513049368715095284050230093075694014884142059538$$

I am unable to simplify it past the polylogs. Running full simplify we can get to

$$\frac{1}{384} \left(48 \pi C+3 \left(-64 \text{Li}_3\left(-\frac{1}{2}-\frac{i}{2}\right)-64 \text{Li}_3\left(-\frac{1}{2}+\frac{i}{2}\right)+32 (\text{Li}_3(1-i)+\text{Li}_3(1+i))+35 \zeta (3)\right)+\text{Li}_2\left(-\frac{1}{2}-\frac{i}{2}\right) (-48 \log (2)-72 i \pi )+24 \text{Li}_2\left(-\frac{1}{2}+\frac{i}{2}\right) (-\log (4)+3 i \pi )-4 \log ^3(2)+7 \pi ^2 \log (2)\right)$$ Where $C$ is Catalan's constant.

I think this is about as closed form as it is going to get.

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  • $\begingroup$ These polylogarithms of complex arguments are very much simplifiable. Unfortunately, Mathematica cannot do it automatically. $\endgroup$ – Vladimir Reshetnikov Aug 11 '16 at 20:39
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In here we will find the anti-derivative of the integrand. It appears that the result can be expressed in terms of elementary functions and the poly-logarithm of order two. In order to arrive at the result the only thing we need to do is to use integration by parts and the definition of the poly-logarithm. We have: \begin{eqnarray} &&\int \arctan(x) \mbox{arccot}(x+1) \frac{1}{x} dx \\ &&= \arctan(x) \mbox{arccot}(x+1) \log(x) - \int \left( \frac{\mbox{arccot}(1+x)}{1+x^2} - \frac{\arctan(x)}{1+(1+x)^2}\right) \log(x) dx \\ &&= \arctan(x) \mbox{arccot}(x+1) \log(x) -{\mathcal I}_1(x) \mbox{arccot}(x+1) + {\mathcal I}_2(x) \arctan(x) + \\ && \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -\int {\mathcal I}_1(x) \frac{1}{1+(1+x)^2} dx - \int {\mathcal I}_2(x) \frac{1}{1+x^2} dx \end{eqnarray} where \begin{eqnarray} {\mathcal I}_1(x) &:=& \int \log(x) \frac{1}{1+x^2} dx = \frac{\pi}{4} \log(1+x^2) - \frac{1}{2 \imath} \left( Li_2(1+\imath x)-Li_2(1-\imath x)\right)\\ {\mathcal I}_2(x) &:=& \int \log(x) \frac{1}{1+(x+1)^2} dx=\arctan(\frac{x}{x+2})\log(x) + \frac{1}{2 \imath} \left(Li_2(-\frac{1+\imath}{2} x) - Li_2(-\frac{1-\imath}{2} x) \right) \end{eqnarray} The equations above have been derived by expressing the inverse polynomials into simple fractions and then relating the result to the definition of the poly-logarithmic function $Li_2(x) := -\int\log(1-x)/x dx$.

There was a mistake in the old version of my answer.More complicated integrals appeared. I will finished this post later and salvage this approach.

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