Evaluate $\int_0^1 \arctan^3 x\,dx$

Question

I don't want to use the Fourier series. My work \begin{align}J&=\int_0^1 \arctan^3 x\,dx\\ &=[x\arctan^3 x]_0^1 -3\int_0^1 \frac{x\arctan^2 x}{1+x^2}\,dx\\ &=\frac{\pi^3 }{64}-3\int_0^1 \frac{x\arctan^2 x}{1+x^2}\,dx\\ &= \frac{\pi^3 }{64}-\frac{3}{2}\left[\ln(1+x^2)\arctan^2 x\right]_0^1 +3\int_0^1 \frac{\ln(1+x^2)\arctan x}{1+x^2}\,dx\\ &=\frac{\pi^3 }{64}-\frac{3\pi^2\ln 2}{32}+3\int_0^1 \frac{\ln(1+x^2)\arctan x}{1+x^2}\,dx\\ \end{align} How to continue?

Answer 1

Following is an "elementary" solution. \begin{align} J&=\int_0^1 \arctan^3 x\,dx\\ &\overset{\text{IPP}}=\Big[x\arctan^3 x\Big]_0^1 -3\int_0^1 \frac{x\arctan^2 x}{1+x^2}\,dx\\ &=\frac{\pi^3 }{64}-3\int_0^1 \frac{x\arctan^2 x}{1+x^2}\,dx\\ &\overset{\text{IBP}}=\frac{\pi^3 }{64}-\frac{3}{2}\left[\ln(1+x^2)\arctan^2 x\right]_0^1 +3\int_0^1 \frac{\ln(1+x^2)\arctan x}{1+x^2}\,dx\\ &=\frac{\pi^3 }{64}-\frac{3\pi^2\ln 2}{32}+3\int_0^1 \frac{\ln(1+x^2)\arctan x}{1+x^2}\,dx\\ &\overset{x=\tan t}=\frac{\pi^3 }{64}-\frac{3\pi^2\ln 2}{32}-6\int_0^{\frac{\pi}{4}}t\ln(\cos t)\,dt\\ A&=\int_0^{\frac{\pi}{4}}t\ln(\cos t)\,dt\\ B&=\int_0^{\frac{\pi}{4}}t\ln(\sin t)\,dt\\ B-A&=\int_0^{\frac{\pi}{4}}t\ln(\tan t)\,dt\\ &\overset{x=\tan t}=\int_0^1 \frac{\arctan x\ln x}{1+x^2}\,dx\\ \end{align} Define on $[0;\infty]$ the function $R$ by,

for all $x\in [0;\infty]$, $\displaystyle \text{R}(x)=\int_0^x \dfrac{\ln t}{1+t^2}\,dt=\int_0^1 \dfrac{x\ln(tx)}{1+t^2x^2}\,dt$.

Observe that $\text{R}(0)=\text{R}(\infty)=0$ et $\text{R}(1)=-\text{G}$ \begin{align} U&=\int_0^1 \frac{\arctan\left(\frac{1}{x}\right)\ln x}{1+x^2}\,dx\\ V&=\int_0^1 \frac{\arctan x\ln x}{1+x^2}\,dx\\ U+V&=-\frac{1}{2}\pi\text{G}\\ U&\overset{\text{IBP}}=\left[R(x)\arctan\left(\frac{1}{x}\right)\right]_0^1 +\int_0^1 \frac{R(x)}{1+x^2}\,dx\\ &=-\frac{1}{4}\pi\text{G}+\int_0^1 \int_0^1 \frac{x\ln(tx)}{(1+x^2)(1+t^2x^2)}\,dt\,dx\\ &=-\frac{1}{4}\pi\text{G}+\int_0^1 \left(\int_0^1 \dfrac{x\ln(x)}{(1+t^2x^2)(1+x^2)}\,dt\right)\,dx+\\ &\int_0^1 \left(\int_0^1 \dfrac{x\ln(t)}{(1+t^2x^2)(1+x^2)}\,dx\right)\,dt\\ &=-\frac{1}{4}\pi\text{G}+V+\frac{1}{2}\ln 2\int_0^1 \frac{\ln t}{1-t^2}\,dt-\frac{1}{2}\int_0^1 \frac{\ln t\ln(1+t^2)}{1-t^2}\,dt\\ &=-\frac{1}{4}\pi\text{G}+V-\frac{1}{16}\pi^2\ln 2-\frac{1}{2}\int_0^1 \frac{\ln t\ln(1+t^2)}{1-t^2}\,dt\\ Z&=\int_0^1 \frac{\ln x\ln(1+x^2)}{1-x^2}\,dx \end{align} Define on $[0;1]$ the function $S$ by,

for all $x\in [0;1]$, $\displaystyle S(x)=\int_0^x \frac{\ln t}{1-t^2}\,dt=\int_0^1 \frac{x\ln (tx)}{1-t^2x^2}\,dt$

Observe that $\displaystyle S(0)=0,S(1)=-\frac{1}{8}\pi^2$. \begin{align}Z&\overset{\text{IBP}}=\Big[S(x)\ln(1+x^2)\Big]_0^1-\int_0^1 \frac{2xS(x)}{1+x^2}\,dx\\ &=-\frac{1}{8}\pi^2\ln 2-\int_0^1 \int_0^1 \frac{2x^2\ln(tx)}{(1+x^2)(1-t^2x^2)}\,dt\,dx\\ &=-\frac{1}{8}\pi^2\ln 2-\int_0^1 \left(\int_0^1 \frac{2x^2\ln x}{(1+x^2)(1-t^2x^2)}\,dt\right)\,dx-\\ &\int_0^1 \left(\int_0^1 \frac{2x^2\ln t}{(1+x^2)(1-t^2x^2)}\,dx\right)\,dt\\ &=-\frac{1}{8}\pi^2\ln 2+\int_0^1 \frac{x\ln\left(\frac{1-x}{1+x}\right)\ln x}{1+x^2}\,dx-\\ &\left(\int_0^1 \frac{\ln(1+t)\ln t}{t}\,dt+\int_0^1 \frac{t\ln\left(\frac{1-t}{1+t}\right)\ln t}{1+t^2}dt-\int_0^1 \frac{\ln(1-t)\ln t}{t}\,dt-\frac{\pi}{2}\int_0^1 \frac{\ln t}{1+t^2}dt\right)\\ &=-\frac{1}{8}\pi^2\ln 2-\int_0^1 \frac{\ln(1+t)\ln t}{t}\,dt+\int_0^1 \frac{\ln(1-t)\ln t}{t}\,dt-\frac{1}{2}\pi\text{G}\\ &=-\frac{1}{8}\pi^2\ln 2+\frac{7}{4}\zeta(3)-\frac{1}{2}\pi\text{G}\\ \end{align} Therefore, \begin{align}U&=V-\frac{7}{8}\zeta(3)\\ V&=\frac{7}{16}\zeta(3)-\frac{1}{4}\pi\text{G}\\ B-A&=\frac{7}{16}\zeta(3)-\frac{1}{4}\pi\text{G}\\ U&=-\frac{7}{16}\zeta(3)-\frac{1}{4}\pi\text{G}\\ \end{align} On the other hand, \begin{align} A+B&=\int_0^{\frac{\pi}{4}}t\ln\left(\frac{1}{2}\sin(2t)\right)\,dt\\ &=\int_0^{\frac{\pi}{4}}t\ln\left(\sin(2t)\right)\,dt-\frac{\pi^2\ln 2}{32}\\ &\overset{x=2t}=\frac{1}{4}\int_0^{\frac{\pi}{2}}x\ln\left(\sin x\right)\,dx-\frac{\pi^2\ln 2}{32}\\ A_2&=\int_0^{\frac{\pi}{2}}t\ln(\cos t)\,dt\\ B_2&=\int_0^{\frac{\pi}{2}}t\ln(\sin t)\,dt\\ A_2+B_2&=\int_0^{\frac{\pi}{2}}t\ln\left(\frac{1}{2}\sin(2t)\right)\,dt\\ &=\int_0^{\frac{\pi}{2}}t\ln\left(\sin(2t)\right)\,dt-\frac{\pi^2\ln 2}{8}\\ &\overset{x=2t}=\frac{1}{4}\int_0^\pi x\ln(\sin x)\,dx-\frac{\pi^2\ln 2}{8}\\ &\overset{t=\pi-x}=\frac{1}{4}\int_0^\pi (\pi-x)\ln(\sin x)\,dx-\frac{\pi^2\ln 2}{8}\\ 2(A_2+B_2)&=\frac{\pi}{4}\int_0^\pi \ln(\sin x)\,dx-\frac{\pi^2\ln 2}{4}\\ A_2+B_2&=\frac{\pi}{8}\int_0^\pi \ln(\sin x)\,dx-\frac{\pi^2\ln 2}{8}\\ &=-\frac{\pi^2\ln 2}{4}\\ B2-A2&=\int_0^{\frac{\pi}{2}}t\ln(\tan t)\,dt\\ &\overset{x=\tan t}=\int_0^\infty \frac{\ln x\arctan x}{1+x^2}\,dx\\ \end{align} \begin{align} U_2&=\int_0^\infty \frac{\arctan\left(\frac{1}{x}\right)\ln x}{1+x^2}\,dx\\ V_2&=\int_0^\infty \frac{\arctan\left(x\right)\ln x}{1+x^2}\,dx\\ U_2+V_2&=\frac{\pi}{2}\int_0^\infty \frac{\ln x}{1+x^2}\,dx\\ &=0\\ U_2&\overset{\text{IBP}}=\left[R(x)\arctan\left(\frac{1}{x}\right)\right]_0^\infty +\int_0^\infty \frac{R(x)}{1+x^2}\,dx\\ &=\int_0^\infty \left(\int_0^1 \dfrac{x\ln(tx)}{(1+t^2x^2)(1+x^2)}\,dt\right)\,dx\\ &=\int_0^\infty \left(\int_0^1 \dfrac{x\ln(x)}{(1+t^2x^2)(1+x^2)}\,dt\right)\,dx+\\ &\int_0^1 \left(\int_0^\infty \dfrac{x\ln(t)}{(1+t^2x^2)(1+x^2)}\,dx\right)\,dt\\ &=V_2+\int_0^1 \frac{\ln^2 t}{t^2-1}\,dt\\ &=V_2+\int_0^1 \frac{t\ln^2 t}{1-t^2}\,dt-\int_0^1 \frac{\ln^2 x}{1-x}\,dx\\ &\overset{u=t^2}=B+\frac{1}{8}\int_0^1 \frac{\ln^2 t}{1-t}\,dt-\int_0^1 \frac{\ln^2 x}{1-x}\,dx\\ &=V_2-\frac{7}{8}\int_0^1 \frac{\ln^2 x}{1-x}\,dx\\ &=V_2-\frac{7}{8}\times 2\zeta(3)\\ &=V_2-\frac{7}{4}\zeta(3)\\ U_2&=-\frac{7}{8}\zeta(3)\\ V_2&=\frac{7}{8}\zeta(3)\\ B_2-A_2&=\frac{7}{8}\zeta(3)\\ A_2&=-\frac{7}{16}\zeta(3)-\frac{1}{8}\pi^2\ln 2\\ B_2&=\frac{7}{16}\zeta(3)-\frac{1}{8}\pi^2\ln 2\\ A+B&=\frac{7}{64}\zeta(3)-\frac{1}{16}\pi^2\ln 2\\ A&=\frac{1}{8}\pi\text{G}-\frac{21}{128}\zeta(3)-\frac{1}{32}\pi^2\ln 2\\ B&=\frac{35}{128}\zeta(3)-\frac{1}{32}\pi^2\ln 2-\frac{1}{8}\pi\text{G}\\ J&=\frac{\pi^3 }{64}-\frac{3\pi^2\ln 2}{32}-6A\\ J&=\boxed{\frac{1}{64}\pi^3+\frac{3}{32}\pi^2\ln 2-\frac{3}{4}\pi\text{G}+\frac{63}{64}\zeta(3)} \end{align}

NB: I assume following results: \begin{align}x>0,\arctan(x)+\arctan\left(\frac{1}{x}\right)&=\frac{\pi}{2}\\ \int_0^\infty \frac{\ln x}{1+x^2}\,dx&=0\\ \int_0^{\frac{\pi}{2}}\ln(\cos x)\,dx&=-\frac{1}{2}\pi\ln 2\\ \int_0^1 \frac{\ln x}{1-x^2}\,dx=-\frac{1}{8}\pi^2\\ \int_0^1 \frac{\ln^2 x}{1-x}\,dx&=2\zeta(3)\\ \int_0^1 \frac{\ln(1+x)\ln x}{x}\,dx&=-\frac{3}{4}\zeta(3)\\ \int_0^1 \frac{\ln(1-x)\ln x}{x}\,dx&=\zeta(3)\\ \int_0^\pi \ln(\sin x)\,dx&=-\pi\ln 2 \end{align}

Answer 2

I am not sure if this counts as a solution using Fourier series, but let me present my solution anyway: By the substitution $x=\tan\theta$, we get

$$ J = \int_{0}^{\frac{\pi}{4}} \theta^3 \sec^2\theta \, \mathrm{d}\theta. $$

In order to compute this integral, we will utilize the following regularized expansion:

$$ \sec^2\theta = \frac{4e^{2it}}{(1+e^{2it})^2} = \lim_{r \uparrow 1} \frac{4r e^{2it}}{(1+r e^{2it})^2} = 4 \lim_{r \uparrow 1} \sum_{n=1}^{\infty} (-1)^{n-1} n r^n e^{2in\theta} $$

Plugging this back to $J$, we can interchange the order of limit and integration by the uniform convergence, whence we get

\begin{align*} J &= 4 \lim_{r \uparrow 1} \sum_{n=1}^{\infty} (-1)^{n-1} n r^n \int_{0}^{\frac{\pi}{4}} \theta^3 e^{2in\theta} \, \mathrm{d}\theta. \end{align*}

Now by the integration by parts,

$$ \int_{0}^{\frac{\pi}{4}} \theta^3 e^{2in\theta} \, \mathrm{d}\theta = -\frac{3i^n}{8n^4} + \frac{3}{8n^4} + \frac{3\pi i^{n+1}}{16n^3} + \frac{3\pi^2 i^n}{64n^2} - \frac{\pi^3 i^{n+1}}{128n}. $$

Plugging this and taking limit,

\begin{align*} J &= \lim_{r \uparrow 1} \sum_{n=1}^{\infty} (-1)^{n-1} r^n \left( -\frac{3i^n}{2n^3} + \frac{3}{2n^3} + \frac{3\pi i^{n+1}}{4n^2} + \frac{3\pi^2 i^n}{16n} - \frac{\pi^3 i^{n+1}}{32} \right) \\ &= \sum_{n=1}^{\infty} (-1)^{n-1} \left( -\frac{3i^n}{2n^3} + \frac{3}{2n^3} + \frac{3\pi i^{n+1}}{4n^2} + \frac{3\pi^2 i^n}{16n} \right) + \frac{\pi^3}{64}(1-i) \\ &= - \frac{3}{2} \left( \frac{1}{2^3} - \frac{1}{4^3} + \frac{1}{6^3} - \dots \right) + \frac{3}{2} \left( \frac{1}{1^3} - \frac{1}{2^3} + \frac{1}{3^3} - \dots \right) \\ &\quad - \frac{3\pi}{4} \left( \frac{1}{1^2} - \frac{1}{3^2} + \frac{1}{5^2} - \dots \right) + \frac{3\pi^2}{16} \left( \frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \dots \right) + \frac{\pi^3}{64} \\ &\quad + \underbrace{\text{[imaginary term]}}_{=0}. \end{align*}

Simplifying this, we get

$$ J = -\frac{3\pi G}{4} + \frac{63\zeta(3)}{64} + \frac{\pi^3}{64} + \frac{3 \pi^2 \log 2}{32}, $$

where $G$ is the Catalan's constant.

Answer 3

Not an answer, but Wolfram Alpha finds the amazing $$\int (\arctan(x))^3\mathrm{d}x$$ $$=\frac{3}{2}\operatorname{Li}_3(-e^{2i\arctan(x)})-3i\arctan(x)\operatorname{Li}_2(-e^{2i\arctan(x)})+(\arctan(x))^2(x\arctan(x)-i\arctan(x)+3\ln(1+e^{2i\arctan(x)}))+c$$ And the even more beautiful $$\int_0^1 (\arctan(x))^3\mathrm{d}x=\frac{1}{64}(\pi^2(\pi+\ln(64))+63\zeta(3)-48\pi C)$$ With Catalan's constant $C$ and the Riemann zeta function $\zeta$.

But, continuing from your last line, $$\int_0^1 \frac{\ln(1+x^2)\arctan(x)}{1+x^2}\mathrm{d}x$$ We can make the substitution $x=\tan\theta, \mathrm{d}x=\sec^2(\theta)\mathrm{d}\theta$ to get $$\int_0^{\arctan(1)} \theta\ln(\sec^2(\theta))\mathrm{d}\theta$$ Although knowledge of the polylogarithm will still be needed.

Answer 4

Continuing the OP’s work,

$$I=\int_0^1 \frac{\ln(1+x^2)\arctan x}{1+x^2}\,dx$$

$$\underset{\arctan x \to x} \implies -2 \int_0^{\frac{\pi}{4}} x \ln(\cos x)\, dx = 2 \int_0^{\frac{\pi}{4}} x \ln 2 \, dx + 2 \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \int_0^{\frac{\pi}{4}} x \cos(2nx)\, dx$$

$$\implies \frac{\pi^2}{16} \ln 2 + 2 \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \int_0^{\frac{\pi}{4}} x \cos(2nx)\, dx$$

$$\underset{ibp}\implies \frac{\pi^2}{16} \ln 2 + \frac{\pi}{4} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \sin\left(\frac{n\pi}{2}\right) - \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \int_0^{\frac{\pi}{4}} \sin(2nx)\, dx$$

$$\implies \frac{\pi^2}{16} \ln 2 + \frac{\pi}{4} \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n + 1)^2} + \sum_{n=1}^{\infty} \frac{(-1)^n}{2n^3} \cos\left(\frac{n\pi}{2}\right) - \sum_{n=1}^{\infty} \frac{(-1)^n}{2n^3}\tag{*}$$

$$\implies \frac{\pi^2}{16} \ln 2 - \frac{\pi}{4} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n + 1)^2} - \frac{7}{16}\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3} = \frac{\pi^2}{16} \ln 2 - \frac{\pi}{4} G + \frac{21}{64} \zeta(3)$$

Using the definition of Catalan constant

In $(*)$ $\cos\left(\frac{n\pi}{2}\right) \in \{-1,1\}$ only for even valued $n$ using this, we can re-write in terms of the third sum in the same step.

You can now add the above result to your work.

Therefore,

$$\int_0^1 \arctan^3 x\,dx=\frac{\pi^3 }{64}+\frac{3\pi^2}{32} \ln 2 - \frac{3\pi}{4} G + \frac{63}{64} \zeta(3)$$

Answer 5

\begin{align} J&=\int_0^1 \arctan^3 t\,dt \overset{t =\tan x} = \int_0^{\frac\pi4}x^3d(\tan x)\\ &\overset{ibp}=\frac{\pi^3 }{64}+\frac32 \int_ 0^{\frac\pi4}x^2d[\ln (\sqrt2\cos x)] \overset{ibp} = \frac{\pi^3 }{64}-6K\\ \end{align} where \begin{align} K& =\int_{0}^{\frac{\pi}{4}}x\ln(\sqrt2\cos x)dx\\ =&\ \frac12 \int_0^{\frac\pi4 }x\ln(\sin 2x)\overset{2x\to x}{dx }- \frac12\int_0^{\frac\pi4 }x\ln(\tan x) \overset{x\to \frac\pi2-x}{dx }\\ =& \ \frac18\int_0^{\frac\pi2}x\ln(\sin x)dx -\frac14\int_0^{\frac\pi2}x\ln(\tan x) dx -\frac\pi8 \int_0^{\frac\pi4 }\ln(\tan x)dx \end{align} Note that $\int_0^{\frac\pi4 }\ln(\tan x)dx=-G$, $\int_0^{\frac\pi2}x\ln(\tan x) dx=\frac78\zeta(3)$ \begin{align} &\int_{0}^{\frac{\pi}{2}} x\ln(\sin x)dx =\frac12 \int_0^{\frac\pi2 }x\ln\frac{\sin 2x}2 \overset{2x\to x}{dx }+\frac12\int_0^{\frac\pi2 }x\ln(\tan x) dx\\ & =\frac18\int_0^{\pi} \underset{=0}{x\ln(2\sin x)}dx-\frac{\pi^2}{8}\ln2 +\frac12\cdot\frac7{8}\zeta(3)= -\frac{\pi^2}{8} \ln2+\frac7{16}\zeta(3)\\ \end{align} As a result $$J= \frac{63}{64}\zeta(3) -\frac{3\pi}{4}{G} +\frac{3 \pi^2}{32}\ln 2 +\frac{\pi^3}{64} $$