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How to calculate the value of the integrals $$\int_0^1\left(\frac{\arctan x}{x}\right)^2\,dx,$$

$$\int_0^1\left(\frac{\arctan x}{x}\right)^3\,dx $$

and

$$\int_0^1\frac{\arctan^2 x\ln x}{x}\,dx?$$

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  • $\begingroup$ For the indefinite integral, can you try with Integration By Parts to derive a Reduction formula $\endgroup$ – lab bhattacharjee Oct 21 '14 at 10:01
  • $\begingroup$ the result for the first integral is $C-\frac{1}{16} \pi (\pi -4 \log (2))$ says my collection $\endgroup$ – Dr. Sonnhard Graubner Oct 21 '14 at 10:33
  • $\begingroup$ and for the second integral $\frac{1}{64} (96 C+\pi (24 \log (2)-\pi (6+\pi )))$ $\endgroup$ – Dr. Sonnhard Graubner Oct 21 '14 at 10:35
  • $\begingroup$ You might want to improve your question to garner more detailed responses. $\endgroup$ – SuperAbound Oct 21 '14 at 11:17
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For the first one, \begin{align} \int^1_0\frac{\arctan^2{x}}{x^2}{\rm d}x =&\color{#BF00FF}{\int^\frac{\pi}{4}_0x^2\csc^2{x}\ {\rm d}x}\\ =&-x^2\cot{x}\Bigg{|}^\frac{\pi}{4}_0+2\int^\frac{\pi}{4}_0x\cot{x}\ {\rm d}x\\ =&-\frac{\pi^2}{16}+4\sum^\infty_{n=1}\int^\frac{\pi}{4}_0x\sin(2nx)\ {\rm d}x\\ =&-\frac{\pi^2}{16}+\sum^\infty_{n=1}\frac{\sin(n\pi/2)}{n^2}-\frac{\pi}{2}\sum^\infty_{n=1}\frac{\cos(n\pi/2)}{n}\\ =&-\frac{\pi^2}{16}+\color{#E2062C}{\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^2}}\color{#21ABCD}{-\frac{\pi}{2}\sum^\infty_{n=1}\frac{(-1)^n}{2n}}\\ =&\color{#E2062C}{G}+\color{#21ABCD}{\frac{\pi}{4}\ln{2}}-\frac{\pi^2}{16}\\ =&\color{#BF00FF}{G+\frac{\pi}{4}\ln{2}-\frac{\pi^2}{16}} \end{align} A justification for the third line may be found here.


For the second one, \begin{align} \int^1_0\frac{\arctan^3{x}}{x^3}{\rm d}x =&\int^\frac{\pi}{4}_0x^3\cot{x}\csc^2{x}\ {\rm d}x\\ =&-\frac{1}{2}x^3\cot^2{x}\Bigg{|}^\frac{\pi}{4}_0+\frac{3}{2}\int^\frac{\pi}{4}_0x^2\cot^2{x}\ {\rm d}x\\ =&-\frac{\pi^3}{128}-\frac{3}{2}\int^\frac{\pi}{4}_0x^2\ {\rm d}x+\frac{3}{2}\color{#BF00FF}{\int^\frac{\pi}{4}_0x^2\csc^2{x}\ {\rm d}x}\\ =&-\frac{\pi^3}{64}+\frac{3}{2}\left(\color{#BF00FF}{G+\frac{\pi}{4}\ln{2}-\frac{\pi^2}{16}}\right)\\ =&\frac{3}{2}G-\frac{\pi^3}{64}+\frac{3\pi}{8}\ln{2}-\frac{3\pi^2}{32} \end{align}


For the third one, \begin{align} \int^1_0\frac{\arctan^2{x}\ln{x}}{x}{\rm d}x =&-\int^1_0\frac{\arctan{x}\ln^2{x}}{1+x^2}{\rm d}x\\ =&-\sum^\infty_{n=0}\sum^n_{k=0}\frac{(-1)^n}{2k+1}\int^1_0x^{2n+1}\ln^2{x}\ {\rm d}x\\ =&-\frac{1}{4}\sum^\infty_{n=0}\frac{(-1)^n\left(H_{2n+1}-\frac{1}{2}H_n\right)}{(n+1)^3}\\ =&\frac{1}{4}\sum^\infty_{n=1}\frac{(-1)^{n}H_{2n}}{n^3}-\frac{1}{8}\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^3} \end{align} Let us compute the generating function of $\displaystyle \frac{H_n}{n^3}$. \begin{align} \sum^\infty_{n=1}\frac{H_n}{n^3}z^n =&\frac{1}{2}\sum^\infty_{n=1}H_n\int^1_0\frac{(xz)^n\ln^2{x}}{x}\ {\rm d}x\\ =&-\frac{1}{2}\int^1_0\frac{\ln^2{x}\ln(1-xz)}{x(1-xz)}{\rm d}x\\ =&-\frac{1}{2}\int^z_0\frac{\ln^2\left(\frac{x}{z}\right)\ln(1-x)}{x(1-x)}{\rm d}x\\ =&-\frac{1}{2}\int^z_0\frac{\ln^2{x}\ln(1-x)}{x(1-x)}{\rm d}x+\ln{z}\int^z_0\frac{\ln{x}\ln(1-x)}{x(1-x)}{\rm d}x-\frac{\ln^2{z}}{2}\int^z_0\frac{\ln(1-x)}{x(1-x)}{\rm d}x\\ =&-\frac{1}{2}\int^z_0\frac{\ln^2{x}\ln(1-x)}{1-x}{\rm d}x-\frac{1}{2}\int^z_0\frac{\ln^2{x}\ln(1-x)}{x}{\rm d}x\\ &+\ln{z}\int^z_0\frac{\ln{x}\ln(1-x)}{x}{\rm d}x+\ln{z}\int^z_0\frac{\ln{x}\ln(1-x)}{1-x}{\rm d}x\\ &-\frac{\ln^2{z}}{2}\int^z_0\frac{\ln(1-x)}{x}{\rm d}x-\frac{\ln^2{z}}{2}\int^z_0\frac{\ln(1-x)}{1-x}{\rm d}x \end{align} The second integral is \begin{align} -\frac{1}{2}\int^z_0\frac{\ln^2{x}\ln(1-x)}{x}{\rm d}x =&\frac{1}{2}{\rm Li}_2(x)\ln^2{x}\Bigg{|}^z_0-\int^z_0\frac{{\rm Li}_2(x)\ln{x}}{x}{\rm d}x\\ =&\frac{1}{2}{\rm Li}_2(z)\ln^2{z}-{\rm Li}_3(x)\ln{x}\Bigg{|}^z_0+\int^z_0\frac{{\rm Li}_3(x)}{x}{\rm d}x\\ =&{\rm Li}_4(z)-{\rm Li}_3(z)\ln{z}+\frac{1}{2}{\rm Li}_2(z)\ln^2{z}\\ \end{align} The third integral is \begin{align} \ln{z}\int^z_0\frac{\ln{x}\ln(1-x)}{x}{\rm d}x =&-{\rm Li}_2(x)\ln{x}\ln{z}\Bigg{|}^z_0+\ln{z}\int^z_0\frac{{\rm Li}_2(x)}{x}{\rm d}x\\ =&{\rm Li}_3(z)\ln{z}-{\rm Li}_2(z)\ln^2{z} \end{align} The fourth integral is \begin{align} \ln{z}\int^z_0\frac{\ln{x}\ln(1-x)}{1-x}{\rm d}x =&{\rm Li}_2(1-x)\ln(1-x)\ln{z}\Bigg{|}^z_0+\ln{z}\int^z_0\frac{{\rm Li}_2(1-x)}{1-x}{\rm d}x\\ =&\zeta(3)\ln{z}-{\rm Li}_3(1-z)\ln{z}+{\rm Li}_2(1-z)\ln(1-z)\ln{z} \end{align} The fifth integral is \begin{align} -\frac{\ln^2{z}}{2}\int^z_0\frac{\ln(1-x)}{x}{\rm d}x =&\frac{1}{2}{\rm Li}_2(z)\ln^2{z} \end{align} The sixth integral is \begin{align} -\frac{\ln^2{z}}{2}\int^z_0\frac{\ln(1-x)}{1-x}{\rm d}x =&\frac{1}{4}\ln^2{z}\ln^2(1-z) \end{align} Putting all of this together, \begin{align}\sum^\infty_{n=1}\frac{H_n}{n^3}z^n=&-\frac{1}{2}\int^z_0\frac{\ln^2{x}\ln(1-x)}{1-x}{\rm d}x+{\rm Li}_4(z)-{\rm Li}_3(1-z)\ln{z}+\zeta(3)\ln{z}\\&+{\rm Li}_2(1-z)\ln{z}\ln(1-z)+\frac{1}{4}\ln^2{z}\ln^2(1-z)\end{align} By Landen's trilogarithm identity, the remaining integral is \begin{align} -\frac{1}{2}\int^z_0\frac{\ln^2{x}\ln(1-x)}{1-x}{\rm d}x =&\frac{1}{4}\ln^2{z}\ln^2(1-z)-\frac{1}{2}\int^z_0\frac{\ln{x}\ln^2(1-x)}{x}{\rm d}x\\ =&\frac{1}{4}\ln^2{z}\ln^2(1-z)+{\rm Li}_4(z)+\int^z_0\frac{{\rm Li}_3(1-x)}{x}{\rm d}x\\ &+\int^\frac{z}{z-1}_0\frac{{\rm Li}_3(x)}{x(1-x)}{\rm d}x-\frac{1}{6}\int^z_0\frac{\ln^3(1-x)}{x}{\rm d}x\\ &+\frac{\pi^2}{6}{\rm Li}_2(z)-\zeta(3)\ln{z}+\zeta(3)\ln\epsilon \end{align} The first integral is \begin{align} \int^z_0\frac{{\rm Li}_3(1-x)}{x}{\rm d}x =&{\rm Li}_3(1-x)\ln{x}\Bigg{|}^z_0+\int^z_0\frac{{\rm Li}_2(1-x)\ln{x}}{1-x}{\rm d}x\\ =&{\rm Li}_3(1-z)\ln{z}-\zeta(3)\ln{\epsilon}+\frac{1}{2}{\rm Li}_2^2(1-x)\Bigg{|}^z_0\\ =&{\rm Li}_3(1-z)\ln{z}+\frac{1}{2}{\rm Li}_2^2(1-z)-\frac{\pi^4}{72}-\zeta(3)\ln\epsilon \end{align} The second integral is \begin{align} \int^\frac{z}{z-1}_0\frac{{\rm Li}_3(x)}{x(1-x)}{\rm d}x =&{\rm Li}_4\left(\tfrac{z}{z-1}\right)-{\rm Li}_3(x)\ln(1-x)\Bigg{|}^\frac{z}{z-1}_0+\int^\frac{z}{z-1}_0\frac{{\rm Li}_2(x)\ln(1-x)}{x}{\rm d}x\\ =&{\rm Li}_4\left(\tfrac{z}{z-1}\right)+{\rm Li}_3\left(\tfrac{z}{z-1}\right)\ln(1-z)-\frac{1}{2}{\rm Li}_2^2\left(\tfrac{z}{z-1}\right) \end{align} The third integral is \begin{align} -\frac{1}{6}\int^z_0\frac{\ln^3(1-x)}{x}{\rm d}x =&-\frac{1}{6}\ln{x}\ln^3(1-x)\Bigg{|}^z_0-\frac{1}{2}\int^z_0\frac{\ln{x}\ln^2(1-x)}{1-x}{\rm d}x\\ =&-\frac{1}{6}\ln{z}\ln^3(1-z)-\frac{1}{2}{\rm Li}_2(1-x)\ln^2(1-x)\Bigg{|}^z_0\\ &-\int^z_0\frac{{\rm Li}_2(1-x)\ln(1-x)}{1-x}{\rm d}x\\ =&-\frac{1}{6}\ln{z}\ln^3(1-z)-\frac{1}{2}{\rm Li}_2(1-z)\ln^2(1-z)\\ &+{\rm Li}_3(1-x)\ln(1-x)\Bigg{|}^z_0-{\rm Li}_4(1-x)\Bigg{|}^z_0\\ =&-{\rm Li}_4(1-z)+{\rm Li}_3(1-z)\ln(1-z)-\frac{1}{2}{\rm Li}_2(1-z)\ln^2(1-z)\\ &-\frac{1}{6}\ln{z}\ln^3(1-z)+\frac{\pi^4}{90} \end{align} After consolidating all the terms and simplifying using the reflection and Landen formulae, we get \begin{align} \sum^\infty_{n=1}\frac{H_n}{n^3}z^n =&2{\rm Li}_4(z)+{\rm Li}_4\left(\tfrac{z}{z-1}\right)-{\rm Li}_4(1-z)-{\rm Li}_3(z)\ln(1-z)-\frac{1}{2}{\rm Li}_2^2\left(\tfrac{z}{z-1}\right)\\ &+\frac{1}{2}{\rm Li}_2(z)\ln^2(1-z)+\frac{1}{2}{\rm Li}_2^2(z)+\frac{1}{6}\ln^4(1-z)-\frac{1}{6}\ln{z}\ln^3(1-z)\\ &+\frac{\pi^2}{12}\ln^2(1-z)+\zeta(3)\ln(1-z)+\frac{\pi^4}{90} \end{align} By letting $z=-1$, we get \begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n}{n^3}=2{\rm Li}_4\left(\tfrac{1}{2}\right)-\frac{11\pi^4}{360}+\frac{7}{4}\zeta(3)\ln{2}-\frac{\pi^2}{12}\ln^2{2}+\frac{1}{12}\ln^4{2} \end{align} You may sub $z=-1$ and verify this yourself using the inversion formulae. Also observe that in general, \begin{align} {\rm Li}_s(i) =&\sum^\infty_{n=1}\frac{\cos(n\pi/2)}{n^s}+i\sum^\infty_{n=1}\frac{\sin(n\pi/2)}{n^s}\\ =&2^{-s}\sum^\infty_{n=1}\frac{(-1)^n}{n^s}+i\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^s}\\ =&\left(2^{1-2s}-2^{-s}\right)\zeta(s)+i\beta(s) \end{align} So we have \begin{align} {\rm Li}_2(i)&=-\frac{\pi^2}{48}+iG\\ {\rm Li}_3(i)&=-\frac{3}{32}\zeta(3)+i\frac{\pi^3}{32}\\ {\rm Li}_4(i)&=-\frac{7\pi^4}{11520}+i\beta(4) \end{align} Using the inversion (and reflection) formulae, we also get \begin{align} {\rm Li}_2\left(\tfrac{1-i}{2}\right) =&-\frac{\pi^2}{6}-\frac{1}{2}\ln^2(-1-i)-{\rm Li}_2(1+i)\\ =&\frac{5\pi^2}{96}-\frac{1}{8}\ln^2{2}+i\left(\frac{\pi}{8}\ln{2}-G\right) \end{align} and \begin{align} {\rm Li}_4\left(\tfrac{1-i}{2}\right) =&-{\rm Li}_4(1+i)-\frac{1}{24}\ln^4(-1-i)-\frac{\pi^2}{12}\ln^2(-1-i)-\frac{7\pi^4}{360}\\ =&-{\rm Li}_4(1+i)+\frac{1313\pi^4}{92160}+\frac{11\pi^2}{768}\ln^2{2}-\frac{1}{384}\ln^4{2}+i\left(\frac{7\pi^3}{256}\ln{2}+\frac{\pi}{64}\ln^3{2}\right) \end{align} Therefore, (this part alone took me more than one hour, embarassingly) \begin{align} \frac{1}{8}\sum^\infty_{n=1}\frac{(-1)^nH_{2n}}{n^3} =&\Re\sum^\infty_{n=1}\frac{H_n}{n^3}i^n\\ =&-2\Re{\rm Li}_4(1+i)+\frac{29\pi^4}{2304}+\frac{35}{64}\zeta(3)\ln{2}+\frac{\pi^2}{64}\ln^2{2} \end{align} Finally, \begin{align} &\color{#FF4F00}{\int^1_0\frac{\arctan^2{x}\ln{x}}{x}{\rm d}x}\\ =&\color{#FF4F00}{-4\Re{\rm Li}_4(1+i)-\frac{1}{4}{\rm Li}_4\left(\tfrac{1}{2}\right)+\frac{167\pi^4}{5760}+\frac{7}{8}\zeta(3)\ln{2}+\frac{\pi^2}{24}\ln^2{2}-\frac{1}{96}\ln^4{2}} \end{align}

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    $\begingroup$ Nice answer. +1. By the way it would be nice to specify $\Re\operatorname{Li}_4(1\pm i)$. We know that $\Re\operatorname{Li}_0(1\pm i)=\Re\operatorname{Li}_1(1\pm i)=0$. Furthermore $\Re\operatorname{Li}_2(1\pm i)=\pi^2/16$. I've also evaluated $\Re\operatorname{Li}_3(1\pm i)$ here. $$\Re \left[\operatorname{Li}_3(1 \pm i)\right] = \frac{\pi^2}{32} \ln 2 + \frac{35}{64} \zeta(3).$$ I don't know how to go further. $\endgroup$ – user153012 Oct 22 '14 at 12:22
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    $\begingroup$ @user153012 Thanks. It is easy to determine $\Re Li_s(1\pm i)$ for $s\le 3$ due to the existence of functional equations like the reflection formula and the Landen trilogarithm identity. However, I do not think such a functional equation exists for the quadrilogarithm, hence I doubt there is a closed form for $\Re{\rm Li}_4(1+i)$ unless one resorts to, say, hypergeometric functions. Then again, I could jolly well be wrong though. $\endgroup$ – M.N.C.E. Oct 22 '14 at 12:28
  • $\begingroup$ @M.N.C.E. It's easy to determine $\Re\operatorname{Li}_3(1\pm i)$? Since I thought it comes just from the result by Lucian in the linked question. Is there any other way to determine it? I really want to see this proof. $\endgroup$ – user153012 Oct 22 '14 at 14:42
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    $\begingroup$ @user153012 Set $z=i$ in \begin{align}\small{{\rm Li}_3(z)+{\rm Li}_3(1-z)+{\rm Li}_3\left(\frac{z}{z-1}\right)=\frac{1}{6}\ln^3(1-z)-\frac{1}{2}\ln{z}\ln^2(1-z)+\frac{\pi^2}{6}\ln(1-z)+\zeta(3)}\end{align} and $z=1\pm i$ in \begin{align} \small{{\rm Li}_3(z)={\rm Li}_3\left(\frac{1}{z}\right)-\frac{1}{6}\ln^3(-z)-\frac{\pi^2}{6}\ln(-z)} \end{align} then equate both formulae and extract the real part. It should work out ideally if I am not wrong. $\endgroup$ – M.N.C.E. Oct 22 '14 at 14:52
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    $\begingroup$ @user153012 and M.N.C.E. A closed form exists for $\operatorname{Re} \operatorname{Li}_4 (1 + i)$! It is $$\operatorname{Re} \operatorname{Li}_4 (1 + i) = -\frac{5}{16} \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{97}{9216} \pi^4 + \frac{\pi^2}{48} \ln^2 2 - \frac{5}{384} \ln^4 2.$$ For a proof of this result, see here. $\endgroup$ – omegadot Jul 13 at 23:28
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Here is a simple and a nice way to evaluate the first and the second integral.

Evaluation of $1^{\mbox{st}}$ Integral :

Making substitution $x=\tan\theta\,$ followed by integration by parts, we get \begin{align} \int_0^1\left(\frac{\arctan x}{x}\right)^2\,dx&=\color{red}{\int_0^{\Large\frac{\pi}{4}}\frac{\theta^2}{\sin^2\theta}\,d\theta}\\ &=-\theta^2\cot\theta\bigg|_0^{\Large\frac{\pi}{4}}+2\int_0^{\Large\frac{\pi}{4}}\theta\cot\theta\,d\theta\tag{1} \\ &=-\frac{\pi^2}{16}+2\theta\ln(\sin\theta)\bigg|_0^{\Large\frac{\pi}{4}}-2\int_0^{\Large\frac{\pi}{4}}\ln(\sin\theta)\,d\theta\tag{2}\\ &=-\frac{\pi^2}{16}-\frac{\pi}{4}\ln2+G+\frac{\pi}{2}\ln2\tag{3}\\ &=\color{red}{G-\frac{\pi^2}{16}+\frac{\pi}{4}\ln2} \end{align}


Evaluation of $2^{\mbox{nd}}$ Integral :

Again making substitution $x=\tan\theta\,$ followed by integration by parts, we get \begin{align} \int_0^1\left(\frac{\arctan x}{x}\right)^3\,dx&=\int_0^{\Large\frac{\pi}{4}}\frac{\theta^3\cos\theta}{\sin^3\theta}\,d\theta\\ &=-\left.\frac{\theta^3}{2\sin^2\theta}\right|_0^{\Large\frac{\pi}{4}}+\frac{3}{2}\color{red}{\int_0^{\Large\frac{\pi}{4}}\frac{\theta^2}{\sin^2\theta}\,d\theta}\tag{4}\\ &=-\frac{\pi^3}{64}++\frac{3}{2}\left[\color{red}{G-\frac{\pi^2}{16}+\frac{\pi}{4}\ln2}\right]\\ &=\color{blue}{\frac{3G}{2}-\frac{\pi^3}{64}-\frac{3\pi^2}{32}+\frac{3\pi}{8}\ln2} \end{align}

Explanation :

$(1)$ Integration by parts, $u=\theta^2\,\mbox{ and }\,dv=\dfrac{d\theta}{\sin^2\theta}$

$(2)$ Integration by parts, $u=\theta\,\mbox{ and }\,dv=\cot\theta\,d\theta$

$(3)$ The evaluation of $\displaystyle\int_0^{\Large\frac{\pi}{4}}\ln(\sin\theta)\,d\theta$. See Mr. Tunk-Fey's answer, his answer is the best!

$(4)$ Integration by parts, $u=\theta^3\,\mbox{ and }\,dv=\dfrac{\cos\theta}{\sin^3\theta}\,d\theta$

Done! $\,$ (>‿◠)✌

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  • $\begingroup$ No offense, but this seems strikingly similar to my answer... $\endgroup$ – M.N.C.E. Oct 21 '14 at 16:28
  • $\begingroup$ @M.N.C.E. Yes, it's almost similar but I didn't copy your answer & I didn't see it showed up because I was typing my answer. The most important, you tackled this integral $\displaystyle\int_0^{\Large\frac{\pi}{4}}\ln(\sin\theta)\,d\theta$ using Fourier series, but mine used Mr. Tunk-Fey's brilliant approach (❛‿❛✿̶̥̥) $\endgroup$ – Anastasiya-Romanova 秀 Oct 21 '14 at 16:33
  • $\begingroup$ I see... I certainly had no derogatory connotation and I hope you did not take offense to my comment. $\endgroup$ – M.N.C.E. Oct 21 '14 at 16:36
  • $\begingroup$ @M.N.C.E. It's okay. I'd not take that as an offensive comment. See my emo, it shows how I feel when replying people's comment or chat (ɔ◔‿◔)ɔ ♥ $\endgroup$ – Anastasiya-Romanova 秀 Oct 21 '14 at 16:42
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    $\begingroup$ Nice answer. +1. $\endgroup$ – user153012 Oct 21 '14 at 18:38
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The first integral is not that difficult to evaluate. Note that

$$\begin{align}\int_0^1 dx \frac{\arctan^2{x}}{x^2} &= \int_0^{\infty} dx \frac{\arctan^2{x}}{x^2} - \int_1^{\infty} dx \frac{\arctan^2{x}}{x^2}\\ &= \int_0^{\infty} dx \frac{\arctan^2{x}}{x^2} - \int_0^1 dx \left ( \frac{\pi}{2} - \arctan{x} \right )^2\end{align}$$

The first integral may be evaluated by a simple substitution $x=\tan{u}$ to get

$$\int_0^{\infty} dx \frac{\arctan^2{x}}{x^2} = \int_0^{\pi/2} du \frac{u^2}{\sin^2{u}} $$

The latter integral is equal to $\pi \log{2}$; the derivation of this result may be found here.

The second integral is evaluated by expansion and repeated integration by parts, as follows:

$$\begin{align} \int_0^1 dx \left ( \frac{\pi}{2} - \arctan{x} \right )^2 &= \frac{\pi^2}{4} - \pi \int_0^1 dx \, \arctan{x} + \int_0^1 dx \, \arctan^2{x}\end{align} $$

The first integral on the RHS is

$$ \begin{align}\int_0^1 dx \, \arctan{x} &= \left [ x \arctan{x} \right ]_0^1 - \int_0^1 dx \frac{x}{1+x^2}\\ &= \frac{\pi}{4} - \frac12 \log{2} \end{align} $$

The second integral is a little more involved, but it is along similar lines:

$$\begin{align} \int_0^1 dx \, \arctan^2{x} &= \left [ x \arctan^2{x} \right ]_0^1 - 2 \int_0^1 dx \frac{x}{1+x^2} \arctan{x} \\ &= \frac{\pi^2}{16} - \left [\log{(1+x^2)} \arctan{x} \right ]_0^1 + \int_0^1 dx \frac{\log{(1+x^2)}}{1+x^2} \end{align} $$

The latter integral on the RHS may be evaluated by recognizing that

$$\begin{align} \int_0^{\infty} dx \frac{\log{(1+x^2)}}{1+x^2} &= \int_0^1 dx \frac{\log{(1+x^2)}}{1+x^2} + \int_1^{\infty} dx \frac{\log{(1+x^2)}}{1+x^2} \\ &= \int_0^1 dx \frac{\log{(1+x^2)}}{1+x^2} + \int_0^1 dx \frac{\log{(1+x^2)}- 2 \log{x}}{1+x^2}\end{align} $$

The latter integrals are obtained through a mapping $x \mapsto 1/x$ in the previous integral. Thus,

$$\begin{align}\int_0^1 dx \frac{\log{(1+x^2)}}{1+x^2} &= \frac12 \int_0^{\infty} dx \frac{\log{(1+x^2)}}{1+x^2} + \int_0^1 dx \frac{\log{x}}{1+x^2} \\ &= - \int_0^{\pi/2} du \, \log{\cos{u}} - G \\ &= \frac{\pi}{2} \log{2} - G \end{align} $$

where $G$ is Catalan's constant. The source of the first integral on the RHS is the same as that found in the above link (here).

Putting this all together, we get that

$$\begin{align}\int_0^1 dx \frac{\arctan^2{x}}{x^2} &= \pi \log{2} - \frac{\pi^2}{4} + \frac{\pi^2}{4} - \frac{\pi}{2} \log{2} - \frac{\pi^2}{16} + \frac{\pi}{4} \log{2} - \frac{\pi}{2} \log{2} + G \\ &= G + \frac{\pi}{4} \log{2} - \frac{\pi^2}{16} \end{align}$$

which matches up with other people's assertions.

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How to calculate the value of the integrals?

Using Wolfram|Alpha Pro, one may obtain $$\int_0^1\left(\frac{\arctan x}{x}\right)^2\,dx=G-\frac{\pi^2}{16}+\frac{\pi}{4}\ln2$$ and $$\int_0^1\left(\frac{\arctan x}{x}\right)^3\,dx=\frac{3G}{2}-\frac{\pi^3}{64}-\frac{3\pi^2}{32}+\frac{3\pi}{8}\ln2$$ Sorry for the Cleo's style answers, but the answer style is just like the OP style.

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  • $\begingroup$ see my comments $\endgroup$ – Dr. Sonnhard Graubner Oct 21 '14 at 10:40
  • $\begingroup$ and now $\int_{0}^{1}\left(\frac{\arctan(x)}{x}\right)^4dx$ $\endgroup$ – Dr. Sonnhard Graubner Oct 21 '14 at 10:41
  • $\begingroup$ I just don't get it users here. How come your Cleo's style answer gets more upvotes than our answers? (╥︣﹏᷅╥) $\endgroup$ – Anastasiya-Romanova 秀 Oct 21 '14 at 17:08
  • $\begingroup$ @Anastasiya-Romanova Usually, the first person to answer receives the majority of the upvotes, though that is not always the case. $\endgroup$ – Gahawar Oct 21 '14 at 17:19
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    $\begingroup$ @Dr.SonnhardGraubner $$\int_{0}^{1}\left(\frac{\arctan(x)}{x}\right)^4dx$$ equals to $$2G-\frac{3\pi^4}{256}-\frac{\pi^3}{48}-\frac{\pi^2}{8}-\frac{\pi^2G}{8}+\frac{3\pi}{64}\zeta(3)-\frac{\pi^3}{96} \ln2+\frac{\pi}{2} \ln2+\frac{1}{768}\psi_3\left(\frac{1}{4}\right).$$ $\endgroup$ – user153012 Oct 21 '14 at 18:36

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