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This question arose in connection with A hard integral identity on MATH.SE.

Let $$f(x)=\arctan{\left (\frac{S(x)}{\pi+S(x)}\right)}$$ with $S(x)=\operatorname{arctanh} x -\arctan x$, and let $$f(x)=\sum_{n=0}^\infty a_nx^n=\frac{2}{3\pi}x^3-\frac{4}{9\pi^2}x^6+\frac{2}{7\pi}x^7+\frac{16}{81\pi^3}x^9-\frac{8}{21\pi^2}x^{10}+\ldots$$ be its Taylor series expansion at $x=0$. Some numerical evidence suggests the following interesting properties of the $a_n$ coefficients ($b_n$, $c_n$, $d_n$, $\tilde{c}_n$, $\tilde{d}_n$ are some positive rational numbers, $k>0$ is an integer):

1) $a_n=0$, for $n=4k$.

2) $a_n=\frac{2/n}{\pi}-\frac{b_n}{\pi^5}+(\text{maybe other terms of higher order in} 1/\pi)$, for $n=4k+3$.

3) $a_n=-\frac{c_n}{\pi^2}+\frac{d_n}{\pi^6}+(\text{maybe other terms of higher order in} 1/\pi)$, for $n=4k+2$.

4) $a_n=\frac{\tilde{c}_n}{\pi^3}-\frac{\tilde{d}_n}{\pi^7}+(\text{maybe other terms of higher order in } 1/\pi)$, for $n=4k+1$, $k>1$.

How can these properties (if correct) be proved?

P.S. We have $$\arctan{\left(1+\frac{2S}{\pi}\right)}-\frac{\pi}{4}=\arctan{\left(\frac{1+2S/\pi-1}{1+(1+2S/\pi)}\right)}=\arctan{\left(\frac{S}{\pi+S}\right)} .$$ Using $$\arctan(1+x)=\frac{\pi}{4}+\frac{1}{2}x-\frac{1}{4}x^2+\frac{1}{12}x^3-\frac{1}{40}x^5+\frac{1}{48}x^6-\frac{1}{112}x^7+\ldots$$ we get $$\arctan{\left(\frac{S}{\pi+S}\right)}=\frac{S}{\pi}-\frac{S^2}{\pi^2}+\frac{2S^3}{3\pi^3}-\frac{4S^5}{5\pi^5}+\frac{4S^6}{3\pi^6}-\frac{8S^7}{7\pi^7}+\ldots$$ This proves 2), 3) and 4), because $$S=2\left(\frac{x^3}{3}+\frac{x^7}{7}+\frac{x^{11}}{11}+\ldots\right)=2\sum_{k=0}^\infty \frac{x^{4k+3}}{4k+3} .$$ To prove 1), we need to prove the analogous property for $\arctan(1+x)$ and the proof can be based on the formula $$\frac{d^n}{dx^n}(\arctan x)=\frac{(-1)^{n-1}(n-1)!}{(1+x^2)^{n/2}}\sin{\left (n\,\arcsin{\left(\frac{1}{\sqrt{1+x^2}}\right)}\right )}$$ proved in K. Adegoke and O. Layeni, The Higher Derivatives Of The Inverse Tangent Function and Rapidly Convergent BBP-Type Formulas For Pi, Applied Mathematics E-Notes, 10(2010), 70-75, available at http://www.math.nthu.edu.tw/~amen/2010/090408-2.pdf. This formula enables us to get a closed-form expression $$\arctan{\left(\frac{S}{\pi+S}\right)}=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\,2^{n/2}\,\sin{\left(\frac{n\pi}{4}\right)}\,\frac{S^n}{\pi^n} .$$ So the initial questions are not actual now. However I'm still interested to know whether one can calculate in a closed-form the integral $$\int\limits_0^1 S^n(x)\frac{dx}{x} .$$

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  • $\begingroup$ Did you search OEIS for the numerators and denominators (possibly 4k + i): oeis.org/classic.html $\endgroup$ – joro Jan 21 '14 at 14:07
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    $\begingroup$ Maybe do $$f(x)=\arctan{\left (\frac{S(x)}{z+S(x)}\right)}$$ instead, since you use nothing special about $\pi$. $\endgroup$ – Gerald Edgar Jan 21 '14 at 15:29
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    $\begingroup$ $S(x)$ has a nice Taylor series expansion: $\displaystyle2\cdot\sum_{n=0}^\infty\frac{x^{4n+3}}{4n+3}$ $\endgroup$ – Lucian Jan 21 '14 at 21:08
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$\newcommand{\Catalan}{\operatorname{Catalan}}$

I made an attempt for $\int_0^1 S^2\;dx/x$, but with limited success.

Let $$ q_1 := \frac{1}{16}\left( \ln \left( 1-i \right) {\pi }^{2}+16\,\zeta \left( 3 \right) -4\,i \ln \left( 1+i \right) \pi \,\ln \left( 2 \right) +i{\pi }^{3}+4\,i \ln \left( 1-i \right) \pi \,\ln \left( 2 \right) \\ -2\,\ln \left( 1+ i \right) {\pi }^{2}-4\,i \left( \ln \left( 1-i \right) \right) ^{2} \pi -16\,i\ln \left( 1+i \right) {\it \Catalan}+10\,i \left( \ln \left( 2 \right) \right) ^{2}\pi \\ +8\,\pi \,{\it \Catalan}-2\,{\pi }^{2}\ln \left( 2 \right) +{\pi }^{2}\ln \left( i\sqrt {2}+\sqrt {2}+2 \right) +{\pi }^{2}\ln \left( 2-\sqrt {2}-i\sqrt {2} \right) \\ +20\,\ln \left( 2 \right) {\rm Li}_2 \left(2 \right) -20\,{\rm Li}_3 \left(2 \right) -8\,{\rm Li}_3 \left(-i \right) -8\, {\rm Li}_3 \left(i \right) +16\,i\ln \left( 1-i \right) {\it \Catalan} \\ +4\,i \left( \ln \left( 1+i \right) \right) ^{2}\pi -2\,{ \pi }^{3}\right) \approx −0.4990969 $$ and $$ q_2 := \sum _{m=1}^{\infty }{\frac {\Psi \left( (m+1)/2\right) -\Psi \left(m/2 \right) }{ 2\left( 2\,m-1 \right) ^{2}}} \approx 0.7416483, $$ Then $$ \int_0^1({\rm arctanh}\; x - \arctan x)^2\frac{dx}{x} = q_1+q_2 \approx 0.2425514 $$

added

Combining the above with Eckhard's alternate version, we get the interesting equation $$ q_2 = \frac{\left( \ln \left( 2 \right) \right) ^{2}\pi}{16} +{\frac {5}{64}}\,{\pi }^{3} -{ \Catalan}\,\ln \left( 2 \right) -2 \,{\rm Im} \; {\rm Li_3} \left( \frac{1+i}{2} \right) $$

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  • $\begingroup$ I checked by numerical integration that the final answer is correct. Unfortunately, it seems these integrals do not have simple closed-form answers. $\endgroup$ – Zurab Silagadze Jan 23 '14 at 8:06
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    $\begingroup$ I obtained the slightly easier expression $\frac{1}{192} \left(96 \operatorname{Catalan} (\pi -\log{4})-2 \mathrm{i} \left(-192 \operatorname{Li}_3\left(\frac{1}{2}+\frac{\mathrm{i}}{2}\right)+105 \zeta (3)+4 \log ^3{2}\right)+3 \pi ^3+12 \pi \log ^2{2}+10 \mathrm{i} \pi ^2 \log{2}\right)$ for $\int_0^1{S(x)^2/x\mathrm{d}x}$. $\endgroup$ – Eckhard Jan 23 '14 at 12:18
  • $\begingroup$ The inverse symbolic calculator (and the fact that $\int_0^1{S^2 dx/x}$) is real suggest that the real part of $\operatorname{Li}_3(1/2+i/2)$ is equal to $\frac{1}{192} \left(105 \zeta(3)+4 \log ^3{2}-5 \pi ^2 \log {2}\right)$. If a similar formula could be found for the imaginary part, the expression for the integral might simplify even further. Unfortunately, the ISC doesn't find such a formula. $\endgroup$ – Eckhard Jan 23 '14 at 12:52
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Here is a possible approach toward proving (1). Let $i=\sqrt{-1}$. Let $g(x)=f(x) +f(ix) +f(-x)+f(-ix)$. We need to show that $g(x)=0$. Unfortunately, Maple is unable to do this directly. Thus write each $\arctan u$ as $\frac i2\log\frac{1-iu}{1+iu}$ and $\mathrm{arctanh}\,u$ as $\frac 12\log\frac{1+u}{1-u}$ and simplify. Every time you see a $\log \frac ab$, replace with $\log a-\log b$. Perhaps the 16 terms will cancel out in pairs. I could not get Maple to do this. You also might need to replace expressions like $\log(-i(1-x))$ with $\log(-i)+\log(1-x)$. Perhaps Mathematica will be smarter than Maple in showing $g(x)=0$.

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    $\begingroup$ Maple 16 has no trouble simplifying both $g(0)$ and $g'(x)$ to 0. I guess that is enough. $\endgroup$ – Brendan McKay Jan 24 '14 at 1:22
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Here is a simple dérivation of the Taylor expansion of $~\arctan(\frac{x}{1+x})~:$
Since $$~\arctan(u)=\dfrac{i}{2}\ln\big(\dfrac{1-iu}{1+iu}\big),~~\arctan(\frac{x}{1+x})=\dfrac{i}{2}\ln\Big(\dfrac{1-i\frac{x}{1+x}}{1+i\frac{x}{1+x}}\Big)$$ $$= \dfrac{i}{2}\ln\Big(\dfrac{1+(1-i)x}{1+(1+i)x}\Big)=\dfrac{i}{2}\ln\Big(\dfrac{1+\sqrt{2}\exp(-i\frac{\pi}{4}) x}{1+\sqrt{2}\exp(i\frac{\pi}{4})x}\Big).~$$ So, for $~|x|<\frac{1}{\sqrt{2}},$ $$\arctan(\frac{x}{1+x})=\dfrac{i}{2}\sum\limits_{n=1}^{\infty}\frac{(-1)^{n-1}2^{n/2}}{n}(e^{-in\frac{\pi}{4}} -e^{in\frac{\pi}{4}})x^n$$ $$=\sum\limits_{n=1}^{\infty}\frac{(-1)^{n-1}2^{n/2}}{n}\sin(n\frac{\pi}{4})x^n,$$ and, for $~|w|<\frac{\pi}{\sqrt{2}},$ $$(1)~~~~~~\arctan(\frac{w}{\pi+w})=\sum\limits_{n=1}^{\infty}\frac{(-1)^{n-1}2^{n/2}}{n}\sin(n\frac{\pi}{4})(\frac{w}{\pi})^n.$$ But I'm a bit worried about the following detail : if $$~S(x)=2\sum\limits_{n=0}^{\infty}\dfrac{x^{4n+3}}{4n+3}, ~~~~~~~~~~~~\lim\limits_{x\rightarrow1^{-}}S(x)=+\infty,~$$ so it seems a bit dangerous to substitute $~S(x)~$ for $~w~$ into $~(1)~$ (since this series diverges for $~|w|>\frac{\pi}{\sqrt{2}})$ (and even more dangerous to interchange summation and integral...)

Am I wrong ?

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  • $\begingroup$ It seems you are right. $\endgroup$ – Zurab Silagadze Feb 17 '14 at 9:48

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