5
$\begingroup$

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 $$\int\limits_0^1 S^n(x)\frac{dx}{x} .$$

$\endgroup$
3
  • $\begingroup$ Did you search OEIS for the numerators and denominators (possibly 4k + i): oeis.org/classic.html $\endgroup$
    – joro
    Jan 21, 2014 at 14:07
  • 2
    $\begingroup$ Maybe do $$f(x)=\arctan{\left (\frac{S(x)}{z+S(x)}\right)}$$ instead, since you use nothing special about $\pi$. $\endgroup$ Jan 21, 2014 at 15:29
  • 1
    $\begingroup$ $S(x)$ has a nice Taylor series expansion: $\displaystyle2\cdot\sum_{n=0}^\infty\frac{x^{4n+3}}{4n+3}$ $\endgroup$
    – Lucian
    Jan 21, 2014 at 21:08

3 Answers 3

3
$\begingroup$

$\newcommand{\Catalan}{\operatorname{Catalan}}$

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 $$ q_2 = \frac{\left( \ln \left( 2 \right) \right) ^{2}\pi}{16} +{\frac {5}{64}}\,{\pi }^{3} -{ \Catalan}\,\ln \left( 2 \right) -2 \,{\rm Im} \; {\rm Li_3} \left( \frac{1+i}{2} \right) $$

$\endgroup$
3
  • $\begingroup$ I checked by numerical integration that the final answer is correct. Unfortunately, it seems these integrals do not have simple closed-form answers. $\endgroup$ Jan 23, 2014 at 8:06
  • 1
    $\begingroup$ 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, 2014 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, 2014 at 12:52
1
$\begingroup$

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 $g(x)=0$.

$\endgroup$
1
  • 1
    $\begingroup$ Maple 16 has no trouble simplifying both $g(0)$ and $g'(x)$ to 0. I guess that is enough. $\endgroup$ Jan 24, 2014 at 1:22
1
$\begingroup$

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 ?

$\endgroup$
1
  • $\begingroup$ It seems you are right. $\endgroup$ Feb 17, 2014 at 9:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.