Consider the following integral: $$\mathcal{I}=\int_1^\infty\operatorname{arccot}\left(1+\frac{2\,\pi}{\operatorname{arcoth}x\,-\,\operatorname{arccsc}x}\right)\frac{\mathrm dx}{\sqrt{x^2-1}}\,,$$ where $\operatorname{arccsc}$ is the inverse cosecant, $\operatorname{arccot}$ is the inverse cotangent and $\operatorname{arcoth}x$ is the inverse hyperbolic cotangent.

Approximate numerical integration suggests a possible closed form: $$\mathcal{I}\stackrel?=\frac{\pi\,\ln\pi}4-\frac{3\,\pi\,\ln2}8.$$ I was not able to rigorously establish the equality, but the value is correct up to at least $900$ decimal digits.

Is it the correct exact value of the integral $\,\mathcal{I}$?

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Vladimir Reshetnikov
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  • Do you plan on putting it up for bounty ? – Lucian Dec 10 '13 at 05:12
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    To clarify is it $\left(1+\frac{2\,\pi}{\operatorname{\color{red}{arccot}}x\,-\,\operatorname{arccsc}x}\right)$ inside the integral? You have written, $\left(1+\frac{2\,\pi}{\operatorname{\color{red}{arcoth}}x\,-\,\operatorname{arccsc}x}\right)$ –  Dec 11 '13 at 12:33
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    As I mentioned in the question, $\operatorname{arcoth}x=\tfrac12\ln\left(\frac{x+1}{x-1}\right)$ is the [inverse hyperbolic cotangent](http://en.wikipedia.org/wiki/Inverse_hyperbolic_cotangent#Logarithmic_representation), sometimes also denoted as $\operatorname{arccoth}x,\,\operatorname{arcth}x$ or $\operatorname{coth}^{-1}x$. – Vladimir Reshetnikov Dec 11 '13 at 20:25
  • I recommend the substitution u = 1/x. After that try integration by parts ( do not touch du ). And after that simplify the expression. I think that helps... Maybe then quadrature ? Or meijer G perhaps. My 50 cents. – mick Dec 15 '13 at 22:57
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    Out of interest, what method of numerical approximation did you use to reach such a precise (and possibly correct) closed form? – gone Dec 17 '13 at 20:06
  • 2 remarks : 1) Where did you get this integral from Vladimir ?? Well assuming it is not just an integral made up to be hard. Maybe it relates to physics ? 2) I notice that exp(8I/pi) is a fraction of zeta(2) , maybe that can help to find the solution !? – mick Jan 07 '14 at 12:09
  • Is this one a reformulation of http://math.stackexchange.com/questions/464769/how-to-prove-int-01-tan-1-left-frac-tanh-1x-tan-1x-pi-tanh-1#comment1200602_464769 ? – Jack D'Aurizio Jan 10 '14 at 19:50
  • @JackD'Aurizio by some substitution ?? – mick Jan 10 '14 at 21:06
  • Im I the only who is considering trig identities to solve this ... – mick Jan 10 '14 at 21:07
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    @JackD'Aurizio Yes, basically it is. I knew about that integral before posting this (through a personal communication), but I did not know it had been posted on M.SE – Vladimir Reshetnikov Jan 11 '14 at 02:40
  • I think it would help (a lot) if you told us as much as you could about the source of this integral. More often than not, as you know, definite integrals become harder when one specializes various parameters to special values, and it's usually easier to prove the more "general" formulation. So either: 1. You know of a number of similar integrals, and thought to guess that this one would have a similar shape. (In which case, it would be useful to see those integrals). 2. You were given the integral (or an equivalent one) by someone else, in which case you don't have that information. –  Jan 21 '14 at 09:18
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    Perhaps you are already aware of this, but that other famous and similar-looking integral has been solved [here](http://arxiv.org/pdf/1402.3830v1.pdf) only recently. – Lucian Feb 19 '14 at 01:10
  • I do not understand. When $x \to +\infty$, arccsc$(x) \to \pi/2$, arcoth$(x) \to 0$, the integrand is equivalent to arccot(-3)/$x$ and the integral DIVERGES! Am I wrong? – Christophe Leuridan May 10 '22 at 11:29

4 Answers4


I will attempt to simplify the integral. It might take me a while and I might edit this answer a few times, so this is not its final form yet.

Let $D$ be the differential operator.

We know $\displaystyle\int \dfrac{dx}{\sqrt{x^2-1}}=\ln(2(x+\sqrt{x^2-1}))$.

So by using integration by parts we reduce the problem to solving $$\displaystyle\mathcal{A}=\int_1^\infty D\left[\operatorname{arccot}\left(1+\frac{2\,\pi}{\operatorname{arcoth}x\,-\,\operatorname{arccsc}x}\right)\right] \ln(2(x+\sqrt{x^2-1}))\, dx.$$

We know that $D[\mathrm{arccot}(x)] = \dfrac{-1}{1+x^2} $ hence by the chain rule for derivatives we get

$$\mathcal{A}=\int_1^\infty \dfrac{D\left[\left(\frac{-2\,\pi}{\operatorname{arcoth}x\,-\,\operatorname{arccsc}x}\right)\right] \ln(2(x+\sqrt{x^2-1}))}{1+\left(1+\dfrac{2\, \pi}{\mathrm{arcoth}(x) - \mathrm{arccsc}(x)}\right)^2}\,dx.$$

Because we know that $D[\mathrm{arccot}(x)] = \dfrac{-1}{1+x^2} $ and also that $ln(2z)=ln(2)+ln(z)$ we can use integration by parts again and arrive at

$$\mathcal{B}=\int_1^\infty \dfrac{D\left[\left(\frac{-2\,\pi}{\operatorname{arcoth}x\,-\,\operatorname{arccsc}x}\right)\right] \ln(x+\sqrt{x^2-1})}{1+\left(1+\dfrac{2\, \pi}{\mathrm{arcoth}(x) - \mathrm{arccsc}(x)}\right)^2}\,dx.$$

Now we can turn this into an infinite sum because we set $U=\dfrac{2\, \pi}{\mathrm{arcoth}(x) - \mathrm{arccsc}(x)}$ and use the Taylor expansion for $\frac{z}{1+(1+z)^2}$.

In other words we substitute $z=U$. ( Remember that $\int \Sigma = \Sigma \int$ )

Finally the core problem is reduced mainly to solving:

$$\displaystyle\mathcal{C_n}=\int_1^\infty \frac{\ln(x+\sqrt{x^2-1})}{(\mathrm{arcoth}(x)-\mathrm{arccsc}(x))^n}\, dx.$$

by induction we get the need to solve for $C_1$ and then are able to get the others.

At this point, I must admit that I have ignored convergeance issues. Those issues can be solved by taking limits.

For instance $C_1$ does not actually converge by itself.

For all clarity the problem is not resolved.

In fact it might require a new bounty. Still thinking about it ...

(in progress)

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By taking the substitution $u=1/x$ and $\theta=\operatorname{arccos}u$, the integral is transformed into: $$ \int^{\pi/2}_{0}\sec\theta\operatorname{arccot}\left(1+\frac{2\pi}{\log\cot(\theta/2)-\pi/2+\theta}\right)\,d\theta. $$

and then integrate by parts to get:

$$ -\int^{\pi/2}_{0}\log(\tan\theta+\sec\theta)\frac{d}{d\theta}\left(\operatorname{arccot}\left(1+\frac{2\pi}{\log\cot(\theta/2)-\pi/2+\theta}\right)\right)\,d\theta $$

which further simplifies to (according to Mathematica)

$$ \int^{\pi/2}_{0}\frac{4\pi(\csc\theta-1)\log\left(\frac{\cot(\theta/2)+1}{\cot(\theta/2)-1}\right)}{4\theta^2+4\theta\pi+5\pi^2+4\log\cot(\theta/2)(2\theta+\pi+\log\cot(\theta/2))}d\theta $$

We now substitute $y=\cot(\theta/2)$ to get:

$$ \int^{\infty}_{1}\frac{4\pi(y-1)^2\log\left(\frac{y+1}{y-1}\right)}{y(y^2+1)\left(4\pi^2+(\pi+2\log y+4\operatorname{arccot}y)^2\right)}\,dy $$

After another substitution $z=(y-1)/(y+1)$, we are able to simplify it further:

$$ 4\pi\int^{1}_{0}\frac{z^2\log z}{(z^4-1)\left(\pi^2+(\log\frac{1+z}{1-z}+2\operatorname{arccot}z)^2\right)}\,dz $$

Therefore, the task now is to prove that $$ I_0=4\pi\int^{1}_{0}\frac{z^2\log z}{(z^4-1)\left(\pi^2+(\log\frac{1+z}{1-z}+2\operatorname{arccot}z)^2\right)}\,dz\stackrel?=\frac{\pi}{8}\log\frac{\pi^2}{8}. $$

Edit: Let $S(z)=2\operatorname{arctanh}z+2\operatorname{arccot}z-\pi=4\sum^{\infty}_{j=1}\frac{z^{4j-1}}{4j-1}$ so that $S$ maps $[0,1)$ to $[0,+\infty)$, and $S'(z)=\frac{4z^2}{1-z^4}$. The integral $I_0$ becomes $$ \int^{1}_0\frac{-\pi\log zS'(z)}{\pi^2+(\pi+S(z))^2}dz=\int^{1}_0-\log z\,d\left(\operatorname{arctan}\left(1+\frac{S(z)}{\pi}\right)\right)=\int^{+\infty}_0\frac{-\log S^{-1}(w)}{\pi^2+(\pi+w)^2}dw. $$

Chen Wang
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  • I appreciate your effort, but I wonder what exactly Mathematica did ... – mick Jan 08 '14 at 21:12
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    It looks a lot like the troublesome: http://math.stackexchange.com/questions/464769/how-to-prove-int-01-tan-1-left-frac-tanh-1x-tan-1x-pi-tanh-1#comment1200602_464769 – Jack D'Aurizio Jan 10 '14 at 19:50
  • I was thinking about differentiation under the integral sign ? Not sure how though. – mick Jan 16 '14 at 00:42

I shall transform the given integral into the previously-proven identity: $$\int_0^1\frac{\operatorname{arccot}\left(1+\frac\pi{\operatorname{arctanh}x-\arctan x}\right)}x\,dx=\frac{\pi\ln\pi}4-\frac{3\pi\ln2}8$$

As was shown by @Chen Wang, the given integral is equal to $$\begin{align}\mathcal{I}&=4\pi \int_{0}^{1} \frac{z^2 \ln (z)}{(z^4-1)(\pi^2+(2\operatorname{arctanh}(z)+2\operatorname{arccot}(z))^2)} \, dz \\ &=\Im \int_{0}^{1} \frac{4z^2\ln (z)}{(z^4-1)(2\operatorname{arctanh} (z)+2\operatorname{arccot} (z)-i\pi)} \, dz \\ &\stackrel{\text{IBP}}{=}\int_{0}^{1} \frac{1}{z} \Im \left(\ln(2\operatorname{arctanh} (z)+2\operatorname{arccot}(z)-i\pi)+\frac{\pi i}{4}\right) \, dz \end{align}$$

Since $\displaystyle\small\Im \left(\ln(2\operatorname{arctanh} (z)+2\operatorname{arccot}(z)-i\pi)+\frac{\pi i}{4}\right)=\Im \left(\ln\left(1+\frac{1+i}{\pi} (\operatorname{arctanh} (z)-\operatorname{arctan}(z))\right)\right)$, this implies that

$$\mathcal{I}=\int_{0}^{1} \frac{1}{z} \Im \left(\ln\left(1+\frac{1+i}{\pi} (\operatorname{arctanh} (z)-\operatorname{arctan}(z))\right)\right)\, dz$$ which is identically the same as the previously-proven identity, and so the conjectured closed-form is true. $\square$

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This note discusses a possible route to prove that $$I=\int_1^\infty\frac{\operatorname{arccot}\left(1+\frac{2\pi}{\operatorname{arccoth}x-\operatorname{arccsc}x}\right)}{\sqrt{x^2-1}}\,dx=\frac{\pi\log\pi}4-\frac{3\log2}8$$ and is based on the approach of de Reyna (2014) who proved the similar identity $$\int_0^1\frac{\operatorname{arccot}\left(1+\frac\pi{\operatorname{arctanh}x-\arctan x}\right)}x\,dx=\frac{\pi\log\pi}4-\frac{3\log2}8.$$

We first substitute $u=1/x$ to get$$I=\int_0^1\frac{\arctan\frac{\operatorname{arctanh}u-\arcsin u}{\operatorname{arctanh}u-\arcsin u+2\pi}}{u\sqrt{1-u^2}}\,du=\Im\int_0^1\frac{\log\left(1+\frac{1+i}{2\pi}(\operatorname{arctanh}u-\arcsin u)\right)}{u\sqrt{1-u^2}}\,du.$$ As we will find later, the $\log$ term on the numerator explains the appearances of $\log\pi$ and $\log2$ in the conjectured result. Another substitution $u=\sin\theta$ gives us $$I=\Im\int_0^{\pi/2}\log\left(1+\frac{1+i}{2\pi}(\operatorname{arctanh}(\sin\theta)-\theta)\right)\csc\theta\,d\theta$$ where we note that the term inside the logarithm has real part $>1$ since $\operatorname{arctanh}\sin\theta-\theta>0$ on $(0,\pi/2)$; Bagul and Chesneau (2018) have determined optimal bounds for this inequality. Therefore, the only singularities on the first quadrant are at integer and half-integer multiples of $\pi$.

In a similar approach to de Reyna, we take a quarter-circular contour on the first quadrant and by letting the radius $R$ tend to infinity. Define $\gamma_0$ to be the line from $0$ to $\pi/2-\epsilon$, and $\gamma_k$ to be the line from $k\pi/2+\epsilon$ to $(k+1)\pi/2-\epsilon$ for each $1\le k\le\lfloor2R/\pi\rfloor$ as $\epsilon\to0^+$. Consequently, define $E_k$ to be the semicircular arc parametrised by $$p_k=\frac{k\pi}2+\epsilon e^{i\varphi};\quad\varphi\in[0,\pi]$$ taken clockwise to avoid each singularity. Define $C$ to be quarter-circular arc of radius $R$ from $\phi=0$ to $\pi/2$, and finally, define $\gamma_{-1}$ to be the line from $iR$ to $0$ to close the contour. Then by Cauchy's theorem, $$\left(\int_{\gamma_0}+\int_{\bigcup_{k\ge1}\gamma_k}+\int_{\bigcup_{k\ge1}E_k}+\int_C-\int_{\gamma_{-1}}\right)G(z)=0$$ where $$G(z)=\log\left(1+\frac{1+i}{2\pi}(\operatorname{arctanh}(\sin z)-z)\right)\csc z.$$

Conjecture 1. The integrals on $\gamma_0$ and $\gamma_{-1}$ are conjugates. In particular, $$2iI=\int_0^{\pi/2}G(z)\,dz-i\lim_{R\to\infty}\int_0^RG(iz)\,dz.$$ I do not have a proof of this yet, but it is extremely similar to $(3.8)$ of de Reyna, where the author exploited the identity $$\operatorname{arctanh}iz-\arctan iz=-i(\operatorname{arctanh}z-\arctan z)$$ to immediately arrive at the conclusion. In our case we are not so fortunate since $\sin iz=\sinh z$ has no immediate relation to $\tanh z$.

Proposition 2. As $R\to\infty$, $$\int_CG(z)\,dz\sim-\left(\pi+i\log\left(\frac{1+i}{2\pi}R\right)\right)\left(\frac\pi2+i\Re\log\cot\frac R2\right).$$

Proof: Substituting $z=Re^{i\phi}$ gives \begin{align}\int_CG(z)\,dz&=i\int_0^{\pi/2}\frac{Re^{i\phi}}{\sin(Re^{i\phi})}\log\left(1+\frac{1+i}{2\pi}(\operatorname{arctanh}(\sin Re^{i\phi})-Re^{i\phi})\right)\,d\phi\\&\sim i\int_0^{\pi/2}\frac{Re^{i\phi}}{\sin(Re^{i\phi})}\log\left(\frac{3+i}4-\frac{1+i}{2\pi}Re^{i\phi}\right)\,d\phi\end{align} as $\operatorname{arctanh}(\sin Re^{i\phi})\to\pi i/2$. To obtain its asymptotic behaviour, we expand the logarithm as a Taylor series \begin{align}\log\left(\frac{3+i}4-\frac{1+i}{2\pi}Re^{i\phi}\right)&=\log\left(\frac{1+i}{2\pi}Re^{i\phi}\right)-i\pi+\log\left(1-\frac{3+i}4\frac{2\pi}{1+i}\frac1{Re^{i\phi}}\right)\\&=\log\left(\frac{1+i}{2\pi}R\right)-i\pi+i\phi+{\cal O}(R^{-1})\end{align} so that \begin{align}\int_CG(z)\,dz&\sim\left(\pi+i\log\left(\frac{1+i}{2\pi}R\right)\right)\int_0^{\pi/2}\frac{Re^{i\phi}}{\sin(Re^{i\phi})}\,d\phi-i\int_0^{\pi/2}\frac{R\phi e^{i\phi}}{\sin(Re^{i\phi})}\,d\phi.\end{align} The first integral can be computed directly as $i\log\cot(Re^{i\phi}/2)\vert_0^{\pi/2}$ whereas the second integral tends to zero after integrating by parts. $\quad\square$

Proposition 3. As $\epsilon\to0^+$, $\int_{\bigcup_{k\,\text{odd}}E_k}G(z)\,dz\to0$. This accounts for all the singularities from the $\operatorname{arctanh}$ term.

Proof: The steps are similar to the proof of Proposition 2 so only an outline will be sketched. When $k$ is odd, we have $\sin p_k\to\pm1$ (where $\pm$ depends on whether $k\equiv1,3\pmod4$) and $$\operatorname{arctanh}\sin p_k=\pm\operatorname{arctanh}\cos(\epsilon e^{i\varphi})\sim\pm\log\epsilon$$ as $\epsilon\to0^+$. Thus for each odd $k$, \begin{align}\int_{E_k,k\,\text{odd}}G(z)\,dz&\sim\pm\int_0^\pi i\epsilon e^{i\varphi}\log\left(1+\frac{1+i}{2\pi}\left(\pm\log\epsilon-\frac{k\pi}2-\epsilon e^{i\varphi}\right)\right)\,d\varphi\to0\end{align} since the linear term $\epsilon$ dominates the logarithmic term multiplicatively. $\quad\square$

Proposition 4. As $\epsilon\to0^+$, $$\int_{\bigcup_{k\,\text{even}}E_k}G(z)\,dz\to i\pi\sum_{n=1}^{\lfloor R/\pi\rfloor}(-1)^n\log\left(1-\frac{1+i}2n\right).$$ This accounts for all the singularities from the $\csc$ term.

Proof: When $k=2n$ is even, we have \begin{align}\int_{E_k}G(z)\,dz&=\small i\int_0^{\pi}\frac{\epsilon e^{i\varphi}}{\sin(n\pi+\epsilon e^{i\varphi})}\log\left(1+\frac{1+i}{2\pi}(\operatorname{arctanh}(\sin(n\pi+\epsilon e^{i\varphi}))-n\pi-\epsilon e^{i\varphi})\right)\,d\varphi\\&\to i\int_0^{\pi}(-1)^n\log\left(1-\frac{1+i}2n\right)\,d\varphi\end{align} as $\epsilon\to0^+$ since $\operatorname{arctanh}\sin\epsilon-\epsilon\to0$. We immediately obtain the desired result upon noting that there are $\lfloor R/\pi\rfloor$ singularities that are integer multiples of $\pi$ for a fixed contour radius $R$. $\quad\square$

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