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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    $\begingroup$ To clarify is it $\left(1+\frac{2\,\pi}{\operatorname{\color{red}{arccot}}x\,-\,\operatorname{arccsc}x}\right)$ inside the integral? You have written, $\left(1+\frac{2\,\pi}{\operatorname{\color{red}{arcoth}}x\,-\,\operatorname{arccsc}x}\right)$ $\endgroup$
    – user112535
    Dec 11, 2013 at 12:33
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    $\begingroup$ As I mentioned in the question, $\operatorname{arcoth}x=\tfrac12\ln\left(\frac{x+1}{x-1}\right)$ is the inverse hyperbolic cotangent, sometimes also denoted as $\operatorname{arccoth}x,\,\operatorname{arcth}x$ or $\operatorname{coth}^{-1}x$. $\endgroup$ Dec 11, 2013 at 20:25
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    $\begingroup$ Out of interest, what method of numerical approximation did you use to reach such a precise (and possibly correct) closed form? $\endgroup$
    – pshmath0
    Dec 17, 2013 at 20:06
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    $\begingroup$ @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 $\endgroup$ Jan 11, 2014 at 2:40
  • 1
    $\begingroup$ Perhaps you are already aware of this, but that other famous and similar-looking integral has been solved here only recently. $\endgroup$
    – Lucian
    Feb 19, 2014 at 1:10

4 Answers 4


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)


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 $$ \int^{1}_0\frac{-\pi\log zS'(z)}{\pi^2+(\pi+S(z))^2}dz=\int^{1}_0-\log z\,d\left(\operatorname{arctan}\left(1+\frac{S(z)}{\pi}\right)\right)=\int^{+\infty}_0\frac{-\log S^{-1}(w)}{\pi^2+(\pi+w)^2}dw. $$

  • $\begingroup$ I appreciate your effort, but I wonder what exactly Mathematica did ... $\endgroup$
    – mick
    Jan 8, 2014 at 21:12
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    $\begingroup$ It looks a lot like the troublesome: math.stackexchange.com/questions/464769/… $\endgroup$ Jan 10, 2014 at 19:50
  • $\begingroup$ I was thinking about differentiation under the integral sign ? Not sure how though. $\endgroup$
    – mick
    Jan 16, 2014 at 0:42

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

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

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

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

  • $\begingroup$ What does IBP stand for ? $\endgroup$
    – mick
    Feb 1, 2023 at 21:02
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    $\begingroup$ @mick integration by parts. $\endgroup$
    – KStarGamer
    Feb 1, 2023 at 21:42

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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