Consider the following integral: $$\mathcal{I}=\int_1^\infty\frac{\operatorname{arccot}\left(1+\frac{2\,\pi}{\operatorname{arcoth}x\,-\,\operatorname{arccsc}x}\right)}{\sqrt{x^2-1}}\mathrm dx,$$ where $\operatorname{arccsc}$ is the inverse cosecant, $\operatorname{arccot}$ is the inverse contangent 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 '13 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 '13 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? $\endgroup$
    – Pixel
    Dec 17 '13 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 '14 at 2:40
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    $\begingroup$ 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 '14 at 1:10

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 '14 at 21:12
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    $\begingroup$ It looks a lot like the troublesome: math.stackexchange.com/questions/464769/… $\endgroup$ Jan 10 '14 at 19:50
  • $\begingroup$ I was thinking about differentiation under the integral sign ? Not sure how though. $\endgroup$
    – mick
    Jan 16 '14 at 0:42

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