# Integral ∫∞1arccot(1+2πarcothx−arccscx)√x2−1dx\int_1^\infty\frac{\operatorname{arccot}\left(1+\frac{2\pi}{\operatorname{arcoth}x-\operatorname{arccsc}x}\right)}{\sqrt{x^2-1}}\mathrm dx

Consider the following integral:
$$I=∫∞1arccot(1+2πarcothx−arccscx)√x2−1dx,\mathcal{I}=\int_1^\infty\frac{\operatorname{arccot}\left(1+\frac{2\,\pi}{\operatorname{arcoth}x\,-\,\operatorname{arccsc}x}\right)}{\sqrt{x^2-1}}\mathrm dx,$$
where $$arccsc\operatorname{arccsc}$$ is the inverse cosecant, $$arccot\operatorname{arccot}$$ is the inverse cotangent and $$arcothx\operatorname{arcoth}x$$ is the inverse hyperbolic cotangent.

Approximate numerical integration suggests a possible closed form:
$$I?=πlnπ4−3πln28.\mathcal{I}\stackrel?=\frac{\pi\,\ln\pi}4-\frac{3\,\pi\,\ln2}8.$$
I was not able to rigorously establish the equality, but the value is correct up to at least $$900900$$ decimal digits.

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

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

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

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

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:

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)