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+2πarcothxarccscx)x21dx,
where arccsc is the inverse cosecant, arccot is the inverse cotangent and arcothx is the inverse hyperbolic cotangent.

Approximate numerical integration suggests a possible closed form:
I?=πlnπ43πln28.
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 I?

Answer

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 dxx21=ln(2(x+x21)).

So by using integration by parts we reduce the problem to solving A=1D[arccot(1+2πarcothxarccscx)]ln(2(x+x21))dx.

We know that D[arccot(x)]=11+x2 hence by the chain rule for derivatives we get

A=1D[(2πarcothxarccscx)]ln(2(x+x21))1+(1+2πarcoth(x)arccsc(x))2dx.

Because we know that D[arccot(x)]=11+x2 and also that ln(2z)=ln(2)+ln(z) we can use integration by parts again and arrive at

B=1D[(2πarcothxarccscx)]ln(x+x21)1+(1+2πarcoth(x)arccsc(x))2dx.

Now we can turn this into an infinite sum because we set U=2πarcoth(x)arccsc(x) and use the Taylor expansion for z1+(1+z)2.

In other words we substitute z=U.
( Remember that Σ=Σ )

Finally the core problem is reduced mainly to solving:

Cn=1ln(x+x21)(arcoth(x)arccsc(x))ndx.

by induction we get the need to solve for C1 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 C1 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)

Attribution
Source : Link , Question Author : Vladimir Reshetnikov , Answer Author : mick

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