Closed-form of $\int_0^{\pi/2} \arctan(x)\cot(x)\,dx$

I’m looking for a closed-form of the following integral problem.

$$I = \int_0^{\pi/2} \arctan(x)\cot(x)\,dx.$$

The numerical approximation of $I$ is

$$I \approx 0.96644524676637380447182915131032699868606574138656587245691342\dots$$

I’ve found nothing with Maple or Mathematica.

Solutions Collecting From Web of "Closed-form of $\int_0^{\pi/2} \arctan(x)\cot(x)\,dx$"

This is not an answer. This is just something to offer some ideas. Actually, this more of a comment on steroids.
$$\int_{0}^{\pi/2} \arctan(x)\cot(x) \text{d}x=I$$

Now, what we do is rewrite the equation in terms of $i$. A nice way to do that is to do two things. We first substitute $x=ia$, then multiply and divide $-i$ to the argument of the integral. (And rewrite $\cot$ as $1/\tan$)

$$\int_{0}^{i\pi/2} \frac{-i\arctan ia}{-i\tan ia} \text{d}a=I$$

Recognizably, we note that we can transform the numerator and denominator to hyperbolic functions. We do so like the following:

$$\int_{0}^{i\pi/2} \frac{\text{artanh}\space a}{\tanh a} \text{d}a=I$$

Now, we use something known as the Gudermannian function. That allows us change between hyperbolic functions and real functions with only having to use the real plane. Assume $a=\text{gd}^{-1}\space b $

$$\int_0^{\log\tan(\frac{\pi}{4}(1+i))} \frac{\text{artanh artanh sin}\space b}{\sin b}\text{d}b=$$

Now substitute $b=\arcsin c$ to get a really disgusting result.

$$\int_0^{\arcsin\log\tan(\frac{\pi}{4}(1+i))} \frac{\text{artanh artanh c}}{c}\text{d}c=$$

You can continue for a few more steps due to the rules behind $\tan$ and some hyperbolic properties. (I’m sorry, I used W|A to check that step.)

$$\int_0^{\arcsin\log(\text{sech}(\pi/2)+i\tanh(\pi/2))} \frac{\text{artanh artanh c}}{c}\text{d}c=$$

Something recognizable begins to occur. Remember, for now, we just want to take down the difficulty of the integral to something simpler. Substitute $c=2di$.
Hmmmm…

$$\int_0^{\arcsin\log(\text{sech}(i\pi)+i\tanh(i\pi))} \frac{\text{artanh artanh 2i}d}{2di}\text{d}d=$$

Simplify:

$$\frac{1}{2}\int_0^{\arcsin(i\pi)} \frac{\text{artanh artanh 2}d}{d}\text{d}d=$$

I want to reiterate a few major points here.

a) This not an answer. This is just a few ideas I had to create a solution all in one huge comment. Maybe some of the ideas will inspire you. I don’t know.

b) This answer is, like I said, not full or correct. If you found a solution by yourself using the above methods, that is fantastic. This is an inspirational answer.

c) I probably messed up somewhere.

d) A few sources

-1. https://en.wikipedia.org/wiki/Gudermannian_function

-2. http://mathworld.wolfram.com/InverseHyperbolicTangent.html

-3. http://www.wolframalpha.com/input/?i=gd%5E%28-1%29%28ipi%2F2%29

-4. http://www.wolframalpha.com/input/?i=tan%28pi%2F4+%281%2Bi%29%29

e) Happy Thanksgiving!