Does this integral have a closed form? \[\newcommand{arctanh}{\mathbin{\text{arctanh}}} \int_{-1}^1\frac{\arctan x}{\arctanh\, x}\,dx\]

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Does this integral have a closed form? \[\newcommand{arctanh}{\mathbin{\text{arctanh}}} \int_{-1}^1\frac{\arctan x}{\arctanh\, x}\,dx\]

Mathematics
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(Quick note: I've defined `\arctanh` in LaTeX, so anyone in this thread should be able to use it just fine)
Here's a preliminary numerical result due to Mathematica
1 Attachment
I want to say yes cause I've never heard of an integral being multivalued before but I'm suspicious haha. http://www.wolframalpha.com/input/?i=integral+from+-1+to+1+arctan+x+%2F+arctanh+x++dx

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What I meant by "exact value" was "closed form" :)
\[\large \int_{-1}^1\frac{\arctan x}{\arctanh\, x}\,dx = 2\int\limits_{-\pi/4}^{\pi/4}\dfrac{u\sec^2u}{\ln|\frac{1+\tan u}{1-\tan u}|}du\]
Ohhhh ok, sorry it's still early for me here haha.
Is there a way to calculate an integral of a ratio as the ratio of the integrals or something like that?
\[\large \int_{-1}^1\frac{\arctan x}{\arctanh\, x}\,dx = 2\int\limits_{-\pi/4}^{\pi/4}\dfrac{u\sec^2u}{\ln|\frac{1+\tan u}{1-\tan u}|}du = 4\int\limits_{0}^{\pi/4}\dfrac{u\sec^2u}{\ln|\frac{1+\tan u}{1-\tan u}|}du\] idk if it can be worked using simple definite integral props, but im giving it a try first
I wonder if there's a geometric way to solve this integral
Here's an idea, can we make this substitution does it help us? \[-i \tan^{-1}(i x) = \tanh^{-1}(x)\]
Or perhaps rewriting as: \[2i \int\limits_0^1 \frac{\ln \frac{1-ix}{1+ix}}{\ln \frac{1+x}{1-x}} dx = 2i \int\limits_0^1 \log_{\frac{1+x}{1-x}} \left( \frac{1-ix}{1+ix} \right) dx\]
Ok ok ok how about from here can we turn this into a double integral perhaps? \[2i \int\limits_0^1 \frac{\ln \frac{1-ix}{1+ix}}{\ln \frac{1+x}{1-x}} dx\] I want to make the substitution \[1+ix = r e^{i \theta}\]\[ idx = e^{i\theta} dr + ir e^{i \theta} d \theta \] and looking at this \[1+ix = r e^{i \theta}\] I change the bounds from \[0 \le x \le 1\] to \[0 \le \theta \le \frac{\pi}{2}\]\[1 \le r \le \sqrt{2} \] And then plug this in somehow I guess? Just an idea.
|dw:1433263922151:dw|
\[a = \tanh^{-1}(1)\]\[1=\tanh(a) = \frac{\sinh(a)}{\cosh(a)} \implies a = \infty\]
|dw:1433264415984:dw|
I'm beginning to think some kind of contour might help us out here. Like a rectangle with two vertices fixed at \((-1,0)\) and \((1,0)\) and the other two extending to infinity in the upper half of the complex plane, but my complex analysis skills have admittedly gather a bit of dust.
For anyone who has more recently dealt with contour integrals, it's possible this form might be of some use. Substituting \(x=\tanh u\) gives us \[\newcommand{sech}{\text{sech }} \int_{-\infty}^\infty \frac{\arctan(\tanh u)\sech^2u}{u}\,du\]
|dw:1433303732363:dw|
|dw:1433352325544:dw|

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