Question

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

Answer 1

(Quick note: I've defined `\arctanh` in LaTeX, so anyone in this thread should be able to use it just fine)

Answer 2

Here's a preliminary numerical result due to Mathematica

Answer 3

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

Answer 4

What I meant by "exact value" was "closed form" :)

Answer 5

\[\large \int_{-1}^1\frac{\arctan x}{\arctanh\, x}\,dx = 2\int\limits_{-\pi/4}^{\pi/4}\dfrac{u\sec^2u}{\ln|\frac{1+\tan u}{1-\tan u}|}du\]

Answer 6

Ohhhh ok, sorry it's still early for me here haha.

Answer 7

Is there a way to calculate an integral of a ratio as the ratio of the integrals or something like that?

Answer 8

\[\large \int_{-1}^1\frac{\arctan x}{\arctanh\, x}\,dx = 2\int\limits_{-\pi/4}^{\pi/4}\dfrac{u\sec^2u}{\ln|\frac{1+\tan u}{1-\tan u}|}du = 4\int\limits_{0}^{\pi/4}\dfrac{u\sec^2u}{\ln|\frac{1+\tan u}{1-\tan u}|}du\]

idk if it can be worked using simple definite integral props, but im giving it a try first

Answer 9

I wonder if there's a geometric way to solve this integral

Answer 10

Here's an idea, can we make this substitution does it help us?

\[-i \tan^{-1}(i x) = \tanh^{-1}(x)\]

Answer 11

Or perhaps rewriting as: \[2i \int\limits_0^1 \frac{\ln \frac{1-ix}{1+ix}}{\ln \frac{1+x}{1-x}} dx = 2i \int\limits_0^1 \log_{\frac{1+x}{1-x}} \left( \frac{1-ix}{1+ix} \right) dx\]

Answer 12

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.