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

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(Quick note: I've defined `\arctanh` in LaTeX, so anyone in this thread should be able to use it just fine)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Here's a preliminary numerical result due to Mathematica

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3I 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What I meant by "exact value" was "closed form" :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\large \int_{1}^1\frac{\arctan x}{\arctanh\, x}\,dx = 2\int\limits_{\pi/4}^{\pi/4}\dfrac{u\sec^2u}{\ln\frac{1+\tan u}{1\tan u}}du\]

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3Ohhhh ok, sorry it's still early for me here haha.

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3Is there a way to calculate an integral of a ratio as the ratio of the integrals or something like that?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\large \int_{1}^1\frac{\arctan x}{\arctanh\, x}\,dx = 2\int\limits_{\pi/4}^{\pi/4}\dfrac{u\sec^2u}{\ln\frac{1+\tan u}{1\tan u}}du = 4\int\limits_{0}^{\pi/4}\dfrac{u\sec^2u}{\ln\frac{1+\tan u}{1\tan u}}du\] idk if it can be worked using simple definite integral props, but im giving it a try first

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3I wonder if there's a geometric way to solve this integral

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3Here's an idea, can we make this substitution does it help us? \[i \tan^{1}(i x) = \tanh^{1}(x)\]

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3Or perhaps rewriting as: \[2i \int\limits_0^1 \frac{\ln \frac{1ix}{1+ix}}{\ln \frac{1+x}{1x}} dx = 2i \int\limits_0^1 \log_{\frac{1+x}{1x}} \left( \frac{1ix}{1+ix} \right) dx\]

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3Ok ok ok how about from here can we turn this into a double integral perhaps? \[2i \int\limits_0^1 \frac{\ln \frac{1ix}{1+ix}}{\ln \frac{1+x}{1x}} dx\] 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.

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3\[a = \tanh^{1}(1)\]\[1=\tanh(a) = \frac{\sinh(a)}{\cosh(a)} \implies a = \infty\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For 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 \[\newcommand{sech}{\text{sech }} \int_{\infty}^\infty \frac{\arctan(\tanh u)\sech^2u}{u}\,du\]

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433303732363:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433352325544:dw
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