I want to evaluate $$ \int \frac{\arctan(x)}{x} dx $$ One way of doing it is to use power-series. I can re-write $$ \arctan(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1} }{2n+1} \ \ |x|<1 $$

$$ \int \frac{\arctan(x)}{x} dx = \int \sum_{n=0}^\infty (-1)^n \frac{x^{2n} }{2n+1} dx = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1} }{(2n+1)^2} + C$$

But now I am curious whether there is any other way to evaluate this integral?

  • 2
    $\begingroup$ Not really. Depending on what you consider a reasonable answer, it's possible to integrate by parts and split the resulting integral of shape $\ln(x)/(1 + x^2)$ into to logarithmic-type integrals (with an imaginary component from writing $x^2 + 1 = (x + i)(x - i)$). $\endgroup$ – davidlowryduda Jun 9 at 23:56

Unfortunately no, there isn't really a nice way to do this other than writing your integral in terms of dilogarithms. But you may be interested to know that this is a well known integral, called the inverse tangent integral, and it was studied heavily by Ramanujan.


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