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#### DreamWeaver

##### Well-known member

- Sep 16, 2013

- 337

I've been trying to evaluate the parametric integral

[tex]\int_0^{\theta}\tan^{-1}(a\tan x)\,dx[/tex]

But I keep getting stuck... For [tex]0 < a < 1\,[/tex], [tex]0 < z < 1\,[/tex], and [tex]z=\tan\theta\,[/tex] I manage to get as far as

[tex]\int_0^{\theta}\tan^{-1}(a\tan x)\,dx =[/tex]

[tex]\theta\tan^{-1}(\tan \theta)+\frac{1}{2}\text{Li}_2\left(\frac{1}{1-a}\right)-\frac{1}{2}\text{Li}_2\left(\frac{1}{1+a}\right)[/tex]

[tex]-\frac{1}{4}\text{Li}_2\left(\frac{1-iaz}{(1-a)(1+a^2z^2)}\right)

+\frac{1}{4}\text{Li}_2\left(\frac{1+iaz}{(1+a)(1+a^2z^2)}\right)[/tex]

[tex]+\frac{1}{4}\text{Li}_2\left(\frac{1-iaz}{(1+a)(1+a^2z^2)}\right)

-\frac{1}{4}\text{Li}_2\left(\frac{1+iaz}{(1-a)(1+a^2z^2)}\right)[/tex]

Try as I might, I just cant seem to find a way to reduce those last four polylogarithmic terms into 'well-behaved' real parts... I'm sure there's a way to do it - undoubtedly in Lewin's book, which I don't have - but I just can't seem to find the solution.

Any ideas?

Cheers!

Gethin

ps. The partial-result shown above was due to splitting the integral into complex partial fraction terms using:

\(\displaystyle \int_0^z\frac{\tan ^{-1}x}{(1+a^2x^2)}\,dx=\)

\(\displaystyle \frac{1}{2i}\int_0^z\log\left(\frac{1+ix}{1-ix}\right)\frac{dx}{(1+iax)(1-iax)}\,dx\)