MHB Evaluating Integral with Polylogarithms: Help Needed!

[DreamWeaver]

Hi all! :D

I've been trying to evaluate the parametric integral

\int_0^{\theta}\tan^{-1}(a\tan x)\,dx

But I keep getting stuck... For 0 < a < 1\,, 0 < z < 1\,, and z=\tan\theta\, I manage to get as far as\int_0^{\theta}\tan^{-1}(a\tan x)\,dx =

\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)

-\frac{1}{4}\text{Li}_2\left(\frac{1-iaz}{(1-a)(1+a^2z^2)}\right)<br /> +\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)<br /> -\frac{1}{4}\text{Li}_2\left(\frac{1+iaz}{(1-a)(1+a^2z^2)}\right)
Try as I might, I just can't 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:

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

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

 

[alyafey22]

In case you haven't seen my response in the other forum look up the following http://mathhelpboards.com/calculus-10/generalized-fractional-logarithm-integral-5467.html.

Mainly we can prove that

$$\overline{\text{Li}_2(z)}=\text{Li}_2(\overline{z})$$

 

[DreamWeaver]

Thanks Z! :D:D:D

I saw your post on t'other forum... I'll tickle this one tonight and see what real results fall out...

Once again, many thanks! (Rock)