# Tricky Polylogs

#### DreamWeaver

##### Well-known member
Hi all! 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

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

$$\theta\tan^{-1}(\tan \theta)+\frac{1}{2}\text{Li}_2\left(\frac{1}{1-a}\right)-\frac{1}{2}\text{Li}_2\left(\frac{1}{1+a}\right)$$

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

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:

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

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

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
In case you haven't seen my response in the other forum look up the following thread.

Mainly we can prove that

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

#### DreamWeaver

##### Well-known member
Thanks Z!   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! 