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Tricky Polylogs

DreamWeaver

Well-known member
Sep 16, 2013
337
Hi all! :D

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:

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

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

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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
Sep 16, 2013
337
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)