- #1
DreamWeaver
- 297
- 0
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:
$$\int_0^z\frac{\tan ^{-1}x}{(1+a^2x^2)}\,dx=$$
$$\frac{1}{2i}\int_0^z\log\left(\frac{1+ix}{1-ix}\right)\frac{dx}{(1+iax)(1-iax)}\,dx$$
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:
$$\int_0^z\frac{\tan ^{-1}x}{(1+a^2x^2)}\,dx=$$
$$\frac{1}{2i}\int_0^z\log\left(\frac{1+ix}{1-ix}\right)\frac{dx}{(1+iax)(1-iax)}\,dx$$