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How do I integrate $\int \log(1+2m\cos x+m^2) dx $ ?

I tried 2 things. First, I tried complex numbers. Putting $\cos x = \frac{e^{ix}+e^{-ix}}{2} $ which led me to $\int \log((me^{ix} +1)(me^{-ix}+1))dx $ and finally $\int \log(me^{ix} +1)+\log(me^{-ix}+1)dx $ , which I expanded using taylor expansion of $ \log(1+x)$ and then transformed it back to sine and cos, and then integrated but the answer was pretty ugly. And I'm not even sure it is correct.

Another thing I tried was to differentiate the integral with respect to m and then integrate it with respect to x and then with respect to m. But I couldn't do the last step, id est, integrating with respect to m.

Is there any other method which can be applied for this integration ?

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By exploiting the Taylor series of $\log(1+z)$ in a neighbourhood of the origin, $$\int \log(1+me^{ix})\,dx = C+i\cdot\text{Li}_2\left(-me^{ix}\right).$$ An interesting possibility is given by differentiation under the integral sign:

$$ \int\frac{2m+2\cos x}{1+m^2+2m\cos x}\,dx = \frac{x}{m}+\frac{2}{m}\,\arctan\left(\frac{m-1}{m+1}\,\tan\frac{x}{2}\right)$$ The integral of $\log(1+m^2+2m\cos x)$ is just the integral of the previous line with respect to the $m$-variable.

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  • $\begingroup$ Yes, I did everything that you did but the last line ("the integral of the previous line with respect to the $m$ -variable"), is what I can't seem to do. $\endgroup$ – Aritra Das Jun 11 '15 at 19:40
  • $\begingroup$ @AritraDas: you may substitute $u=\frac{m-1}{m+1}$. In any case, you have to reach the real part of a dilogarithm. Are you sure the original problem was requesting to compute a primitive for $\log(1+m^2+2m\cos x)$, for any $|m|<1$? $\endgroup$ – Jack D'Aurizio Jun 11 '15 at 19:47
  • $\begingroup$ Not even $ |m| < 1 $ but for $ m > 0 $. I think we need 2 cases, one for $ |m| < 1 $ and the other for $ |m| > 1 $ . $\endgroup$ – Aritra Das Jun 11 '15 at 20:03
  • $\begingroup$ This quesiton is from a book, which also lists some hints. There they have mentioned using $ \log(1+x) = x - x^2/2 + x^3/3 - .... $ But is this formula applicable for complex values of x ? $\endgroup$ – Aritra Das Jun 11 '15 at 20:07
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    $\begingroup$ That series is applicable in a region for which $\log z$ is analytic. $\endgroup$ – Mark Viola Jun 11 '15 at 20:14

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