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I'm trying to evaluate

$$\int_0^{2\pi}e^{2it}\ln(a^2-2a \cos(t) + 1)dt$$

for $a \in (0, 1)$.

I keep getting different answers depending on the method. First, if I split up the integrals in to real and imaginary parts I get:

$$ = \int_0^{2\pi} (\cos(2t) + i \sin(2t)) \ln(a^2-2a \cos(t) + 1)dt\\ = \int_0^{2\pi} \cos(2t) \ln(a^2-2a \cos(t) + 1)dt + i \int_0^{2\pi} \sin(2t) \ln(a^2-2a \cos(t) + 1)dt\\ = - \frac{1}{4 a^2} \left[\begin{array}& (a^4 + 1) t \\ - 2 (a^4 - 1) \arctan \frac{(a + 1) \tan(t/2)}{a - 1} \\ + 2 (a^3 + a) \sin(t) \\ + a^2 \sin(2 t) (1 - 2 \sin(2 t) \ln(a^2 - 2 a \cos(t) + 1)) \end{array} \right]_0^{2\pi} + i * 0 \\ = - \pi \frac{a^4 + 1}{2 a^2}$$

Where I got $\int \cos(2t) \ln(a^2-2a \cos(t) + 1) dt$ from Wolfram Alpha. This other online integral calculator gives a different but equivalent antiderivative.

Second, if I do a contour integral:

$$ = \int_0^{2\pi} e^{2it} \ln((1-ae^{it}) (1 - a e^{-it})) dt \\ = \int_0^{2\pi} e^{2it} (\ln(1-ae^{it}) + \ln(1 - a e^{-it})) dt \\ = \int_0^{2\pi} e^{2it} \ln(1-ae^{it}) + \int_0^{2\pi} e^{2it} \ln(1 - a e^{-it}) dt \\ = \oint \frac{z \ln(1-az)}{i}dz - \oint \frac{\ln(1 - a z)}{iz^3} dz \\ = 2 \pi i (0 - \frac{a^2}{2i}) \\ = -\pi a^2 $$

Third, if I ask an online integral calculator, I just get 0.

I have no idea which, if any, answer is correct, or what mistakes I made in any of the methods. Any help would be appreciated.

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    $\begingroup$ Your second approach is correct. The same result should be obtained by integration with $\cos(2t)$. I wonder how did you get a different result. You did not write how it was obtained. $\endgroup$
    – user
    Apr 12, 2020 at 18:14
  • $\begingroup$ @user I've added the additional steps in the integral. I didn't think they were suspect since I got an equivalent antiderivative from two independent sources. Though maybe I'm missing something really obvious. $\endgroup$
    – Jay Lemmon
    Apr 12, 2020 at 20:39

1 Answer 1

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HINT.

Your second approach is correct. In the first one the expression used for the antiderivative is discontinuous at the point $t=\pi$ if the standard definition of the real $\arctan x$ function is used. To avoid this issue use the symmetry and calculate the integral as: $$ 2\int_0^{\pi}\cos(2t)\ln(a^2-2a \cos(t) + 1)dt. $$

You will obtain an additional summand $-\frac{\pi(a^4-1)}{2a^2}$.

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  • $\begingroup$ Thanks! I hadn't even thought to check the antiderivative for a discontinuity except at the endpoints. $\endgroup$
    – Jay Lemmon
    Apr 12, 2020 at 21:42

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