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I want to evaluate an integral featuring the functions $\ln, \ \cos,$ and $e$. Due to the nature of those functions, I was wondering if taking a Fourier transform of it might lead to some simplification and give an explicit answer. The integral is:

$$I(a, b, c) = \int_0^{2 \pi} \ln(a + \cos(x - b) + \cos(x - c) + \cos(c - b)) e^{i x} dx.$$

Can this be Fourier transformed into a simpler form? Is it possible to evaluate it by any other methods?

Edit1: (Writing in terms of $e^{ix}$) $$I(a, b, c) = \int_0^{2 \pi} \ln(a + \frac{1}{2}(e^{i(x - b)} + e^{-i(x - b)} + e^{i(x - c)} + e^{-i(x - c)} + e^{i(b - c)} + e^{-i(b - c)})) e^{i x} dx.$$

Edit 2: (Fourier Transformed integrand) $$\int_0^{2 \pi} \int_{-\infty}^\infty \ln(a + \frac{1}{2}(e^{ix}(e^{-i b} +e^{-i c}) + e^{-ix}(e^{i b} +e^{i c}) + e^{i(b - c)} + e^{-i(b - c)})) e^{i x(1 - 2 \pi \omega) }d\omega dx .$$

Edit 3: (Fourier Transformed with respect to parameter $b$) $$F[I(a, b, c)](\omega) = \\ \int_{-\infty}^\infty \int_0^{2 \pi} \ln(a + \frac{1}{2}(e^{ix}(e^{-i b} +e^{-i c}) + e^{-ix}(e^{i b} +e^{i c}) + e^{i(b - c)} + e^{-i(b - c)})) e^{i x} dx e^{-i2 \pi b \omega} d\omega .$$

Edit 4: (Using substitution $z = e^{ix}$) $$I(a, b, c) = \\ -i \int_{|z| = 1} \ln(a + \frac{1}{2}(z(e^{-i b} +e^{-i c}) + \frac{1}{z}(e^{i b} +e^{i c}) + e^{i(b - c)} + e^{-i(b - c)})) dz .$$

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    Writing everything in terms of $e^{ix}$ and expanding the log judiciously looks like a reasonable first shot. – 2017-02-19
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    @Chappers I have edit the post to include $f(x)$ written in terms of $e^{i x}$. However I don't see how to expand log. It would be straightforward if the terms were being summed, but they are being summed here, so I don't see how we can expand it? – 2017-02-19
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    My point is that it seems possible to find the Fourier series for the log in some way, and then just pick out the correct term. This may require a rather subtle factorisation of the argument to make sure that the terms end up in the right places. Alternatively, one may try using the substitution $z=e^{ix}$ or $z=e^{-ix}$ and using the residue theorem. – 2017-02-19
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    @Chappers Ok I've gathered some terms now and applied the Fourier Transform. Hmmm, its beginning to look like a big mess now. I thought because the Fourier transform of functions like $\ln$ and complex exponentials are related to the dirac delta function that applying the FT could somehow lead to huge simplification of the original integral. But maybe it is only suitable for simpler integrals, it seems to have made this one a lot worse! – 2017-02-19
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    Your original integral is not a function of x though, it is a constant. Do you mean to take the fourier transform of the integrand itself? – 2017-02-19
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    @Triatticus Yes, I meant the fourier transform/series of the integrand. I have edited the post to fix that. – 2017-02-19
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    I am interested in whatever it takes to evaluate the original integral $I(a, b, c)$ so I can get an explicit expression for it. I have also made an edit to include the Fourier transform with respect to the parameter $b$. I still cannot see how to go any further though, using either of the Fourier transformed versions. – 2017-02-19
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    I have also now looked included the complex analysis approach suggested by Chappers. This initially looks promising as it gets rid of the $e^{ix}$ term that gets multiplied against the $\ln$ term. However I'm left with a pole that is 'caught' inside the $\ln$ term so I come to another dead end. – 2017-02-19
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    @eurocoder What assumptions (if any) can be made about the parameters $a,b,c$? Can we assume $a\ge\frac32$ and that $b,c$ are real? – 2017-02-19
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    @DavidH Yes, $b$ and $c$ are real and are in the range $[0, 2\pi]$ as they represent the angles of certain objects with respect to the positive real axis. $a \ge \frac{3}{2}$ also seems like an ok assumption. I know $a$ has to be $\ge \frac{1}{2}$, and I don't think it invalidates anything to require it instead to be $a \ge \frac{3}{2}$. – 2017-02-20

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