calculus and analysis - Analytic result for integral?

I would like to do the integral

$$I=\int_0^{2\pi}d\phi\frac{\ln(e^{i\phi}+e^{-i\phi}-\frac{5}{2})}{e^{i\phi}+e^{-i\phi}-\frac{5}{2}}.$$

Numerically, we readily find that it has a specific finite value:

fun = -(5/2) + E^(-I \[Phi]) + E^(I \[Phi]);
NIntegrate[ Log[fun]/fun, {\[Phi], 0, 2 \[Pi]}]

-0.493368 - 13.1595 I

Now, if we want to consider the integral analytically, we could substitute for instance

$$e^{i\phi}=z~~~,~~~d\phi=\frac{-i}{z}dz$$

which leads to

$$I=-i\oint_{|z|=1}\frac{\ln\left[\frac{1}{z}(z - \frac{1}{2}) (z - 2)\right]}{(z - \frac{1}{2}) (z - 2)}$$

This looks like there is a pole at z=1/2 within the unit circle. So I tried to get the residue:

Residue[-(( I Log[((-2 + z) (-(1/2) + z))/z])/((-2 + z) (-(1/2) + z))), {z, 1/2}]

Residue[-(( I Log[((-2 + z) (-(1/2) + z))/z])/((-2 + z) (-(1/2) + z))), {z, 1/2}]

which just gave back the input. Also, the 1/z term inside the logarithm seems to blow up inside the unit circle as well. This integral is confusing and does not seem to be accessible via straightforward analytical methods. Is there a way to evaluate it exactly using Mathematica?

Answer

Noting that $e^{\text{i}\phi} + e^{-\text{i}\phi} = 2\cos\phi$, simply write:

f[t_] := Log[t Cos[fi] - 5/2]/(2 Cos[fi] - 5/2)

g[t_] := Integrate[f[t], {fi, 0, 2 Pi}]

h[t_] := Integrate[Integrate[f'[t], {fi, 0, 2 Pi}], t]

Clear[c]; {{c}} = {c} /. Solve[g[0] == (h[t] /. t -> 0) + c, c];


Limit[h[t] + c, t -> 2, Direction -> 1] // Expand

I get:

$-\frac{2}{3}\pi\log\left(\frac{81}{64}\right) - \frac{4}{3}\pi^2\,\text{i} \approx -0.493368 - 13.1595\,\text{i}$

without knowing anything about Complex Analysis.

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Theorems on the derivation under the integral sign

Let $\Gamma \subseteq \mathbb{C}$ a smooth curve arc and $f : \Gamma \to \mathbb{C}$ is a continuous function. Called $F : \mathbb{C}\backslash \Gamma \to \mathbb{C}$ the function defined by placing $F(\zeta) := \int_{\Gamma} \frac{f(z)}{z-\zeta}\text{d}z $, it is holomorphic in $\mathbb{C}\backslash \Gamma$. In addition, $F$ possesses in $\mathbb{C}\backslash \Gamma$ derivatives of still higher order (which are therefore all holomorphic functions in $\mathbb{C}\backslash \Gamma$) and, for each $n \in \mathbb{N}$, you have $F^{(n)}(\zeta) = n!\int_{\Gamma} \frac{f(z)}{(z-\zeta)^{n+1}}\text{d}z$, i.e. the derivatives of $F$ can be calculated by deriving under the integral sign.

Wanting to simplify things, let $\Omega \subset \mathbb{R}^n$ an open limited, $I \subset \mathbb{R}$ an open interval and $f : I \times \bar{\Omega} \to \mathbb{R}$ a continuous function. If $\frac{\partial f}{\partial t}(t,x)$ continues exists for each $(t,x) \in I \times \bar{\Omega}$, then you have $\frac{\text{d}}{\text{d}t}\int_{\Omega} f(t,x)\text{d}x = \int_{\Omega} \frac{\partial f}{\partial t}(t,x)\text{d}x$.