I would like to do the integral


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


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?

  • $\begingroup$ That guy got so many, which one are you referring to? Introduce parameter and take derivative, and integrate parameter after? $\endgroup$
    – Kagaratsch
    Commented Feb 16, 2017 at 23:40
  • 1
    $\begingroup$ By expanding the function inside the closed loop integral using Series at z=1/2 doesn't have the term (z-1/2)^-1. $\endgroup$ Commented Feb 16, 2017 at 23:49

3 Answers 3


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.

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 $\frac{\text{d}}{\text{d}t}\int_{\Omega} f(t,x)\text{d}x = \int_{\Omega} \frac{\partial f}{\partial t}(t,x)\text{d}x$.

  • $\begingroup$ Wait, so you get a wrong sign in the imaginary part, compared to numerical integration? Wonder why that is... $\endgroup$
    – Kagaratsch
    Commented Feb 17, 2017 at 0:18
  • $\begingroup$ But you know about the uniform convergence :). Maybe you should explain why it is possible to exchange the order of two operations... $\endgroup$
    – yarchik
    Commented Feb 17, 2017 at 0:27
  • $\begingroup$ Should f really be on both sides in the first equation? $\endgroup$
    – Felix
    Commented Feb 17, 2017 at 1:08
  • $\begingroup$ I would really like to know more about the uniform convergence and the criteria for when it is allowed to exchange the operations! Maybe you can direct me to some resource online where I can read more about it? $\endgroup$
    – Kagaratsch
    Commented Feb 17, 2017 at 2:44

The whole thing is much simple. Only apply 'ExpToTrig' and you get the right analytical definite integral.

     {int = 
     Integrate[Log[fun]/fun // ExpToTrig,{Phi, 0, 2 Pi}] // 
     FullSimplify, N[int]}

     (*   {-(4/3) I Pi (Pi - I Log[9/8]), -0.493368 - 13.1595 I}   *)

This is probably cheating, but it seems to work.

First we get the anti-derivative

int = -I Integrate[Log[1/z - 5 /2 + z]/(1 - 5 /2 z + z^2), z] /. z -> Exp[I \[Phi]];

It is important to do this in the z variable, since in the $\phi$ Mathematica does not return a result.

Then we evaluate the upper and lower boundaries:

 Assuming[ep>0,Series[(int/.\[Phi]->(2\[Pi]-ep))-(int/.\[Phi]->(ep)),{ep, 0, 0}]//FullSimplify]

enter image description here

Setting ep to zero or some value very close to zero, and checking numerically, we get the correct imaginary part of the result. The real part does not agree though. Still, we can assume that the real part is similar to what comes out, so we consider -4\[Pi] Log[x]/3==-0.49336842604379194 which has the solution x=9/8. In this way we deduce

result = -((4 I \[Pi]^2)/3) - 4/3 \[Pi] Log[9/8];

And confirm

fun = -(5/2) + E^(-I \[Phi]) + E^(I \[Phi]);
nint=NIntegrate[ Log[fun]/fun, {\[Phi], 0, 2 \[Pi]},WorkingPrecision->300];

0.*10^-300 + 0.*10^-300 I

That was not entirely analytic though. I'll try Feynman trick next.


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