# How to prove $\int^{\pi/2}_0 \log{\cos{x}} \, \mathrm{d}x = \frac{\pi}{2}\log\left(\frac12\right)$

I was trying to prove the result that the OP of this question is given as a hint.

That is to say: imagine that you are not given the hint and you need to evaluate:

$$I = \int^{\pi/2}_0 \log{\cos{x}} \, \mathrm{d}x \color{red}{\overset{?}{=} }\frac{\pi}{2} \log{\frac{1}{2}} \tag{1}$$

How would you proceed?

Well, I tried the following steps and, despite it seems that I am almost there, I have found some troubles:

• Taking advantage of the fact: $$\cos{x} = \frac{e^{ix}+e^{-ix}}{2}, \quad \forall x \in \mathbb{R}$$
• Plugging this into the integral and performing the change of variable $z = e^{ix}$, so the line integral becomes a contour integral over a quarter of circumference of unity radius centered at $z=0$, i.e.: $$I = \frac{1}{4i} \oint_{|z|=1}\left[ \log{ \left(z+\frac{1}{z}\right)} - \log{2} \right] \, \frac{\mathrm{d}z }{z}$$

$\color{red}{\text{We cannot do this because the integrand is not holomorphic on } |z| = 1 }$

• Note that the integrand has only one pole lying in the region enclosed by the curve $\gamma : |z|=1$ and it is holomorphic (is it?) almost everywhere (except in $z =0$), so the residue theorem tells us that:

$$I = \frac{1}{4i} \times 2\pi i \times \lim_{z\to0} \color{red}{z} \frac{1}{\color{red}{z}} \left[ \underbrace{ \log{ \left(z+\frac{1}{z}\right)} }_{L} - \log{2} \right]$$

• As I said before, it seems that I am almost there, since the result given by eq. (1) follows iff $L = 0$, which is not true (I have tried L'Hôpital and some algebraic manipulations).

Where did my reasoning fail? Any helping hand?

Thank you in advance, cheers!

Please note that I'm not much of an expert in either complex analysis or complex integration so please forgive me if this is trivial.

Notation: $\log{x}$ means $\ln{x}$.

A graph of the function $f(z) = \log{(z+1/z)}$ helps to understand the difficulties:

where $|f(z)|$, $z = x+i y$ is plotted and the white path shows where $f$ is not holomorphic.

• This is the same since $\ln a=-\ln\frac{1}{a}$ Commented Aug 26, 2014 at 13:32
• @Matthias: Could you develop your argument ? I don't understand.
– idm
Commented Aug 26, 2014 at 13:41
• $e^{-\ln \frac{1}{a}}=\frac{1}{e^{\ln\frac{1}{a}}}=\frac{1}{\frac{1}{a}}=a=e^{\ln a}$ Commented Aug 26, 2014 at 13:47
• I'm sorry, I thought it was written $\frac{\pi\ln 2}{2}$, I've just not read correctly. I erased my comment :-)
– idm
Commented Aug 26, 2014 at 14:03
• You cannot integrate $$z\mapsto \frac{\ln\left(z+\frac{1}{z}\right)}{z}$$ on $\{z\mid |z|=1\}$ because the function is not holomorphic in $\{z\mid |z|\leq 1\}$, indeed, the function is not holomorphic for $\Re\left(z+\frac{1}{z}\right)<0$
– idm
Commented Aug 26, 2014 at 14:11

An other way:

Firstly $$\int_0^{\pi/2}\ln(\cos t)dt\underset{t=\frac{\pi}{2}-u}{=}\int_{0}^{\pi/2}\ln\left(\cos\left(\frac{\pi}{2}-u\right)\right)du=\int_0^{\pi/2}\ln(\sin u)du \tag 1$$

Then, $$\int_0^{\pi/2}\ln(\sin t)dt=\frac{1}{2}\left(\int_{0}^{\pi/2}\ln(\sin t)dt+\int_0^{\pi/2}\ln(\cos t)dt\right)=\frac{1}{2}\int_0^{\pi/2}\ln\left(\frac{\sin(2t)}{2}\right)dt\underset{r=2t}{=}\frac{1}{4}\int_0^\pi\ln\left(\frac{\sin r}{2}\right)dr=\frac{1}{4}\int_{0}^\pi\ln(\sin r)dr-\frac{\pi\ln 2}{4}\underset{Chasles}{=}\frac{1}{4}\int_0^{\pi/2}\ln(\sin r)dr+\int_{\pi/2}^\pi\ln(\sin t)dt-\frac{\pi\ln 2}{4}\underset{t=r+\frac{\pi}{2}}{=}\frac{1}{4}\int_0^{\pi/2}\ln(\sin r)dr+\frac{1}{4}\int_0^{\pi/2}\ln(\sin t)dt=\frac{1}{2}\int_0^{\pi/2}\ln(\sin t)dt-\frac{\pi\ln 2}{2}$$

And thus $$\int_0^{\pi/2}\ln(\sin t)dt=\frac{1}{2}\int_0^{\pi/2}\ln(\sin t)dt-\frac{\pi\ln 2}{4}\iff\int_0^{\pi/2}\ln(\sin t)dt=-\frac{\pi\ln 2}{2}.$$

By $(1)$ we conclude that $$\int_0^{\pi/2}\ln(\cos t)dt=-\frac{\pi\ln 2}{2}$$

• As I said to the other user, I would like to know where I made a mistake in my reasoning. Anyway, +1 for the effort. Commented Aug 26, 2014 at 14:01
• Ok, I answered in your post.
– idm
Commented Aug 26, 2014 at 14:11

Hint: with $\cos x=u$ $$\int_0^{\pi/2}\log\cos x\mathrm{d}x=-\int_0^1\frac{\log u}{\sqrt{1-u^2}}\mathrm{d}u$$

• $$\int_0^1\frac{\ln u}{\sqrt{1-u^2}}du$$ doesn't look easy to calculate.
– idm
Commented Aug 26, 2014 at 13:24
• @idm this can be easily integrated by parts. Commented Aug 26, 2014 at 13:30
• By part: $$\int_0^1\frac{\ln u}{\sqrt{1-u^2}}du=[\ln(u)\arcsin(u)]_0^1-\int_0^1\frac{\arcsin u}{u}du$$ and $\int \frac{\arcsin x}{x}dx$ doesn't look calculable by hand like you can see in the link wolframalpha.com/input/?i=integrate+arcsin%28x%29%2Fx
– idm
Commented Aug 26, 2014 at 13:37
• This is great and I can deal with it (I hope) but I would like to know where I made a mistake in my way to find the value of the integral. +1 for the hint. Commented Aug 26, 2014 at 14:01

As shown by @idm, we have that $$\int_0^{\pi/2}\ln(\cos(x))dx = \int_0^{\pi/2}\ln(\sin(x))dx.$$ We can exploit this identity for another one and use Feynman's method (differentiating under the integral). Consider $$\int_0^{\pi/2}x\cot(x)dx.\tag{1}$$ By integration by parts, we have $$\int_0^{\pi/2}x\cot(x)dx = x\ln(\sin(x))\Bigr|_0^{\pi/2} - \int_0^{\pi/2}\ln(\sin(x))dx = - \int_0^{\pi/2}\ln(\sin(x))dx$$ since $\lim_{x\to 0}x\ln(\sin(x)) = 0$. Therefore, we can evaluate the negative of equation $(1)$. \begin{align} I(\alpha) &= \int_0^{\pi/2}\arctan(\alpha\tan(x))\cot(x)dx\tag{2}\\ I'(\alpha) &= \int_0^{\pi/2}\frac{\partial}{\partial\alpha}\Bigl[\arctan(\alpha\tan(x))\cot(x)\Bigr]dx\\ &= \int_0^{\pi/2}\frac{dx}{\alpha^2\tan^2(x) + 1}\\ &= \frac{\pi}{2(\alpha +1)}\\ I(\alpha) &= \frac{\pi}{2}\ln(\alpha + 1) + C \end{align} Thus, $I(0)\Rightarrow C=0$ so $$I(\alpha) = \frac{\pi}{2}\ln(\alpha + 1)$$ We recover equation $(1)$ from equation $(2)$ when $\alpha = 1$ so $I(1) = \frac{\pi}{2}\ln(2)$ and since $$\int_0^{\pi/2}\ln(\sin(x))dx = -\int_0^{\pi/2}x\cot(x)dx = -\frac{\pi}{2}\ln(2),$$ we have $$\int_0^{\pi/2}\ln(\sin(x))dx = -\frac{\pi}{2}\ln(2),$$

Since this post was tagged with complex analysis, I can provide a contour integration solution as well. Again, I will exploit the identity given to you by @idm. $$\int_0^{\pi/2}\ln(\sin(\theta))d\theta = \frac{1}{2}\int_0^{\pi}\ln(\sin(\theta))d\theta$$ Consider $1 - e^{2iz} = -2ie^{iz}\sin(z)$. We can write $1 - e^{2iz}$ as $$1 - e^{-2y}(\cos(2x) + i\sin(2x)) < 0\text{ when } x=\pi n, \ y\leq 0$$ Now let's consider the contour from $0$ to $\pi$ to $\pi + iA$ to $iA$ where we take a quarter of a circle around $0$ and $\pi$ with radius $\epsilon$. From the periodicity of the function, the vertical line segments cancel each other since they have opposite signs. Additionally, as $A\to\infty$, the top integral of the top line goes to zero and as $\epsilon\to 0$, the integral around $0$ and $\pi$ go to zero. \begin{align} \ln(-2ie^{ix}\sin(z)) &= \ln(-2i) + \ln(e^{ix}) + \ln(\sin(\theta))\\ &= \ln|-2i| + i\arg(-2i) + ix + \ln(\sin(\theta))\\ &= \ln(2) - i\frac{\pi}{2} + \ln(\sin(\theta)) + i\frac{\pi}{2} \end{align} where $\ln(2i) = \ln(2) + i\arg(-2i)$ and we take the principle argument to be $-\frac{\pi}{2}$ and the imaginary part of $ix$ is between $0$ and $\pi$. Since there are no poles in them contour, by the Cauchy integral formula, the integral is equal to zero. \begin{alignat}{2} \int_0^{\pi/2}\ln(\sin(\theta))d\theta &=\frac{1}{2}\int_0^{\pi}\ln(\sin(\theta))d\theta\\ &= \frac{\ln(2)}{2}\int_0^{\pi}d\theta - \frac{i\pi}{4}\int_0^{\pi}d\theta + \frac{1}{2}\int_0^{\pi}\ln(\sin(\theta))d\theta + \frac{i\pi}{4}\int_0^{\pi}d\theta &&{}= 0\\ &= \frac{\pi\ln(2)}{2} - \frac{i\pi^2}{4} + \frac{1}{2}\int_0^{\pi}\ln(\sin(\theta))d\theta + \frac{i\pi^2}{4} &&{}=0\\ \frac{1}{2}\int_0^{\pi}\ln(\sin(\theta))d\theta &= -\frac{\pi\ln(2)}{2}\\ \int_0^{\pi/2}\ln(\sin(\theta))d\theta &= -\frac{\pi\ln(2)}{2} \end{alignat}