Show That

$$\int_{0}^{\pi}\ln (\sin x) \cos(2nx) \, dx = -\dfrac{\pi}{2n}$$

I tried to use substitutions and taylor series of $\ln (\sin x)$, but to no avail.

Any help will be appreciated.

  • $\begingroup$ I think $I_n-I_{n-1}$ would be of help.. $I_n$ is the integral on the LHS $\endgroup$ – Qwerty Dec 19 '16 at 19:09
  • $\begingroup$ Hints: ln() calls for integration by parts (to get rid of log). Second hint: if log has complicated stuff in it, usually you want to substitute variables (but not before halving the integration domain due to symmetry around pi/4). $\endgroup$ – orion Dec 19 '16 at 19:23

The first step in these cases is to get rid of nested transcendental functions, which we otherwise have no idea how to integrate. Integrate by parts:

$$u=\ln\sin x \Rightarrow du=\frac{\cos x}{\sin x}\,dx$$ $$dv = \cos (2nx)\,dx\Rightarrow v=\frac{1}{2n}\sin (2nx)$$

$$\int_0^\pi \ln(\sin x)\cos (2nx)\,dx = \ln\sin x\frac{1}{2n}\sin (2nx)|_0^\pi - \frac1{2n}\int_0^\pi \frac{\cos x}{\sin x}\sin (2nx)\,dx$$ The first term is 0, if you recall $\lim_{x \to 0}x\ln x=0$ and consider linear behaviour of $\sin x$ around $0$ and $\pi$. Notice also, that all the interesting dependence on $n$ is now a prefactor. The second term remains to compute (and should be independent on $n$, based on the expression we are trying to prove):

$$- \frac1{2n}\int_0^\pi \frac{\cos x}{\sin x}\sin (2nx)\,dx$$ This can be resolved in many ways. One way is by taking a shortcut through complex series (or recalling from physics the formula for finite diffraction grating): $$2i\sin (2nx)=e^{2nxi}-e^{-2nxi}$$ $$2i\sin x=e^{xi}-e^{-xi}$$ For less writing, use $q=e^{xi}$. $$\frac{\sin 2nx}{\sin x}=\frac{q^{2n}-q^{-2n}}{q-q^{-1}}= \frac{q^{-2n+1}-q^{2n+1}}{1-q^2}= q^{-2n+1}\frac{1-q^{4n}}{1-q^2} $$ Recognize finite geometric sum: $$=q^{-2n+1}\sum_{k=0}^{2n-1} q^{2k}=\sum_{k=0}^{2n-1} q^{2k-2n+1}=q^{-2n+1}+\cdots+q^{-1}+q+\cdots q^{2n-1}$$ which is a symmetric sum of every second power. Recall also $\cos x = \frac12(q+q^{-1})$. This just makes two copies of the upper sum, shifted by two, symmetric again... the left and rightmost term are counted only once. $$\frac{\cos x}{\sin x}\sin(2nx)=\frac12 q^{-2n}+\cdots+q^{-2}+1+q^2+\cdots \frac12q^{2n}$$ All terms come in pairs $q^{2k}+q^{-2k}=2\cos 2kx$ (for every $k\neq 0$), and when integrated on a whole number of periods (from $0$ to $\pi$), amount to zero. The only term to survive is: $$\int_0^\pi\frac{\cos x}{\sin x}\sin(2nx)dx=\int_0^\pi 1\cdot dx=\pi$$ which means we are done.

Of course I'm sure I missed much more obvious ways of proving that integral equals $\pi$ regardless of $n$.


Hint. One may write, for $x \in (0,\pi)$, $$ \begin{align} \log\left(\sin x \right)&=\log\left(\frac{e^{ix}-e^{-ix}}{2i} \right) \\&=\log\left(\frac{e^{ix}(1-e^{-2ix})}{2i} \right) \\&=\log\left(\frac{e^{i(x-\pi/2)}(1-e^{-2ix})}{2} \right) \\&=i(x-\pi/2)-\log 2+\log\left(1-e^{-2ix} \right) \\&=i(x-\pi/2)-\log 2-\sum_{n=1}^\infty\frac{e^{-2nix}}{n} \\&=-\log 2-\sum_{n=1}^\infty\frac{\cos(2nx)}{n}+i(x-\pi/2)+i\sum_{n=1}^\infty\frac{\sin(2nx)}{n} \end{align} $$ then by the uniqueness of Fourier coefficients one gets that

$$ \frac2\pi\int_{0}^{\pi}\ln (\sin x) \cos(2nx) \, dx=-\frac1n,\quad n\ge1, $$

as wanted.


$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{\pi}\ln\pars{\sin\pars{x}}\cos\pars{2nx}\,\dd x = -\,{\pi \over2n}:\ {\large ?}\,,\qquad n \in \mathbb{Z}\setminus\braces{0}}$.

\begin{align} &\int_{0}^{\pi}\ln\pars{\sin\pars{x}}\cos\pars{2nx}\,\dd x = 2\,\Re\int_{0}^{\pi/2}\ln\pars{\sin\pars{x}}\expo{2\verts{n}x\ic}\,\dd x \\[5mm] = &\ \left.2\,\Re\int_{\theta\ =\ 0}^{\theta\ =\ \pi/2} \ln\pars{z - 1/z \over 2\ic}z^{2\verts{n}} \,{\dd z \over \ic z}\,\right\vert_{\ z\ =\ \exp\pars{\ic\theta}} \\[5mm] = &\ \left.2\,\Im\int_{\theta\ =\ 0}^{\theta\ =\ \pi/2} \ln\pars{{1 - z^{2} \over 2z}\,\ic}z^{2\verts{n} - 1}\,\,\dd z\, \right\vert_{\ z\ =\ \exp\pars{\ic\theta}} \\[1cm] = &\ -2\,\ \overbrace{\Im\int_{1}^{\epsilon}\ln\pars{1 + y^{2} \over 2y}y^{2\verts{n} - 1}\exp\pars{\ic\,{\pi \over 2}\bracks{2\verts{n} - 1}}\ic\,\dd y} ^{\ds{=\ 0}} \\[5mm] - &\ 2\,\Im\int_{\pi}^{-\pi}\ln\pars{\ic\expo{-\ic\theta} \over 2\epsilon} \epsilon^{2\verts{n} - 1}\exp\pars{\bracks{2\verts{n} - 1}\theta\,\ic}\epsilon\expo{\ic\theta}\ic\,\dd\theta \\[5mm] - &\ 2\,\Im\int_{0}^{1}\bracks{\ln\pars{1 - x^{2} \over 2x} + {\pi \over 2}\,\ic}x^{2\verts{n} - 1}\,\dd x \end{align}

In the limit $\ds{\epsilon \to 0^{+}}$: \begin{align} &\int_{0}^{\pi}\ln\pars{\sin\pars{x}}\cos\pars{2nx}\,\dd x = -\,\pi\int_{0}^{1}x^{2\verts{n} - 1}\,\dd x = \bbx{\ds{-\,{\pi \over 2\verts{n}}\,,\quad n \in \mathbb{Z}\setminus\braces{0}}} \end{align}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.