# Ways to prove $\int_0^\pi dx \frac{\sin^2(n x)}{\sin^2 x} = n\pi$

In how many ways can we prove the following theorem?

$$I(n):= \int_0^\pi dx \frac{\sin^2(n x)}{\sin^2 x} = n\pi$$ Here $n$ is a nonnegative integer.

The proof I found is by considering $I(n+1)-I(n)$, which can be reduced to $$g(n):= \int_0^\pi dx \frac{\sin(2 n x) \cos x}{\sin x}.$$ I then proved that $g(n)=g(n+1)$, whence $g(n) = g(1) = \pi$. This completes the proof. I was wondering if there is a more direct way to prove it. By 'direct' I mean without deriving auxiliary recursions.

• You could probably use complex analysis to evaluate it too. – dustin Nov 22 '14 at 21:07
• Any statement on natural numbers needs an inductive proof. So I do not see any other proof, which will by pass your "recursion". Of course, you can always be clever enough to hide it in your proof. – Adhvaitha Nov 22 '14 at 21:14
• Note: you can also get down to $g(n)$ simply by integrating by parts. – CuriousGuest Nov 22 '14 at 22:05
• I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. – GNUSupporter 8964民主女神 地下教會 Mar 12 '18 at 17:01

This statement is the same as proving $$\int_0^{2\pi} \frac{\sin^2 nx}{\sin^2 x} dx = 2\pi n.$$

Put $z=e^{ix}$ so that $dz = i e^{ix} \; dx = iz \; dx$ and use $$\sin x = \frac{e^{ix}-e^{-ix}}{2}$$ to get $$\int_{|z|=1} \frac{(z^n-1/z^n)^2}{(z-1/z)^2} \frac{1}{iz} dz = \int_{|z|=1} \frac{1}{z^{2n}} \frac{(z^{2n}-1)^2}{(z-1/z)^2} \frac{z}{iz^2} dz \\ = \int_{|z|=1} \frac{1}{z^{2n}} \frac{(z^{2n}-1)^2}{(z^2-1)^2} \frac{z}{i} dz = \frac{1}{i} \int_{|z|=1} \frac{1}{z^{2n-1}} \frac{(z^{2n}-1)^2}{(z^2-1)^2} dz.$$

Note that $z=\pm 1$ is not really a pole here because it is canceled by the numerator of the rational term.

Observe that $$\frac{(z^{2n}-1)^2}{(z^2-1)^2} = n z^{2n-2} + \sum_{q=0}^{n-2} (q+1) \left(z^{2q} + z^{4n-4-2q}\right).$$

To verify this multiply both sides by $(z^2-1)^2$ to get $$n z^{2n-2} (z^2-1)^2 + \sum_{q=0}^{n-2} (q+1) \left(z^{2q} + z^{4n-4-2q}\right) (z^2-1)^2 \\ = n z^{2n+2} - 2 n z^{2n} + n z^{2n-2} \\ + \sum_{q=0}^{n-2} (q+1) \left(z^{2q+4} - 2 z^{2q+2} + z^{2q} + z^{4n-2q} - 2 z^{4n-2-2q} + z^{4n-4-2q}\right).$$

The first part of the sum is $$\sum_{q=0}^{n-2} (q+1) z^{2q} - 2 \sum_{q=1}^{n-1} q z^{2q} + \sum_{q=2}^n (q-1) z^{2q}$$ which telescopes to give $$1 + 2z^2 - 2z^2 - 2(n-1)z^{2n-2} + (n-2)z^{2n-2} + (n-1) z^{2n} \\ = 1 - n z^{2n-2} + (n-1) z^{2n}.$$ By symmetry we get for the second part of the sum the term $$z^{4n} ( 1 -n z^{2-2n} + (n-1) z^{-2n} ) = z^{4n} - n z^{2n+2} + (n-1) z^{2n}.$$

Adding all three contributions we get $$n z^{2n+2} - 2 n z^{2n} + n z^{2n-2} + 1 - n z^{2n-2} + (n-1) z^{2n} + z^{4n} - n z^{2n+2} + (n-1) z^{2n} \\ = z^{4n} - 2 z^{2n} + 1 = (z^{2n}-1)^2,$$ which concludes the proof.

Returning to the integral we obtain the value $$\frac{1}{i} \times 2\pi i \times \mathrm{Res}_{z=0} \frac{1}{z^{2n-1}} \frac{(z^{2n}-1)^2}{(z^2-1)^2}$$ which is $$\frac{1}{i} \times 2\pi i \times [z^{2n-2}] \frac{(z^{2n}-1)^2}{(z^2-1)^2} = 2\pi n,$$ QED.

Addendum. Actually the above admits considerable simplification. Note that $$\left(\frac{z^n-1}{z-1}\right)^2 = \left(1+z+z^2+\cdots+z^{n-1}\right)^2$$ and therefore $$[z^{n-1}]\left(\frac{z^n-1}{z-1}\right)^2 = \sum_{q=0}^{n-1} 1 \times 1 = n.$$ This immediately yields $$[z^{2n-2}]\left(\frac{z^{2n}-1}{z^2-1}\right)^2= n.$$

• You can use the following to get the coefficient of $\frac{1}{z}$ and it is much easier:\begin{eqnarray} \frac{(1-z^{2n})^2}{(1-z^2)^2}\frac{1}{z^{2n-1}}&=&\frac{1}{z^{2n-1}}\sum_{k=1}^\infty kz^{2(k-1)}(1-2z^{2n}+z^{4n})\\ &=&\sum_{k=1}^\infty kz^{2(k-n)-1}(1-2z^{2n}+z^{4n}). \end{eqnarray} Now you can see that the coefficient of $\frac{1}{z}$ is $n$. Thus – xpaul Dec 3 '14 at 3:02

Here is a cannon to shoot a fly.

Rewrite the integral as $$I(n):= \int_0^\pi\frac{1-\cos2nx}{1-\cos2x}\,dx\stackrel{2x\,\mapsto\, x}\Longrightarrow \frac{1}{2}\int_0^{2\pi}\frac{1-\cos nx}{1-\cos x}\,dx\tag{1}$$ From my answer here, we have $$\int_0^{2\pi}\frac{\cos mx}{p-q\cos x}\, dx=\frac{2\pi}{\sqrt{p^2-q^2}}\left(\frac{p-\sqrt{p^2-q^2}}{q}\right)^m\qquad\hbox{for}\qquad |q|<p\tag{2}$$ Now, we will treat the integral $(1)$ as though it is separated using $(2)$. We must be careful here because each integrals diverge. We set $m=0$, $m=n$, $p=1$, and take the limit as $q\to1^-$, then \begin{align} I(n)&:= \lim_{q\to1^-}\left[\;\frac{\pi}{\sqrt{1-q^2}}-\frac{\pi}{\sqrt{1-q^2}}\left(\frac{1-\sqrt{1-q^2}}{q}\right)^n\;\right]\tag{3} \end{align} The limit above succumbs to apply L'Hôpital's once, then it follows $$I(n):= \int_0^\pi\frac{\sin^2nx}{\sin^2x}\,dx=n\pi$$ which is the announced result.$\qquad\square$

I can post my solution , from my document. It can be found on page 83. If one truly want to shoot this problem with a canon one can use the following generalization $$\int _{0}^{\pi }\! \left( {\frac {\sin \left( nx \right) }{\sin \left( x \right) }} \right)^{m}{dx}=\pi \sum _{l=0}^{\large\left\lfloor {\frac{m\left(n-1\right)}{2n}} \right\rfloor }\left( -1 \right) ^{l}{m\choose l}{\dfrac{m}{2}\left( n+1\right) -ln-1\choose m-1}\tag{10}$$ Which is proved by Graham Hesketh, see equation 10 here. Set $m=1$ use the properties of the floor function and one is is done. Another way to prove $(n)$ can be found in the same answer

Proof

Lemma: Let $k \in \mathbb{Z}$ then \begin{align*} \int_0^{\pi}\frac{\sin 2kx}{\sin x}\mathrm{d}x & = 0 \tag{1} \\ \int_0^{\pi}\frac{\sin (2k-1)x}{\sin x}\mathrm{d}x & = \pi \tag{2} \end{align*}

Proof: We first define the following function $\displaystyle I_n = \int_0^\pi \frac{\sin 2k x}{\sin x} \mathrm{d}x$. Note that we now have $I_0 = 0$ and $I_1=\pi$. since $\sin 0 = 0$ and $\sin 2x = 2\cos x \sin x$. We have the following relation for all $n$ $$I_n-I_{n-2} =\int_0^{\pi}\frac{\sin{nx}-\sin{(n-2)x}}{\sin x}\mathrm{d}x =2\int_0^{\pi}\cos(n-1)x\mathrm{d}x =2\left[\frac{\sin{(n-1)x}}{n-1}\right]_0^{\pi} =0$$ For $|n|\geq 3$. This means $I_{2k}=I_{2k-2}=\cdots=I_{2}=\pi$ and simmilarly $I_{2k+1} = I_{2k-1}=\cdots I_1=0$. Which is what we wanted to show.

Proposition: Let $k \in \mathbb{Z}$ then $$\ell_2(k) = \int_0^\pi \left( \frac{\sin kx}{\sin x} \right)^2\mathrm{d}x = \int_0^\pi \frac{1 - \cos kx}{1 - \cos x}\,\mathrm{d}x = |k|\pi \tag{3}$$

Proof 1: Here we will follow in the footsteps of the other answer and use the lemma above. We define $\displaystyle J_n=\int_0^{\pi}\left(\frac{\sin nx}{\sin x}\right)^2\mathrm{d}x$. We can note that $J_n - J_{n-1}$ is constant eg \begin{align*} J_k - J_{k-1} & = -\frac{1}{2}\int_0^{\pi}\frac{\cos{2kx}-\cos{(2n-2)x}}{(\sin x)^2} = -\frac{1}{2}\int_0^{\pi}\frac{-2\sin\left(\frac{4k-2}{2}x\right) \sin\left(\frac{2x}{2}\right)}{(\sin x)^2} \\ & = \int_0^{\pi}\frac{\sin{(2k-1)x}}{\sin x} = I_{2k-1}=\pi. \end{align*} Then we can write $$J_n=J_{n-1}+\pi=J_{n-2}+2\pi=\cdots=J_1+(n-1)\pi=n\pi.$$ This completes the proof.

Proof 2: Here is another way to solve this problem. I got the idea from chat, but the tecnique is much older. We want to prove that $(I_{n+1} + I_{n-2}) / 2 = I_n$. Eg that $I_n$ is the average of the next and previous term

Some calculations show that

\begin{align*} \frac{I_{n+1}+I_{n-1}}{2} & = \frac{1}{2}\int_0^\pi \frac{1 - \cos(n+1)x }{1 - \cos x} + \frac{1-\cos(n-1)x}{1-\cos x} \,\mathrm{d}x \\ & = \frac{1}{2}\int_0^\pi \frac{2 -\bigl[ \cos(n+1)x + \cos(n-1)x\bigr]}{1-\cos x} \,\mathrm{d}x \\ & = \int_0^\pi \frac{1 - \cos nx \cos x}{1-\cos x} \,\mathrm{d}x \\ & = \int_0^\pi \frac{(1-\cos nx) + (1-\cos x) \cos nx}{1-\cos x} \,\mathrm{d}x \\ & = \int_0^\pi \frac{1 - \cos nx}{1-\cos x} \,\mathrm{d}x = I_n \end{align*} Several things was used here. Like $\cos (n+1)x + \cos(n-1)x = 2 \cos nx \cos x$ and $\int_0^\pi \cos n x\,\mathrm{d}x = 0 \ \forall \ n \in\mathbb{Z} \backslash \{0\}$. Now we have shown that $$I_n = \frac{I_{n+1}+I_{n-1}}{2} \ \Rightarrow \ 2I_n = I_{n+1}+I_{n-1} \ \Rightarrow \ I_n - I_{n-1} = I_{n+1} - I_n$$ Which is just an arithmetric sequence because the difference between two terms is constant (same as the previous answer). Hence $I_n = I_0 + (n-0)d = nd$, where $d$ is the difference between two terms. We have $d = I_1 - I_0 = \pi - 0 = \pi$. This completes the second proof. $\qquad\square$

Recall that the Féjer kernel $F_n(t)$ is defined to be the mean of the Dirichlet kernel $$D_n(t)=\sum_{|k|\leqslant n}e^{ikt}$$

It is known $F_n(2t)=\dfrac 1 n\dfrac{\sin^2 nt}{\sin^2 t}$. This is a simple use of the geometric series. It is also known that for any $2\pi$-periodic function $f:S^1\to\Bbb R$, the means of the Fourier partial sums of $f$ are given by $$F_n\star f(s)=\frac{1}{2\pi}\int_0^{2\pi} F_n(t)f(s-t)dt$$

This is verified by using the definition of the $n$-th Fourier coefficient.

If we let $f$ be the function constantly equal to $1$, we get the partial sums all equal to $1$, so $$1=\frac{1}{2\pi}\int_0^{2\pi}F_n(t)dt$$

Here I prove $\displaystyle\int_0^{2\pi}\frac{\sin^2(2n+1)x}{\sin^2x}dx=2\pi(2n+1).$

Since $\displaystyle\sum_{j=1}^n \cos 2jx = -\frac{1}{2}+\frac{\sin(2n+1)x}{2\sin(x)}$, we have that $\displaystyle1+\sum_{j=1}^n \cos 2jx = \frac{\sin(2n+1)x}{\sin(x)}$.

Now take both sides to the power of $2$ and integrate term-by-term. \begin{align}I&=\int_0^{2\pi}\left(1+2\sum_{j=1}^n \cos 2jx\right)^2dx\\&=2\pi+4\sum_{j=1}^n \int_0^{2\pi}\cos^2 2jx\,dx+4\sum\sum_{j\neq k} \int_0^{2\pi}\cos 2jx\cos 2kx\,dx+4\sum_{j=1}^n \int_0^{2\pi}\cos 2jx\,dx\end{align} Therefore $\displaystyle\int_0^{2\pi}\frac{\sin^2(2n+1)x}{\sin^2x}dx=2\pi(2n+1).$


$\ds{\,{\rm I}\pars{n}\equiv\int_{0}^{\pi}{\sin^{2}\pars{nx} \over \sin^{2}\pars{x}}\,\dd x = \verts{n}\pi:\ {\large ?}\,,\qquad n \in {\mathbb Z}}$.

\begin{align} &\bbox[10px,#ffe]{\ds{\int_{0}^{\pi}{\sin^{2}\pars{nx} \over \sin^{2}\pars{x}}\,\dd x}} =\int_{0}^{\pi}{1 - \cos\pars{2nx} \over 1 - \cos\pars{2x}}\,\dd x =\half\int_{0}^{2\pi}{1 - \cos\pars{nx} \over 1 - \cos\pars{x}}\,\dd x \\[5mm]&=\half\,\Re\int_{0}^{2\pi} {1 + \ic\verts{n} x- \expo{\ic\verts{n}x} \over 1 - \cos\pars{x}}\,\dd x =\half\,\Re\oint_{\verts{z}\ =\ 1} {1 + \verts{n}\ln\pars{z} - z^{\verts{n}} \over 1 - \pars{z^{2} + 1}/\pars{2z}} \,{\dd z \over \ic z} \\[5mm]&=-\,\Im \oint_{\verts{z}\ =\ 1\atop{\vphantom{\Huge A}0\ <\ \,{\rm Arg}\pars{z}\ <\ \pi}} {1 + \verts{n}\ln\pars{z} - z^{\verts{n}} \over \pars{1 - z}^{2}}\,\dd z \\[5mm]&=\Im\braces{\!\!% \int_{-1}^{0}\!{1 + \verts{n}\bracks{\ln\pars{-x} + \ic\pi} - x^{\verts{n}} \over \pars{x - 1}^{2}}\,\dd x +\int_{0}^{-1}\!{1 + \verts{n}\bracks{\ln\pars{-x} - \ic\pi} - x^{\verts{n}} \over \pars{x - 1}^{2}}\,\dd x\!\!} \\[5mm]&=\Im\bracks{2\verts{n}\pi\ic\int_{-1}^{0}{\dd x \over \pars{x - 1}^{2}}} =2\verts{n}\pi\bracks{-\,{1 \over x - 1}}_{-1}^{0} =2\verts{n}\pi\pars{1 - \half} =\ \bbox[10px,#ffe,border:1px solid #000]{\ds{\verts{n}\pi}} \end{align}

As $\sin^2(n+1)x-\sin^2nx=\sin x\sin(2n+1)x,$

$$J(n)=I(n+1)-I(n)=\int_0^\pi\frac{\sin(2n+1)x}{\sin x}dx$$

As $\sin(2m+1)x-\sin(2m-1)x=2\sin x\cos2m x,$

Again, $$J(m)-J(m-1)=2\int_0^\pi\cos2mx\ dx$$

Now for $m\ne0,$ $$\int_0^\pi\cos2mx\ dx=\frac{\sin2mx}{2m}\mid_0^\pi=0$$

$$\implies J(m)=J(m-1)=\cdots=J(0)=\int_0^\pi\frac{\sin(2\cdot0+1)x}{\sin x}dx=(\pi-0)$$

$\implies I(n+1)-I(n)=\pi$ for $n>0$

and $I(1)=\cdots=\pi$

• Wish I know the mistake here that has caused the down-vote – lab bhattacharjee Nov 23 '14 at 9:55
• The downvote (not me) is probably because the method is identical to the one the OP purposed, asking for other methods. – nbubis Nov 23 '14 at 13:04