Compute $\lim_{n\to\infty} \left(\int_0^{\pi} \frac{\sin^2 n x}{\sin x} \ dx-\sum_{k=1}^n \frac{1}{k}\right)$

Question

Compute the limit $$\lim_{n\to\infty} \left(\int_0^{\pi} \frac{\sin^2 n x}{\sin x} \ dx-\sum_{k=1}^n \frac{1}{k}\right)$$

Answer 1

Let's suppose that $\ \displaystyle f(n):=\int_0^{\pi} \frac{\sin^2 n x}{\sin x} \ dx\ $ then : \begin{align} f(n+1)-f(n)&=\int_0^{\pi} \frac{\sin^2((n+1) x)-\sin^2(n x)}{\sin x} \ dx\\ &=\int_0^{\pi} \frac{\cos(2n x)-\cos(2(n+1) x)}{2\sin x} \ dx\\ &=\int_0^{\pi} \frac{\cos(2n x)(1-\cos(2x))+\sin(2nx)\sin(2x)}{2\sin x}dx\\ &=\int_0^{\pi} \frac{\cos(2n x)2\sin(x)^2)+\sin(2nx)2\sin(x)\cos(x)}{2\sin x} dx\\ &=\int_0^{\pi} \cos(2n x)\sin(x)+\sin(2nx)\cos(x)\;dx\\ &=\int_0^{\pi} \sin((2n+1) x)\;dx\\ &=\frac 2{2n+1} \\ \end{align}

So that your limit (as $\ (n+1)\to +\infty$) is the series : $$f(0)+\sum_{n=0}^\infty \left(\frac 2{2n+1}-\frac 1{n+1}\right)=2\sum_{n=0}^\infty \left(\frac 1{2n+1}-\frac 1{2n+2}\right)=2\,\log(1+1)=\log(4)$$ (using the expansion of $\;\log(1+x)\,$ at $\,x=1$)

Answer 2

Lemma: $\int_{0}^{\pi} e^{(2n+1)ix} dx = \frac{-2}{i}(2n+1)$.

Now let $t=e^{ix}$. We'll simplify the integrand: $$\frac{\sin^2(nx)}{\sin x} = (\frac{t^{n}-t^{-n}}{2i})^2 (\frac{t-t^{-1}}{2i})^{-1}$$ $$=\frac{i}{-2} \frac{t^{2n+1}+t^{1-2n}-2t}{t^2-1} = \frac{i}{-2} t \frac{t^{2n}-1}{t^2-1}(1-t^{-2n})$$ $$=\frac{i}{-2} \sum_{i=0}^{n-1} (t^{2i+1} - t^{2(i-n)+1})$$

Now, by applying the lemma, we find that the integral is $$\frac{i}{-2}\frac{-2}{i} \sum_{i=0}^{n-1} \frac{1}{2i+1} - \frac{1}{2(i-n)+1}=$$ $$=2 \sum_{i=0}^{n-1} \frac{1}{2i+1} =2(H_{2n}-\frac{1}{2}H_n)$$

So the limit in question is $\lim_{n \to \infty} 2(H_{2n}-\frac{1}{2}H_n) - H_{n} = \lim 2H_{2n}-2H_{n}$. Now we can either use $H_{n} \sim \log n + \gamma + O(n^{-1})$ or compare $H_{2n}-H_{n}$ to the integral $\int_{n}^{2n} \frac{dt}{t} = \ln 2$ to conclude that the limit is $\ln 4$.

Answer 3

The point is that you can write $\sum_{k=1}^nsin(2k-1)x=\frac{cos(2nx)-1}{-2sinx}=\frac{sin^2(nx)}{sinx}$, so the integral is equal to $I=lim_{n\to\infty}\sum_{k=1}^n(\int_{0}^\pi(sin(2k-1)xdx-\frac{1}{k})=lim_{n\to\infty}\sum_{k=1}^n(\frac{2}{2k-1}-\frac{1}{k})$, then using $\sum_{k=1}^{n}\frac{1}{k}\sim log(n)+c$, $c$ is the Euler constant, so $I=lim_{n\to\infty}2[\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{2k}]-\sum_{k=1}^n\frac{1}{k}=lim_{n\to\infty}(2log2n+2c-logn-c-logn-c)=2log2$.

Answer 4

By classical methods (although they are considered known sums) we find: $$\displaystyle{\left\{ \begin{array}{l} \cos 2x + \cos 4x + .. + cos2nx = \dfrac{{\sin nx \cdot \cos \left( {n + 1} \right)x}}{{\sin x}}\\ \\ \sin 2x + \sin 4x + .. + \sin 2nx = \dfrac{{\sin nx \cdot \sin \left( {n + 1} \right)x}}{{\sin x}} \end{array} \right\}}$$ Then $$\displaystyle{\frac{{{{\sin }^2}nx}}{{{{\sin }^2}x}} = {\left( {\cos 2x + \cos 4x + .. + cos2nx} \right)^2} + {\left( {\sin 2x + \sin 4x + .. + \sin 2nx} \right)^2}}$$

Consequently $$\displaystyle{\frac{{{{\sin }^2}nx}}{{{{\sin }^2}x}} = n + 2\sum\limits_{k = 1}^{n - 1} {\sum\limits_{m = k + 1}^n {\left( {\cos 2kx \cdot \cos 2mx + \sin 2kx \cdot \sin 2mx} \right)} } = n + 2\sum\limits_{k = 1}^{n - 1} {\sum\limits_{m = k + 1}^n {\cos \left( {2k - 2m} \right)x} } }$$

consequently $$\displaystyle{\frac{{{{\sin }^2}nx}}{{\sin x}} = n \cdot \sin x + 2\sum\limits_{k = 1}^{n - 1} {\sum\limits_{m = k + 1}^n {\left( {\sin x \cdot \cos \left( {2k - 2m} \right)x} \right)} } = n \cdot \sin x + }$$ $$\displaystyle{\sum\limits_{k = 1}^{n - 1} {\sum\limits_{m = k + 1}^n {\left( {\sin \left( {2k - 2m + 1} \right)x - \sin \left( {2k - 2m - 1} \right)x} \right)} } }$$

Then $$\displaystyle{\int\limits_0^\pi {\frac{{{{\sin }^2}nx}}{{\sin x}}dx} = n \cdot \int\limits_0^\pi {\sin x\;dx} + \sum\limits_{k = 1}^{n - 1} {\sum\limits_{m = k + 1}^n {\left( {\int\limits_0^\pi {\sin \left( {2k - 2m + 1} \right)x\;dx} - \int\limits_0^\pi {\sin \left( {2k - 2m - 1} \right)x\;dx} } \right)} } = }$$

$$\displaystyle{ = 2n + \sum\limits_{k = 1}^{n - 1} {\sum\limits_{m = k + 1}^n {\left( { - \frac{{\left[ {\cos \left( {2k - 2m + 1} \right)x} \right]_0^\pi }}{{2k - 2m + 1}} + \frac{{\left[ {\cos \left( {2k - 2m - 1} \right)x} \right]_0^\pi }}{{2k - 2m - 1}}} \right)} } } $$ $$\displaystyle{ = 2n + 2\sum\limits_{k = 1}^{n - 1} {\sum\limits_{m = k + 1}^n {\left( {\frac{1}{{2k - 2m + 1}} - \frac{1}{{2k - 2m - 1}}} \right)} } = }$$

$$\displaystyle{ = 2n + 2\sum\limits_{k = 1}^{n - 1} {\sum\limits_{m = 1}^{n - k} {\left( {\frac{1}{{2m + 1}} - \frac{1}{{2m - 1}}} \right)} } = 2n + 2\sum\limits_{k = 1}^{n - 1} {\left( {\frac{1}{{2n - 2k + 1}} - 1} \right)} = 2 + \sum\limits_{k = 1}^{n - 1} {\frac{1}{{n - k + \frac{1}{2}}}} = }$$

$$\displaystyle{ = 2 + \sum\limits_{k = 1}^{n - 1} {\frac{1}{{k + \frac{1}{2}}}} = 2 + 2\sum\limits_{k = 1}^{n - 1} {\frac{1}{{2k + 1}}} = 2 - \frac{2}{{2n + 1}} + 2\sum\limits_{k = 1}^n {\frac{1}{{2k + 1}}} }$$

Eventually $$\displaystyle{\int\limits_0^\pi {\frac{{{{\sin }^2}nx}}{{\sin x}}dx} - {H_n} = 2 - \frac{2}{{2n + 1}} + 2\sum\limits_{k = 1}^n {\left( {\frac{1}{{2k + 1}} - \frac{1}{{2k}}} \right)} = 2\sum\limits_{k = 1}^{2n} {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} }$$

Consequently $$\displaystyle{\mathop {\lim }\limits_{n \to \infty } \left( {\int\limits_0^\pi {\frac{{{{\sin }^2}nx}}{{\sin x}}dx} - {H_n}} \right) = 2\sum\limits_{k = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} = 2 \cdot \log 2}$$ . :)

P.S. Maybe the result $$\displaystyle{\int\limits_0^\pi {\frac{{{{\sin }^2}nx}}{{\sin x}}dx} = 2 - \frac{2}{{2n + 1}} + 2\sum\limits_{k = 1}^n {\frac{1}{{2k + 1}}} }$$ lies in an easier way.. I didn't look at it..