$\lim_{n \to \infty} \int_0^{\frac{\pi}{2}} \sum_{k=1}^{n} \left( \frac {\sin kx}{k} \right)^2 \, \mathrm{d}x$

Here is a problem in calculus shared by a friend. Compute $$\lim_{n \to \infty} \displaystyle\int_{0}^{\frac{\pi}{2}} \displaystyle\sum_{k=1}^{n} \left( \frac {\sin kx}{k} \right)^2 \, \mathrm{d}x.$$ It is not trivial to find the partial sum. Can we pull the limit inside to get $$\displaystyle\int_{0}^{\frac{\pi}{2}} \displaystyle\sum_{k=1}^{\infty} \left( \frac {\sin kx}{k} \right)^2 \, \mathrm{d}x?$$Still, this is hard.

• You can express $\sin^2 kx$ in terms of $\cos 2kx$ and a constant term. Separate them, and use integration by parts. – Math.StackExchange Dec 17 '14 at 12:19
• Very nice problem; answer posted and +1! @AranKomatsuzaki No need for integration by parts, although it'd be nice to see an alternate solution. If you find another solution, please feel free to post as an answer. :) – Ahaan S. Rungta Dec 17 '14 at 12:27
• The question is not clear. Is it $x$ or $k$ in the denominator? – Pranav Arora Dec 17 '14 at 12:31
• @PranavArora Good point. Hopefully it is $k$ on the denominator. – Ahaan S. Rungta Dec 17 '14 at 12:47
• Yes, it is k on the denominator. Sorry about that! – Spinning Turntable Dec 17 '14 at 13:14

Lemma: $\displaystyle\int_{0}^{\frac{\pi}{2}} \sin^2 kx \, \mathrm{d}x = \frac {\pi}{4}$, for any integer $k$.

Proof. First note that $\sin^2 (kx) = \frac {1 - \cos (2kx)}{2}$, so our integral is \begin {align*} \frac {1}{2} \cdot \displaystyle\int_{0}^{\frac {\pi}{2}} \left( 1 - \cos (2kx) \right) \, \mathrm{d}x &= \frac {1}{2} \cdot \left[ x - \frac {\sin (2kx)}{2k} \right]_{0}^{\frac {\pi}{2}} \\&= \frac {1}{2} \cdot \left( \frac {\pi}{2} - \frac {\sin \left( \pi k x \right)}{2k} \right) \\&= \frac {1}{2} \cdot \left( \frac {\pi}{2} - 0 \right) \\&= \frac {\pi}{4}, \end {align*}as desired. $\Box$

\begin {align*} \displaystyle\int_{0}^{\frac{\pi}{2}} \displaystyle\sum_{k=1}^{n} \left( \frac {\sin kx}{k} \right)^2 \, \mathrm{d}x &= \displaystyle\sum_{k=1}^{n} \displaystyle\int_{0}^{\frac{\pi}{2}} \left( \frac {\sin kx}{k} \right)^2 \, \mathrm{d}x \\&= \displaystyle\sum_{k=1}^{n} \left( \frac {1}{k^2} \cdot \displaystyle\int_{0}^{\frac{\pi}{2}} \sin^2 kx \, \mathrm{d}x \right) \\&= \frac {\pi}{4} \cdot \displaystyle\sum_{k=1}^{n} \frac {1}{k^2}. \end {align*}Taking the limit, our answer is \begin {align*} \frac {\pi}{4} \cdot \lim_{n \to \infty} \displaystyle\sum_{k=1}^{n} \frac {1}{k^2} &= \frac {\pi}{4} \cdot \displaystyle\sum_{k=1}^{\infty} \frac {1}{k^2} \\&= \dfrac {\pi}{4} \cdot \dfrac {\pi^2}{6} \\&= \boxed {\dfrac {\pi^3}{24}}. \end {align*}

EDIT: obsolete. The limit I was discussing was not the one intended by OP.

This diverges to infinity. Let $M \in \mathbb{R}_+$ and $I = \int_0^{\pi/2} \frac{\sin \, x}{x}dx$:

\begin{eqnarray} \int_0^{\pi/2} \sum_{k=1}^n \left(\frac{\sin\, kx}{x} \right)^2 dx & = & \sum_{k=1}^n \int_0^{\pi/2} \left(\frac{\sin\, kx}{x} \right)^2 dx \; \;\mbox{because finite sum} \\ & = & \sum_{k=1}^n k^2 \int_0^{\pi/2} \left(\frac{\sin\, kx}{kx} \right)^2 dx \\ & = & \sum_{k=1}^n k \int_0^{k\pi/2} \left(\frac{\sin\, x}{x} \right)^2 dx \\ & \geq & \sum_{k=1}^n k I \end{eqnarray}

So, with $k$ large enough, your integral of sum is superior to $M$

• +1 although the OP meant the denominator to be $k$, not $x$. – Ahaan S. Rungta Dec 17 '14 at 13:38

Let $$I_n(a)=\int_0^{\frac{\pi}{2}}\sum_{k=1}^n\left(\frac{\sin(akx)}{k}\right)^2dx.$$ Then $I_n(0)=0$ and \begin{eqnarray} I_n'(a)&=&\int_0^{\frac{\pi}{2}}\sum_{k=1}^n\frac{x\sin(2akx)}{k}dx\\ &=&\sum_{k=1}^n\frac{-ak\pi\cos(ak\pi)+\sin(akx)}{4a^2k^3}\\ &=&-\frac{d}{da}\sum_{k=1}^n\frac{\sin(ak\pi)}{4ak^3} \end{eqnarray} and hence \begin{eqnarray} \lim_{n\to\infty}I_n(1)&=&-\sum_{k=1}^n\frac{\sin(ak\pi)}{4ak^3}\bigg|_{0}^1\\ &=&\frac{\pi}{4}\sum_{k=1}^n\frac{1}{k^2}\\ &=&\frac{\pi^3}{24}. \end{eqnarray}