Question

First of a series of trig integrals: Show that
\[\lim_{n\to\infty}\int_{\pi/4}^{3\pi/4}\sum_{k=1}^n\sin(2k-1)x\,dx=\ln\left(\cot\frac{\pi}{8}\right)\]

Answer 1

The next one looks a bit tougher, I'll post it once this one's resolved.

Answer 2

By the way, this problem's inspired by the observation that the sum of sines is bounded by a single cosecant:
http://www.wolframalpha.com/input/?i=Plot%5B%7BSum%5BSin%5B%282k-1%29x%5D%2C%7Bk%2C1%2C15%7D%5D%2C+Csc%5Bx%5D%7D%2C+%7Bx%2C0%2CPi%7D%5D

Answer 3

(I suppose that qualifies as a hint)

Answer 4

*area under the sum of sines

Answer 5

Oooh interesting this reminds me of how phase velocity is related to group velocity.

Answer 6

Oh so wait is this finding the coefficients on the fourier series representation of this function?

Answer 7

Fourier series of which function? I'm not seeing the connection, but that's not to say there isn't one. This is just a problem I came up with after working with the limit,
\[\lim_{x\to0}\frac{\sum\limits_{k=1}^n\sin(2k-1)x}{\sum\limits_{k=1}^n\sin 2kx}\]
(a neat problem on its own).

Answer 8

consider \(\sin a\sin b=\frac12\left[\cos(a-b)-\cos(a+b)\right]\) so we have $$\sin x\sin((2k-1)x)=\frac12\left[\cos((2k-2)x)-\cos(2kx)\right]\\\implies\sin x\sum_{k=1}^n\sin((2k-1)x)=\frac12\left[1-\cos(2nx)\right]\\\implies\sum_{k=1}^n\sin((2k-1)x)=\frac{1-\cos(2nx)}{2\sin x}$$

Answer 9

now it's tempting to say we could take the limit now, even though strictly that's not true, but intuitively as \(n\to\infty\) the frequency of \(\cos(2nx)\) grows without bound and we get that it oscillates wildly and on a given compact interval (like \((\pi/4,3\pi/4)\)) will 'average out' in the integral to \(0\), giving $$\lim_{n\to\infty}\int_{\pi/4}^{3\pi/4}\frac{1-\cos(2nx)}{2\sin x}\, dx=\int_{\pi/4}^{3\pi/4}\frac1{2\sin x}\, dx=\frac12\int_{\pi/4}^{3\pi/4}\csc x\, dx$$

Answer 10

where we know $$\int\csc x\, dx=-\log|\csc x+\cot x|+C$$ so $$-\log|\csc(3\pi/4)+\cot(3\pi/4)|+\log|\csc(\pi/4)+\cot(\pi/4)|\\\qquad=-\log|\sqrt2-1|+\log|\sqrt2+1|\\\qquad=\log\frac{\sqrt2+1}{\sqrt2-1}\\\qquad=\log(3+2\sqrt2)$$ so we find $$\int_{\pi/4}^{3\pi/4}\frac1{2\sin x}\, dx=\frac12\log(3+2\sqrt2)\approx 0.88137$$

Answer 11

I really like your approach. My attempt was a bit taxing by comparison. Now, I'm not entirely clear on the necessary conditions for being able to do so, but to start with I tried swapping the order of summation/integration. (I'm not fluent in measure theory, but I *think* this function satisfies Fubini's theorem.) I get
\[\begin{align*}
\lim_{n\to\infty}\int_{\pi/4}^{3\pi/4}\sum_{k=1}^n\sin(2k-1)x&=\sum_{k=1}^\infty \int_{\pi/4}^{3\pi/4}\sin(2k-1)x\,dx\\[1ex]
&=\sum_{k=1}^\infty \frac{\cos\dfrac{(2k-1)\pi}{4}-\cos\dfrac{(2k-1)3\pi}{4}}{2k-1}\\[1ex]
&=\frac{2}{\sqrt2}\sum_{k=1}^\infty \frac{\cos\dfrac{k\pi}{2}+\sin\dfrac{k\pi}{2}}{2k-1}
\end{align*}\]
Alternating terms in the numerator disappear according to this pattern:
\[\begin{array}{c|cccc|cccc|c}
k&1&2&3&4&5&6&7&8&\cdots\\
\hline
\cos&0&-1&0&1&0&-1&0&1&\cdots\\
\sin&1&0&-1&0&1&0&-1&0&\cdots
\end{array}\]
so I can write the sum as
\[\sqrt2\sum_{k=1}^\infty \left(-\frac{1}{8k-5}+\frac{1}{8k-1}+\frac{1}{8k-7}-\frac{1}{8k-3}\right)\]
I'm trying to use to rewrite each term in terms of harmonic numbers, but I'm failing to wrap my mind around the proper manipulation.

Answer 12

you can write the series in a closed form using digamma functions, if that helps:

https://en.wikipedia.org/wiki/Digamma_function#Series_formula

Answer 13

Ah, another thing to learn, cool. Thanks.