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

1. anonymous

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

2. anonymous

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

3. anonymous

(I suppose that qualifies as a hint)

4. anonymous

*area under the sum of sines

5. Empty

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

6. Empty

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

7. anonymous

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).

8. anonymous

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

9. anonymous

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 $$\lim_{n\to\infty}\int_{\pi/4}^{3\pi/4}\frac{1-\cos(2nx)}{2\sin x}\, dx=\int_{\pi/4}^{3\pi/4}\frac1{2\sin x}\, dx=\frac12\int_{\pi/4}^{3\pi/4}\csc x\, dx$$

10. anonymous

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 $$\int_{\pi/4}^{3\pi/4}\frac1{2\sin x}\, dx=\frac12\log(3+2\sqrt2)\approx 0.88137$$

11. anonymous

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.

12. anonymous

you can write the series in a closed form using digamma functions, if that helps: https://en.wikipedia.org/wiki/Digamma_function#Series_formula

13. anonymous

Ah, another thing to learn, cool. Thanks.