Question

Another trig sum/integral:
\[\int_0^{\pi/2}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}\sin(2n-1)x\,dx\]

Answer 1

Admittedly, the integral part isn't that hard, but the result is interesting.

Answer 2

Wait when you say it's not that hard, is the x inside the sine function or outside it? Like is it _that_ easy? xD

Answer 3

Inside, \(\sin((2k-1)x)\).

Answer 4

Some plots with \(n=1,5,10,100,1000\).

Answer 5

\[\int_0^{\pi/2}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}\sin(2n-1)x\,dx\]\[\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n1}\int_0^{\pi/2}\sin(2n-1)x\,dx\]\[\sum_{n=1}^\infty \frac{(-1)^n}{(2n-1)^2}\cos((2n-1)x)|_0^{ \pi/2}\]\[\sum_{n=1}^\infty \frac{(-1)^n}{(2n-1)^2}[\sin(n \pi) - 1]\]\[\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-1)^2}\]

At first I was like, "Ahh I've seen this before, that's just \(\frac{\pi}{4}\)" But then I realized there's a square in the numerator. I'm guessing it's gonna be \(\frac{\pi}{2}\) since you gave the hint but I still haven't officially got it yet!

Answer 6

Actually the answer isn't in terms of \(\pi\)! But a good start nonetheless.

Answer 7

Ooooh right of course that 1.57... was the upper limit, >_> the thing you're looking at is the area under that curve which looks suspiciously like a logarithm. Well I guess it's time to play around and figure this out haha.

Answer 8

I've also been playing around with the idea that
\[\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}\sin(2n-1)x=\sin x-\frac{1}{3}\sin3x+\frac{1}{5}\sin5x-\frac{1}{7}\sin7x\cdots\]
almost resembles a Fourier sine expansion of something, but no progress with that so far...

Answer 9

Oooh yeah that's a good idea interesting.

Answer 10

Unfortunately, I don't see any way to extract the closed form without prior knowledge of the closed form: http://mathworld.wolfram.com/CatalansConstant.html