## anonymous one year ago Another trig sum/integral: $\int_0^{\pi/2}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}\sin(2n-1)x\,dx$

1. anonymous

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

2. Empty

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

3. anonymous

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

4. anonymous

Some plots with $$n=1,5,10,100,1000$$.

5. Empty

$\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!

6. anonymous

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

7. Empty

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.

8. anonymous

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...

9. Empty

Oooh yeah that's a good idea interesting.

10. anonymous

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