Help with fourier transform for special square wave

xanthium
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I know how to describe a square wave with Fourier analysis, but what if I'm looking for a square wave with "peaks" that are longer than the "valleys."
For example, from f(x)=1 {from 0 to 2}, f(x)=-1 {from 2 to 3}, f(x)=1 {from 3 to 5}, f(x)=-1 {from 5 to 6}... and so on in a periodic fashion.
Any ideas?
Thanks
 
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That's not so bad- first, I'll rectify the wave and re-normalize it to make the functions easier to type.

Then, the wavetrain you describe is rect(x/2)*comb(x/4), where '*' is convolution. I'm pretty sure that will reproduce the function you describe. The Fourier transform is then Sinc(2q)*Sinc(4q), where sinc(x) = sin(x)/x, and q is the conjugate variable to x.

Tweak the original function as needed.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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