Solved it Myself to Self-Help Problem Solving

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figured it out myself, thanks
 
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Normally, "the" function is the unit step function- u(x)= the largest integer less than or equal to x. As for finding its Fourier series, it is not periodic. I assume you are restricting it to -\pi to \pi. Now, just use the standard integral formulas for the Fourier coefficients. Because the step function is constant over each integer interval, you will just be integrating sine and cosine over 8 intervals.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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