# Ways to show $\int_0^{\infty}\frac{\sin^2(\pi x)}{x^2}\Big\lvert x-\Big\lfloor x +\frac12 \Big\rfloor \Big\rvert \, \mathrm{d}x = \frac{\pi^2}{8}$?

Whilst reading about Lobachevsky's integral formula I tried constructing some interesting integrals which could be evaluated with said formula. One result I found was

$$\int\limits_{0}^{\infty} \frac{\sin^2(\pi x)}{x^2}\left\lvert x - \left\lfloor x + \frac12 \right\rfloor \right\rvert \, \mathrm{d}x = \frac{\pi^2}{8}$$

which under substitution $$u = \pi x$$ can be evaluated with Lobachevsky's formula since $$\int_0^\frac\pi2 \Big\lvert \frac{x}{\pi} - \Big\lfloor \frac{x}{\pi} + \frac12 \Big\rfloor \Big\rvert \, \mathrm{d}x =\int_0^\frac\pi2 \frac{x}{\pi}\, \mathrm{d}x =\frac{\pi}{8}$$.

However, once I had shown this result I remembered that $$\frac{\pi^2}{8}$$ has the very nice series representation $$\sum_{n\ge0} \frac{1}{(2n+1)^2} = \frac{\pi^2}{8} \tag{1}$$ So my question is

Is there a way to evaluate this integral using $$(1)$$? Or is it just a coincidence that the values match?

## 1 Answer

The triangle wave involved in the integral has a simple Fourier series:

$$\left|x-\left\lfloor x+\frac{1}{2}\right\rfloor\right|=\frac{1}{4}-\frac{2}{\pi^2}\sum_{n\geq 0}\frac{\cos((2n+1)2\pi x)}{(2n+1)^2} \tag{1}$$ and since

$$\int_{0}^{+\infty}\frac{\sin^2(\pi x)}{x^2}\,dx = \frac{\pi^2}{2},\qquad \int_{0}^{+\infty}\frac{\sin^2(\pi x)}{x^2}\,\cos((2n+1)2\pi x)\,dx = 0\tag{2}$$ the integral converges to $$\frac{\pi^2}{2}$$ times the average value of the triangle wave.
On the other hand, since the triangle wave is piecewise linear, such average value is exactly half the value at $$x=\frac{1}{2}$$. Since the RHS of $$(1)$$ is absolutely convergent the evaluation at $$x=\frac{1}{2}$$ leads to $$\sum_{n\geq 0}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}\tag{3}$$ i.e. to a proof of the Basel problem, independent from Lobachevsky's integral formula.