$$ \int_{-\infty}^\infty\frac{1}{\alpha}(1-\cos\alpha\pi)e^{i\alpha x}\,\mathrm{d}\alpha $$
Solving this integral is part of solving for a Fourier transformation problem and I am stuck on this integral. I tried treating this integral like another integration by parts but it wasn't doing much to simplify the integral any further.
I separated the integral into two by distributing the ($1/\alpha$)
I believe we can solve this by consulting a Fourier Transform table and using some basic properties of the Fourier Transform. We rewrite the integral as $$I(\omega) = \sqrt{2\pi}F(\omega) = \int_{-\infty}^{\infty}\frac{1}{t}(1-\cos\pi t)e^{i\omega t}dt.$$ Then we have $$F(\omega) = \mathcal{F}\left\{\frac{1}{t}(1-\cos\pi t)\right\}(\omega) = \mathcal{F}\{f(t)\}(\omega).$$ We can then use the property $$F'(\omega) = \mathcal{F}\{itf(t)\}(\omega)$$ for any Fourier transform pair $f,F$. We then have (From Wolfram Alpha)
$$I'(\omega) = \sqrt{2\pi}i\mathcal{F}\{1-\cos\pi t\}(\omega) = 2\pi i\delta(\omega)-\pi i\delta(\omega-\pi)-\pi i\delta(\omega+\pi).$$
Integrating this and requiring that our function goes to 0 at infinity, we have
$$I(\omega) = \pi i[2H(x)-H(x+\pi)-H(x-\pi)].$$ I'll be the first to admit that this is not my area of expertise so let me know if I messed up.