Prove $\int_{-\infty}^{\infty}{e^{itx}\frac{1-\cos{x}}{x^2}dx}=\frac{1}{2}(|t+1|+|t-1|-2|t|)\int_{-\infty}^{\infty}{\frac{1-\cos{x}}{x^2}dx}$

Question

This is a japanese graduate school’s entrance examination problem. Sorry, I have NO idia at all. Please teach me how to prove the equation.

$\frac{1}{2}(|t+1|+|t-1|-2|t|) = 0,(|t|>1)$

This is what I know about this question.

Answer 1

We use the formula $$ \int_0^\infty \frac{\sin (tx)}{x}dx = \frac \pi 2,\quad t >0. $$ Define $$ g(t) = \int_{-\infty}^{\infty} e^{i tx} \frac{1 -\cos x}{x^2} dx. $$ We have $g(t) = g(-t)$. Hence, it is enough to consider $t \geq 0$. For $t \geq 0$, we have $$ g(t) = \int_{-\infty}^\infty \cos(tx) \frac{1-\cos x}{x^2} dx = 2 \int_0^\infty \cos(tx) \frac{1-\cos x}{x^2} dx.$$ By integration by parts, $$g(t) = -2\int_0^\infty \cos(tx) (1-\cos x) \left(\frac 1x\right)'dx = -2t\int_0^\infty \frac{\sin(tx) (1 -\cos x)}x dx + 2 \int_0^\infty \frac{\cos(tx) \sin x} xdx.$$ Using the formula $$\sin(tx) \cos x = \sin((t+1)x) + \sin((t-1)x), \quad \cos(tx) \sin(x) = \sin((t+1)x) + \sin((1-t)x),$$ and the integral above, we get the result.