Fourier transform of $(1-\cos(tx))/x^2$ I am trying to compute the Fourier transform of $f(x) = \frac{1-\cos(tx)}{x^2}$, $(t > 0)$ directly. I tried contour integration, and could not seem to get it to work. So, I am wondering if it can be done in this way, or, even better, if there is a simpler way to do it. 
 A: There is another way, if you are OK working with a FT that you might know.
Use the half-angle formula to write
$$\hat{f}(\omega) = \int_{-\infty}^{\infty} dx \frac{1-\cos{t x}}{x^2} e^{i \omega x} = 2 \int_{-\infty}^{\infty} dx \frac{\sin^2{(t x/2)}}{x^2} e^{i \omega x}$$
Substitute $u = t x/2$ to get
$$\hat{f}(\omega) = t \int_{-\infty}^{\infty} du \frac{\sin^2{u}}{u^2} e^{i (2 \omega/t) u}$$
So we are looking at the FT of $\sin^2{u}/u^2$.  Let's assume you do not know this, but that you do know that
$$\int_{-\infty}^{\infty} du \frac{\sin{u}}{u} e^{i y u} = \pi \, \text{rect}(y) = \begin{cases}\\ \pi & |y| \lt 1 \\ 0 & |y| > 1 \end{cases}$$
The FT of $\sin^2{u}/u^2$ is evaluated using the convolution theorem.  Here, we convolve the function $\text{rect}(y)$ with itself:
$$\int_{-\infty}^{\infty} du \frac{\sin^2{u}}{u^2} e^{i y u} = \frac{\pi}{2} \int_{-\infty}^{\infty} dy' \, \text{rect}(y')\, \text{rect}(y-y')$$
The latter integral is evaluated directly: take the product of the area of two rectangles.  One rectangle ($\text{rect}(y')$) is fixed about the origin, the other ($\text{rect}(y-y')$) is centered at $y'=y$.  Draw a picture: it should be clear that this product, the convolution, is zero when $|y| \gt 2$.  For the rest, just calculate the area of the overlap between the two rectangles.  The result is, using the original expression for $\hat{f}(\omega)$ above:
$$\hat{f}(\omega) = \begin{cases}\ \pi (t-|\omega|) & |\omega| \lt t\\ 0 &  |\omega| \gt t\end{cases}$$
A: Here is what I suggested, with unsuccessfully as it turned out (see also the comments below): Consider $$\oint_C \frac{1-e^{itz}}{z^2}e^{i\omega z}\,dz=0$$ where the contour $C$ is composed of the real intervals $[-R,-r]$ and $[r,R]$ joined by two semicircles in the upper halfplane centered at $0$ and with radii $r$ and $R$. Provided that $\omega>0$ (which is all you need) you will find that the integral over the large semicircle vanishes in the limit as $R\to\infty$. You need to be more careful about the small semicircle as $r\to0$, but I expect you can fill that in for yourself. (It will not vanish, of course, or else $\hat f$ would vanish.)
Now the integral around the small semicircle works out to be $-\pi t$ in the limit as $r\to0$. Thus we end up with
$$\int_{-\infty}^\infty\frac{1-\cos tx}{x^2}\cos\omega x\,dx+\int_{-\infty}^\infty\frac{\sin tx\sin\omega x}{x^2}\,dx=\pi t$$
for $\omega>0$, which is not as useful as I had thought. (The second integral must be considered a principal value.) Oops.
