TITLE: Fourier transform of $(1-\cos(tx))/x^2$
QUESTION [7 upvotes]: 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.

REPLY [8 votes]: 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:
$$\hat{f}(\omega) = \begin{cases}\ \pi (t-|\omega|) & |\omega| \lt t\\ 0 &  |\omega| \gt t\end{cases}$$