Computing the Fourier transform of $\cos(x/2)$

Question

How can I prove that $\int_{-\pi}^{\pi} \cos(x/2)e^{ixt} dx=\frac{4\cos(\pi t)}{1-4t^2}$?

Answer 1

You can utilize the exponential form of the cosine to combine exponentials

$$\begin{align}\int_{-\pi}^\pi \cos(\frac{x}{2})e^{ixt}dx&=\frac{1}{2}\int_{-\pi}^\pi (e^{\frac{ix}{2}}+e^{-\frac{ix}{2}})e^{ixt}dx\\ \\ &=\frac{1}{2}\int_{-\pi}^\pi e^{ix(t+\frac{1}{2})}dx+\frac{1}{2}\int_{-\pi}^\pi e^{ix(t-\frac{1}{2})}dx\\ \\ &=\frac{1}{2}\left(\frac{e^{i\pi(t+\frac{1}{2})}-e^{-i\pi(t+\frac{1}{2})}}{i(t+\frac{1}{2})}+\frac{e^{i\pi(t-\frac{1}{2})}-e^{-i\pi(t-\frac{1}{2})}}{i(t-\frac{1}{2})}\right)\end{align}$$ Adding the two terms together inside and using the exponential definition of the cosine again yields the answer.

Answer 2

By parts, as hinted by @projectilemotion.

$$I=\int\cos\frac x2e^{ixt}dx=2\sin\frac x2e^{ixt}-i2t\int\sin\frac x2e^{ixt}dx \\=2\sin\frac x2e^{ixt}+i2t\cos\frac x2e^{ixt}dx+4t^2I.$$

When integrating from $-\pi$ to $\pi$, the first term equals

$$2(e^{ixt}+e^{-ixt})$$ and the second vanishes.