Evaluating the integral $\int_{-\infty}^{\infty}{e^{2i\mu t}\frac{{\sin^2{\lambda t}}}{\lambda t^2}}dt$

Question

It is stated that (for $\lambda>0$) $$\frac{1}{\pi}\int_{-\infty}^{\infty}{e^{2i\mu t}\frac{{\sin^2{\lambda t}}}{\lambda t^2}}dt = 1-\frac{|\mu|}{\lambda}$$ for $ 0\leq|\mu|\leq\lambda$, and zero otherwise. I tried to solve it by changing it a little bit to form a sum of integrals in which the integrand is the real or imagine value of a complex functions. Then I used a similar method to here only I used two semi-circles to avoid the pole in z=0. the integral is supposed then to be equal to the value of the integral on the smaller semi-circle, as its radius tends to zero. It didn't work, perhaps because of arithmetic mistakes, but I think it is just not the right way. Can anyone help please?

Answer 1

Denote by

$$\begin{align*} \mathcal{F}f(\mu) := \frac{1}{2\pi} \int e^{i \mu t} f(t) \, dt \\ \mathcal{F}^{-1} f(t) := \int e^{- i \mu t} f(\mu) \, d\mu \end{align*}$$

the Fourier transform and the inverse Fourier transform of $f \in L^1$, respectively. Then your claim is equivalent to

$$ \mathcal{F}f(\mu) = \left( 1- \frac{|\mu|}{2\lambda} \right), \qquad 0 \leq |\mu| \leq 2\lambda,$$

where $f(t) :=2 \frac{\sin^2( \lambda t)}{\lambda t^2}$. By the uniqueness of the Fourier transform, it suffices to show that

$$g(\mu) := \begin{cases} 1- \frac{|\mu|}{2\lambda}, & |\mu| \leq 2\lambda, \\ 0, & \text{otherwise} \end{cases}$$

satisfies $\mathcal{F}^{-1} g(t) = f(t)$, i.e. we have to calculate

$$I := \int_{-2\lambda}^{2\lambda} e^{- i \mu t} \left(1- \frac{|\mu|}{2\lambda} \right) \, d\mu$$

and show that it equals $f(t)$. To do so, show that

$$I = \int_{-2\lambda}^{2\lambda} \cos(\mu t) \, dt - \int_0^{2\lambda} \cos(\mu t) \frac{\mu}{\lambda} \, d\mu$$

and use integration by parts to determine the second integral.

@DanielFischer suggested in his comment an alternative way to prove this result using the theory of Fourier transforms.