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**Laurent** First, you can notice that the change of variable $\displaystyle y=nx$, together with letting $\displaystyle n$ tend to infinity inside the integral gives a "formal" proof, provided you know $\displaystyle \int_0^\infty \frac{\sin x}{x}dx=\frac{\pi}{2}$. Maybe this is enough for your electromagnetic course.

If this were maths, you'd have to justify taking the limit inside the integral. One difficulty is that the limit is an improper integral, hence usual Lebesgue's theorem don't apply directly. Here's a suggestion. On $\displaystyle [a,b]$ containing the support of $\displaystyle f$, write $\displaystyle f(x)=f(0)+x\varphi(x)$ where $\displaystyle \varphi$ is $\displaystyle \mathcal{C}^\infty$ as well (but not necessarily compactly supported). Use linearity to split the integral; for the first part, change of variable, use the above classic integral, ok. For the second part, you can note this is a case of Riemann-Lebesgue lemma. It can be proved for step functions $\displaystyle \varphi$ first and generalized by density, for instance. Or using an integration by part (derivating phi), if I remember well.