$\int_{-\infty}^{+\infty} e^{-j\omega_{1}t} e^{-j\omega t}dt=\delta(\omega_{1}-\omega)$ when $\delta(\omega_{1}-\omega)\rightarrow\infty$ if $\omega_{1}=\omega$, $\delta(\omega_{1}-\omega)=0$ if $\omega_{1}\neq\omega$. Is it true? I have some doubts on the validity of this report, which I found on a book.
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The integral diverges for every $(\omega,\omega_1)$ so your doubts are quite well founded. – Did Mar 10 '12 at 14:33
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Yes it is nearly true (a $2\pi$ factor is missing at the right) and a classical property of the Dirac delta distribution. More exactly we have formally : $$\int_{-\infty}^{+\infty} e^{-2\pi ixt}dt=\delta(x)$$
(set $x:=\frac{\omega_1-\omega}{2\pi}$ with $\delta(\frac xa)=|a|\delta(x)$ to get your answer)
You may object that the integral is not convergent (we are merely turning around the origin at distance $1$ going nowhere...) but from a formal point of view this is simply another way to write $\mathcal{F}(1)=\delta$ : the Fourier transform of the function $1:t\to 1$ is the $\delta$ distribution (for more about this see the Wikipedia link). We may too use formal manipulations to get other useful identities (for example integrating relatively to $x$ under the integral sign to get a formula for the Heaviside step function and so on...).
Another formal (divergent) expression you may encounter is this one (see too the Sokhatsky–Weierstrass theorem) : $$\int_0^{+\infty} e^{-2\pi ixt}dt=\frac 12\left(\delta(x) -\frac i{\pi} P.V. \frac 1x \right)=\frac 12 \delta_+(x)$$ (the notation is from Feynman's book "Quantum Electrodynamics" when only positive energy is considered in scattering)
Raymond Manzoni
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A crucial word is missing here, which is formally. And to explain in what formal sense the relation you wrote holds, seems mandatory. – Did Mar 10 '12 at 14:34
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@DidierPiau: I edited my answer a bit and hope you'll like it more! – Raymond Manzoni Mar 10 '12 at 15:17