Delta function expressed as integral of exponential problem

Question

I've read that the delta function can be expressed as:

δ(t) = $\int_{-\infty}^\infty e^{j2\pi ft} df$.

The exponential inside the integral can be written as:

$(e^{j2\pi})^{ft} = (cos(2\pi) + jsin(2\pi))^{ft} = 1^{ft} = 1$

So how can we derive from $\int_{-\infty}^\infty df$ to the delta function?

Answer 1

In terms of the sinc function, for small $\epsilon>0$ we get$$\int_{-1/\epsilon}^{1/\epsilon}e^{j2\pi ft}df=\left[\frac{1}{j2\pi t}e^{j2\pi ft}\right]_{-1/\epsilon}^{1/\epsilon}=\frac{e^{j2\pi t/\epsilon}-e^{-j2\pi t/\epsilon}}{j2\pi t}=\frac{1}{\pi t}\sin\frac{2\pi t}{\epsilon}=\frac{2}{\epsilon}\operatorname{sinc}\frac{2\pi t}{\epsilon}.$$This function of $\epsilon$:

• Has $\epsilon\to0^+$ pointwise limit equal to $0$ for $t\ne0$ and $\infty$ for $t=0$;
• Integrates via $\int_{-\infty}^\infty dt$ to $1$, as a consequence of $\int_{-\infty}^\infty\operatorname{sinc}xdx=\pi$.

Thus its distributional $\epsilon\to0^+$ limit is $\delta(t)$. In particular, this is a treatment in terms of nascent deltas.

Answer 2

One way to define the Fourier transform of a function $\varphi$ is $$\hat\varphi(f) = \int_{-\infty}^{\infty} \varphi(t) e^{-i2\pi ft} dt.$$

The inverse transform is then given by $$\varphi(t) = \int_{-\infty}^{\infty} \hat\varphi(f) e^{i2\pi ft} dt.$$

Now, strictly speaking, $\delta$ is not a function but a distribution. Still, the above expressions are valid with $\varphi=\delta.$ This gives $$\hat\delta(f) = \int_{-\infty}^{\infty} \delta(t) e^{-i2\pi ft} dt = e^{-i2\pi f\cdot 0} = 1$$ and $$\delta(t) = \int_{-\infty}^{\infty} \hat\delta(f) e^{i2\pi ft} dt = \int_{-\infty}^{\infty} 1 e^{i2\pi ft} dt = \int_{-\infty}^{\infty} e^{i2\pi ft} dt.$$