How to find $\mathcal{L}^{-1}\left\{1\right\}$?

Question

This is probably a really simple question, but I cannot figure it out and it's not mentioned in my books. How do I find \begin{align} \mathcal{L}^{-1}\left\{1\right\}?\tag{1} \end{align} It seems like it should be so simple but I can't find anything. I've checked Wolfram Alpha and it gives this: \begin{align} \mathcal{L}^{-1}\left\{1\right\}=\delta\left(t\right),\tag{2} \end{align} the dirac delta function. Why is this? I've read the chapter on it in my textbook, Fundamentals of Differential Equations, and it states that \begin{align} \int_{-\infty}^{\infty}\delta\left(t\right)\:dt=1,\tag{3} \end{align} which reminds me a lot of a function like the Gaussian distribution. But how, if so, is this related to (1)?

Answer 1

From the inverse Mellin transform, we have \begin{align} \mathcal{L}^{-1}\{1\} &= \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{st}ds\\ &=\frac{1}{2\pi i}\lim_{u\to\infty}\int_{-iu}^{iu}e^{st}ds\\ &=\frac{1}{2\pi i}\lim_{u\to\infty}\frac{e^{iut}-e^{-iut}}{s}\\ &=\lim_{u\to\infty}\frac{\sin(tu)}{t\pi} \end{align} Here we see one of the identities of the Dirac Delta function is $$ \delta(x) = \lim_{\epsilon\to 0}\frac{1}{\pi x}\sin\Bigl(\frac{x}{\epsilon}\Bigr) $$ from equation $37$. Let $\epsilon = \frac{1}{u}$