# Laplace of Sinc? [closed]

How can I find the Laplace transform of a sinc function?

Inserting $f(t) = \frac{\sin(t)}{t}$ into $F(s) = \int_{0}^{\infty}e^{-st}f(t)dt$ gives three functions dependent on variable t within the integral.

Surely there is a less cumbersome way to solve this than by chain rule / u-substitution?

• hm, we generally don't like to do Laplace transforms of functions that are $\ne 0$ for $t<0$... – Marcus Müller May 13 '17 at 14:32
• and, why should there be a "less cumbersome way"? Math can be hard, yo! (notice: the Laplace transform isn't even necessarily existing for just every function of $t$... it does exist here) – Marcus Müller May 13 '17 at 14:32
• Application of L'Hôpital's rule, perhaps? – howland12 May 13 '17 at 15:18
• I'm voting to close this question as off-topic because it belongs on the Math SE – brhans May 13 '17 at 17:55

Let me start a little more general. Suppose:

$$f_t=\frac{g_t}{t}$$

You should be able to see that:

\begin{align*} \mathcal{L}^{-1}\left\{ F^{'}_s\right\} &= -t\cdot f_t \\&= -g_t\\\\ &=\mathcal{L}^{-1}\left\{-G_s\right\}\\\\\therefore F^{'}_s=-G_s\end{align*}

So:

$$F_s = -G_s^{-1}+C$$

(With $G_s^{-1}$ being any anti-derivative of $G_s$.) The Laplace transform vanishes at $s=\infty$, so $F_\infty=0$ and $C=G_\infty^{-1}$. So:

$$F_s=G_\infty^{-1}-G_s^{-1}=\int_s^\infty G_u\:\textrm{d}u$$

Clearly:

$$F_s=\mathcal{L}\left\{f_t\right\}=\mathcal{L}\left\{\frac{g_t}{t}\right\}=\int_s^\infty G_u\:\textrm{d}u$$

A table lookup provides: $G_u=\mathcal{L}\left\{g_t\right\}=\mathcal{L}\left\{\operatorname{sin}\:t\right\}=\frac{1}{s^2+1}$, then:

\begin{align*} \mathcal{L}\left\{\frac{\operatorname{sin} t}{t}\right\}&=\int_s^\infty \frac{\textrm{d}u}{u^2+1}\\\\&=\left[\operatorname{tan}^{-1} u\right]\bigg|_s^\infty\\\\&=\frac{\pi}{2}-\operatorname{tan}^{-1} s\\&=\operatorname{cot}^{-1} s=\operatorname{tan}^{-1} \frac{1}{s} \end{align*}

If I didn't mess up, anyway.