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?

  • 1
    \$\begingroup\$ hm, we generally don't like to do Laplace transforms of functions that are \$\ne 0\$ for \$t<0\$... \$\endgroup\$ – Marcus Müller May 13 '17 at 14:32
  • \$\begingroup\$ 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) \$\endgroup\$ – Marcus Müller May 13 '17 at 14:32
  • \$\begingroup\$ Application of L'Hôpital's rule, perhaps? \$\endgroup\$ – howland12 May 13 '17 at 15:18
  • 4
    \$\begingroup\$ I'm voting to close this question as off-topic because it belongs on the Math SE \$\endgroup\$ – brhans May 13 '17 at 17:55

Let me start a little more general. Suppose:


You should be able to see that:

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


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

(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:

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


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

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.

| improve this answer | |

Not the answer you're looking for? Browse other questions tagged or ask your own question.