I have to calculate the inverse Laplace transform of $\dfrac{1}{s^2+1}$ (which I know is sin(x)) by residue theorem: $\int^{i \infty}_{-i \infty}exp(t\cdot s)\cdot \dfrac{1}{s^2+1}\mathrm{d}s$.

Therefore I want to integrate over a vertical semicircle and show that the integral over the circular arc is zero. However I can only apply the theorem if the path goes around the poles i and -i.

I haven't found something on this topic yet. Thanks for helping!


The integration over The Bromwich Contour does not intersect the poles. The inverse Laplace Transform of $F(s)$ is given by

$$\mathscr{L}^{-1}\{F\}(t)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}F(s)e^{st}\,ds$$

where $c$ is greater than the real part of the singularities of $F(s)$.

For $F(s)=\frac{1}{s^2+1}$, the singularities are at $s=\pm i$. Therefore, we need only take $c>0$ to find

$$\mathscr{L}^{-1}\{F\}(t)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{e^{st}}{s^2+1}\,ds$$

For $t>0$, we close the integration path with (heuristically) an "infintite semi-circle that encloses the left-half plane. It can easily be shown that the integration over the "infinite semi-circle" is zero. Therefore, we can use the residue theorem to show that for $t>0$

$$\mathscr{L}^{-1}\{F\}(t)= \frac{1}{2\pi i}\left(2\pi i \text{Res}\left(\frac{e^{st}}{s^2+1}, s=\pm i\right)\right)=\frac{e^{it}}{2i}+\frac{e^{-it}}{-2i}=\sin(t)$$

For $t<0$, we close the integration path in the right-half plane and find

$$\mathscr{L}^{-1}\{F\}(t)= 0$$

Therefore, putting it all together yields

$$\mathscr{L}^{-1}\{F\}(t)= \sin (t)u(t)$$

and we are done!

  • $\begingroup$ @TrialAndError Thank you! Much appreciative. Happy Holidays. - Mark $\endgroup$ – Mark Viola Dec 30 '15 at 23:03

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