# Using contour integration for the inverse Laplace transform to find the inverse transform of $\dfrac{s}{s^2 + a^2}$

I am trying to use the contour integration formula for the inverse Laplace transform, find the inverse transform of $\dfrac{s}{s^2 + a^2}$.

My textbook says that the solution is $\cos(at)$, but it doesn't show any intermediate steps, and it isn't clear to me how this is done.

Do we somehow use Cauchy's residue theorem?

If $f(z)$ is analytic in a domain $D$ then for every closed path $C$ in $D$,

$$\oint_C f(z) \ dz = 0$$

Can someone please show and explain how this is done so that I can understand it and do it myself for future problems? Thank you.

## 2 Answers

The inverse Laplace transform of $F(s)=\frac{s}{s^2+a^2}$ is given by

$$f(x)=\frac1{2\pi i}\int_{\alpha-i\infty}^{\alpha+i\infty}F(s)e^{sx}ds$$ where $\alpha=\Re (s)$. The integral is carried out by contour integration. The contour from $\alpha-i\infty$ to $\alpha+i\infty$ is referred to as the Bromwich contour, and $\alpha$ is taken to the right of all singularities in order to insure

$$\int_0^\infty e^{-\alpha x}\vert f(x)\vert dx<\infty$$ Hence, using the residue theorem and as $R\to\infty$ and so $T\to\infty$, we get $$f(x)=\sum_{j=1}^2 \operatorname{Res}\left(\frac{se^{sx}}{s^2+a^2},s_j\right),\quad s_1=ia,\quad s_2=-ia$$ so $$f(x)=\frac{ia e^{iax}}{2ia}+\frac{iae^{-iax}}{2ia}=cos(ax)\quad (x>0)$$

We go by the definition here: $$f(t) = \frac{1}{2 \pi i} \oint_{C}^{}F(s)e^{st}ds = \frac{1}{2\pi i} 2\pi i \sum{}ResF(s)e^{st}$$ We can see that we have poles in $s=ia$ and $s = -ia$ and these are complex conjugetes and therefore we have that the sum of these residues is: $$Res_{s=ia} F(s)e^{st} + Res_{s=-ia} F(s)e^{st} = 2Re(Res_{s=ia} F(s)e^{st})$$ So we have: $$Res_{s=ia} = \lim_{s\rightarrow ia} \frac{(s-ia)se^{st}}{(s-ia)(s+ia)} = \frac{e^{iat}}{2} = \frac{1}{2} (cos(at) + i sin(at))$$ therefore: $$f(t) = 2 * \frac{1}{2} Re(cos(at) + i sint(at))$$ $$f(t) = cos(at)$$