Vtyjkyuk

I am trying to use the contour integration formula for the inverse Laplace transform, find the inverse transform of $dfracss^2 + a^2$.

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.

The inverse Laplace transform of $F(s)=fracss^2+a^2$ is given by

$$f(x)=frac12pi iint_alpha-iinfty^alpha+iinftyF(s)e^sxds$$
where $alpha=Re (s)$.
The integral is carried out by contour integration.

The contour from $alpha-iinfty$ to $alpha+iinfty$ is referred to as the Bromwich contour, and $alpha$ is taken to the right of all singularities in order to insure

$$int_0^infty e^-alpha xvert f(x)vert dx<infty$$
Hence, using the residue theorem and as $Rtoinfty$ and so $Ttoinfty$, we get
$$f(x)=sum_j=1^2 operatornameResleft(fracse^sxs^2+a^2,s_jright),quad s_1=ia,quad s_2=-ia$$
so
$$f(x)=fracia e^iax2ia+fraciae^-iax2ia=cos(ax)quad (x>0)$$

We go by the definition here:
$$f(t) = frac12 pi i oint_C^F(s)e^stds = frac12pi i 2pi i sumResF(s)e^st$$
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:
$$Res_s=ia = lim_srightarrow ia frac(s-ia)se^st(s-ia)(s+ia) = frace^iat2 = frac12 (cos(at) + i sin(at))$$
therefore:
$$f(t) = 2 * frac12 Re(cos(at) + i sint(at))$$
$$f(t) = cos(at)$$