# Finding the inverse Laplace transform of $\frac{s}{(s+1)^3}$ using inversion formula

I need to find the inverse Laplace transform of $$F(s) = \frac{s}{(s+1)^3}$$ using Bromwich Integral. The Bromwich contour will look something like this.

Actually you can see this problem on the following link: https://youtu.be/cXjbPsc-Z5w. I would like to know, why should we show the integral along $$L_u$$, $$C_R$$, $$L_D$$ is $$0$$? I mean, i've seen many examples on some books (such Mathematical Methods for Physicists, 3rd ed.) it's just need to show the residue at simple poles for solving the inversion of laplace transform

So, in this case it should be:

\begin{align} \mathcal{L}\bigg\{\frac{s}{(s+1)^3}\bigg\} &= \frac{1}{2\pi i}\int_{\gamma - i\infty}^{\gamma + i\infty} \frac{se^{st}}{(s+1)^3} \Bbb ds \\ &= \mathrm{Res}_{s=-1} \left(\frac{se^{st}}{(s+1)^3}\right) \\ &= \frac 12 \lim_{s=-1} \frac{\Bbb d^2}{\Bbb ds^2} \left[(s+1)^3 \frac{se^{st}}{(s+1)^3}\right]\\ &= \frac 12 \lim_{s=-1} te^{st}(2+st)\\ &= te^{-t} \left(1-\frac t2\right) \end{align}

Can you explain why should we show the integral along $$L_u$$, $$C_R$$, $$L_D$$ is $$0$$ (based on the given link) if the residue theory is enough to evaluate the integral to find the inverse laplace transform of $$F(s)$$?

The Residue Theorem is an extension of Cauchy's Integral Theorem. Both theorems begin with rectifiable closed curves within a simple connected domain in $$\mathbb{C}$$.

The inverse Laplace Transform of $$F(s)$$, $$f(t)=\mathscr{L}^{-1}\{F\}(t)$$, is expressed by

$$f(t)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}F(s) e^{st}\,ds\tag1$$

where $$c$$ is a real number that is greater than all of the singularities of $$F(s)$$.

To apply the Residue Theorem, we evaluate the integral of $$F(s)e^{st}$$ over a closed and rectifiable curve. So, we begin our analysis and write

\begin{align} \oint_C F(s)e^{st}\,ds&=\oint_{L_R+L_u+C_R+L_d}F(s)e^{st\,ds}\\\\ &=\int_{L_R}F(s)e^{st\,ds}+\int_{L_u+C_R+L_d}F(s)e^{st\,ds}\tag2 \end{align}

Given the specific question of the OP, we assume herein that the only singularities of $$F(s)$$ are pole singularities. If $$F(s)$$ has branch point singularities, then we would close the Bromwich path such that the branch points and corresponding branch cuts are excluded from within the closed contour.

Suppose all of the $$N$$ number of poles of $$F(s)$$ are inside the closed contour $$C$$ and denote the location of the $$n$$'th pole by $$s_n$$, where $$n=1,2\cdots N$$. Then, we have from the Residue theorem,

$$\oint_{C}F(s)e^{st}\,ds=2\pi i \sum_{n=1}^N \text{Res}\left(F(s)e^{st}, s=s_n\right)\tag3$$

In addition, as $$R\to \infty$$, the first integral on the right-hand side of $$(2)$$ approaches $$2\pi i f(t)$$ as expressed in $$(1)$$. So, if the integral over $$L_u+C_R+L_d$$ vanishes as $$R\to \infty$$, then from equating $$(2)$$ and $$(3)$$, we find that

$$f(t) = \sum_{n=1}^N \text{Res}\left(F(s), s=s_n\right)\tag4$$

NOTE: The expression in $$(4)$$ was based on the assumption that

$$\lim_{R\to \infty} \int_{L_u+C_R+L_d}F(s)e^{st}\,ds=0\tag5$$

If $$(5)$$ fails to hold, then $$(4)$$ fails to hold likewise.

• Sorry for asking after 4 years but why do we need the horizontal lines? Can't we just construct a closed curve from just the vertical line and the half circle? When R gets arbitrarily large, then won't the contour still enclose the singularities? Commented Aug 20 at 11:09
• @cittee There are a variety of ways to close the curve. What way do you propose? Commented Aug 20 at 13:13
• Couldnt just a closed curve formed by the vertical line and the half circle work? Commented Aug 21 at 17:55
• The vertical line is in the right-half plane. It extends from $a-iR$ to $a+iR$ for $a>0$ and $R>0$. We eventually take the limit as $R\to 0$. So, as stated in the previous comment, one can close the curve in a variety of ways. One way would be to do as you suggest. However, you will need to show that as $R\to \infty$ the integration over those arcs in the right-half plane vanishes. Commented Aug 21 at 18:57
• Yes, i see. In that case, as $R$ grows we need to have the factor $e^{st}$ bounded. So for $t\geq0$ and $s=a+iy$ we get that $|e^{st}|\leq e^{at}$ since $|e^{iyt}|=1$ ? Commented Aug 22 at 19:47