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I'm trying find the inverse Laplace transform $f(t)$ where I have $F(p)=\dfrac{9}{p(p+3)^2}$. I know $f(t)$ already to be $1-3te^{-3t}-e^{-3t}$. I have the integral $$f(t)=\dfrac{9}{2 \pi i}\int_{c-i\infty}^{c+i\infty}\dfrac{9e^{pt}}{p(p+3)^2}dp$$ has a simple pole at $p=0$ and a double pole at $p=-3$ with $$Res(\dfrac{e^{pt}}{p(p+3)^2};0)=\lim_{p \to 0}p\dfrac{e^{pt}}{p(p+3)^2}=\frac{1}{9}$$ and$$Res(\dfrac{e^{pt}}{p(p+3)^2};-3)=\dfrac{d}{dp}(p+3)^2\dfrac{e^{pt}}{p(p+3)^2}=-\dfrac{3te^{-3t}-e^{-et}}{9}$$. After this, I close the Bromwich contour to the left and multiply this sum of residues by $\dfrac{9.2\pi i}{2\pi i}$ and get $f(t)$ as required. The next step is to argue away the circular arc of the Bromwich contour which I've closed to the left. My question is, as $t>0$ in $e^{pt}$ and $\dfrac{1}{p(p+3)^2}\to0$ as $p\to\infty$ (how do I check this?), can I use Jordan's Lemma to argue away the large circular arc when $t>0$ (and also when $t<0$ and we close the contour to the right. Or do I have to use the estimation lemma which would resemble something like $$|\int_c|\leq2\pi R.max|\dfrac{e^{px}}{p(p+3)^2}|$$ and then arguing it away as $R\to\infty$. Can I use both methods to argue away? My lecturer's notes are vague in this regard. Thanks.

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Either metod works.

Since the degree in the denominator is so large, you don't need Jordan's lemma. Let $p=x+iy$ and note that $|e^{pt}| = |e^{(x+iy)t}| = e^{xt} \le e^{ct}$ if $t > 0$ and $x \le c$. In other words, $|e^{pt}|$ is bounded by a constant, independently of $p$ along the semi-circle. (Similarly, $|e^{pt}| \le e^{ct}$ if $t < 0$ and $x \ge c$.) Furthermore $$\left|\frac{1}{p(p+3)^2}\right| \le \frac{2}{|p|^3}$$ for $|p|$ large, and the standard estimation lemma (the length of the curve is $\pi R$, not $2\pi R$) brings you home.

Jordan's lemma is only needed when the degree of the denominator is just one more than the degree of the numerator.

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Thank you. So I could still use Jordan's lemma as the criteria for it is satisfied. – L.oiler Jan 20 '13 at 0:43
Yes, Jordan's lemma works too. – mrf Jan 20 '13 at 7:56

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