# arguing away - complex analysis

Probably a trivial question but I can't understand how to argue away the value of integrals in complex analysis. I am trying to find the inverse Laplace transform of $F(s)=\frac{1}{s(s+1)}$. The integral I therefore have to compute is $f(t)=\dfrac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}\dfrac{e^{st}}{s(s+1)}ds$ and I'm using the 'Bromwich contour'. I have to 'argue away' the contribution from the semi-circular arc of the 'Bromwich contour' but I don't understand the method of how to do so.$$|\int_c|\leq2\pi R.max\dfrac{e^{st}}{s(s+1)}$$ any help in understanding the method would be appreciated.

-
I think you need this: en.wikipedia.org/wiki/Estimation_lemma – DonAntonio Jan 8 '13 at 12:36

Hint: On the arc and assuming $t>0$, $|e^{st}|\leq e^{ct}$ and $\max \left|s^{-1}(s+1)^{-1}\right| \sim R^{-2}$ as $R \to \infty$.
How does one get $s^{-1}(s+1)^{-1}$ to $O(R^{-2})$? Am I right in saying that by the estimation lemma, we have $|\int_c|\leq \dfrac{\pi Re^{ct}}{R^2-R}$ which approaches zero as R goes to $\infty$? – L.oiler Jan 8 '13 at 15:24
@L.oiler, Yes, definitely (don't forget the $2$ :) ). The asymptotic argument is similar. Write $s^{-1}(s+1)^{-1} = s^{-2}(1+s^{-1})^{-1}$. Then, on the arc, you have $|s| = R$, so $$\max |s^{-2}(1+s^{-1})^{-1}| = R^{-2} \max |1+s^{-1}|^{-1} = R^{-2}|1+\hat{s}(R)^{-1}|^{-1},$$ where $\hat{s}(R)$ maximizes the expression $|1+s^{-1}|^{-1}$ on the arc for a given $R$. But $|\hat{s}(R)|=R$, so $1+\hat{s}(R)^{-1} \to 1$ as $R \to \infty$. – Antonio Vargas Jan 8 '13 at 17:01