Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
I think you need this: en.wikipedia.org/wiki/Estimation_lemma –  DonAntonio Jan 8 '13 at 12:36

1 Answer 1

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 $R \to \infty$.

share|improve this answer
    
Have I understood your question correctly? –  Antonio Vargas Jan 8 '13 at 13:23
    
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
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.