Contribution of Semi-Circle in Complex Analysis Integration

[Niles]

Homework Statement

Hi all.

I have the following integral:

<br /> I = \int_{2 - i\infty}^{2+i\infty}{f(s) \exp(st)ds},<br />

where f(s) is some function. In order to perform this integral, I will choose to close the vertical line with a semi-circle in some halfplane (in order to use Cauchy's integral theorem), but this requires that the contribution from the semi-circle is zero.

Question: Now, let us say that for the specific f(s) in this case, then the contribution from the semi-circle in e.g. the right halfplane does go to zero: With this in mind, then will the limit f(s)\exp(st) for s\rightarrow \infty be zero?Thank you very much in advance.Niles.

 

[Office_Shredder]

I don't quite understand your question... the contribution from the semi-circle in the right halfplane going to zero is a statement of integrating the function over the whole semicircle. Are you asking about how the function f(s)exp(st) behaves as s goes to infinity for any constant t (is t real or complex?)? Because if you already know how the integral over the semi-circle behaves you don't really care about that in terms of answering the question as far as I can tell

 

[Niles]

The variable t is real.

What I am asking is that if the following is true

<br /> \int\limits_{semi - circle, right\,\, halfplane} {f(s)e^{st} ds = 0},<br />

then does this imply that

<br /> \mathop {\lim }\limits_{s \to \infty } f(s)e^{st} = 0<br />
?

The reason why I am asking is because I am looking for a method to find the half plane, where the contribution from the semi-circle will be zero. This I can do by the above limit (if what I am asking is correct), because we are guaranteed that one of the half planes will be correct.

 

[Office_Shredder]

Well, you can't just take a limit as s goes to infinity, since obviously we don't expect the function to uniformly decrease in all directions (if it did, we wouldn't have to worry about which half circle to take). The easiest way to make this determination is to note that if s=a+bi then

|f(s)e^{st}| = |f(s)||e^{at+bti}| = |f(s)|e^{at}| as |e^{tbi}|=1 always. For most functions f that you see in this context, f grows and shrinks slowly compared to exp. In these cases, we're interested in the behavior of e^{at} which if t is positive, you want to have a be negative, and vice versa (in order to make it small). It's a pretty quick and dirty way of figuring out which way the circle should probably be pointing

 

[Niles]

Thanks. It is kind of you to help me.