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Improper integral using residues

  • Thread starter cscott
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  • #1
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Homework Statement



integral of sech(x) from -Inf to Inf using residues.

Homework Equations



Calculate using: (2 Pi I) * Res[sech(x), "poles in upper half plane"]

The Attempt at a Solution



I used sech(x) = 2/[exp(x)+exp(-x)] to find a simple pole at z = (I Pi)/2 with a residue of -I. Then,

Result = (2 Pi I)(-I) = 2 Pi

As far as I know the answer should be Pi so I'm off by a factor of two. I can't find the mistake in finding the residue. Can anyone verify that the residue should be -I/2?

Let f(z)=g(z)/h(z), then residue of a simple pole z0 is g(z0)/h'(z0), no? (Assuming g(z0) != 0, h(z0)=0)

Thanks.
 

Answers and Replies

  • #2
Dick
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I can tell you what's wrong. i*pi/2 isn't the only pole in the upper half plane. i*3*pi/2, i*5*pi/2 ... are also poles. There's an infinite number. You need to make a more clever choice of contour.
 
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  • #3
782
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I can tell you what's wrong. i*pi/2 isn't the only pole in the upper half plane. i*3*pi/2, i*5*pi/2 ... are also poles. There's an infinite number. You need to make a more clever choice of contour.
Ah you are right :( I approached this too quickly.

My hint from the prof is 'convert the integrand to a rational function by changing variables'

Either way I'm not sure exactly how to proceed now.
 
  • #4
Dick
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Ah you are right :( I approached this too quickly.

My hint from the prof is 'convert the integrand to a rational function by changing variables'

Either way I'm not sure exactly how to proceed now.
Well, your professor gave you one suggestion, and I gave you another. I think they both will work. Pick one and try it. Here's a hint for my route. How are sech(x) and sech(x+i*pi) related?
 
  • #5
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So if I take a rectangle with height [itex]\pi[/itex] I'd enclose one pole at [itex]i\pi/2[/itex]. Something tells me the two side integrals should go to zero as rectangle width->inf and I'll get a multiple of my integral on the upper part.

So in this case [itex]sech(z) = -sech(z + i\pi)[/itex] and because of the opposite integration directions I'll get double my desired integral, which is why I have [itex]2\pi[/itex] in my OP instead of [itex]pi[/itex]?
 
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  • #6
Dick
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So if I take a rectangle with height 1 I'd enclose one pole at [itex]i\pi/2[/itex]. Something tells me the two side integrals should go to zero as rectangle width->inf and I'll get a multiple of my integral on the upper part.

So in this case [itex]sech(z) = -sech(z + i\pi)[/itex] and because of the opposite integration directions I'll get double my desired integral, which is why I have [itex]2\pi[/itex] in my OP instead of [itex]pi[/itex]?
That's basically it. You mean a rectangle with height pi, right?
 
  • #7
782
1
Yes, I meant Pi, sorry.

Thanks for your help.
 

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