Contour integration

xdrgnh · Dec 17, 2013

1. The problem statement, all variables and given/known data

Integrate [e^(bx)]/(1+e^x)dx from inf to -inf

0<b<1

2. Relevant equations

Sum of the residues times 2pi equals the integral with a path that encloses the residues. That path includes the complex part of path which is above the x axis and the real part which is a straight line on the x axis. It's a semi circle.

3. The attempt at a solution

So I put this into the complex plain and at (pi)i as one of poles and calculate the residue there. However usually the complex curve part goes to zero but in this case it doesn't. Also I never saw an integral in this form before. I know integrals in the form (x^-a)/(1+x) and trig integrals using the Jorden Lemma. Any help will be appreciated.

ShayanJ · Dec 17, 2013

[itex] e^{bx}=(e^b)^x [/itex] and because b<1,if we'll tend to zero when x tends to infinity.Also [itex]1+e^x [/itex] tends to infinity when x tends to infinity.So you can use Jordan's lemma.

vela · Dec 17, 2013

I suggest you start with the substitution ##u=e^x## first.

ShayanJ · Dec 17, 2013

sorry guys,what I said was wrong.b<1 doesn't mean we can use Jordan's lemma.

jackmell · Dec 18, 2013

xdrgnh said: ↑

So I put this into the complex plain and at (pi)i as one of poles and calculate the residue there. However usually the complex curve part goes to zero but in this case it doesn't. Also I never saw an integral in this form before. I know integrals in the form (x^-a)/(1+x) and trig integrals using the Jorden Lemma. Any help will be appreciated.

How about using a rectangle enclosing just the pole at ##\pi i##?

And by the way, you mind showing how you're computing the residue(s) because even the one at ##\pi i## is not in my opinion, too easy to compute.

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Contour integration

xdrgnh

ShayanJ

vela

ShayanJ

jackmell

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