Complex integration and residue theorem.

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peripatein
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Hi,

Homework Statement


I was wondering whether any of you could kindly explain to me how the equation in the attachment was derived.
I mean, how could I have known that it could be separated into these two fractions?

Homework Equations


The attachment also specifies the integration to be performed in the rectangular region whose coordinates are given.


The Attempt at a Solution


I know that the integrand, and integral, are zero for x=R, as R->infinity.
 
Attachments
  • Handwritten-style math: integral from 0 to infinity of dx cos x over cosh x, the rectangle corners (-R, pi), (R, pi), (R, 0), (-R, 0), and the contour identity for dz e^(iz) over cosh z split into two real-line integrals with e^x and e^(x-pi) over cosh x and cosh(x+i pi).
    D.JPG
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But how were these expressions derived? I mean, how could I have obtained them myself?
Is it by using Cauchy's Theorem as shown in the attachment I just added? I have tried using that to obtain these expressions, but was unsuccessful.
 
Attachments
  • Typeset line of complex contour integral identities: integral over C of f(z)dz equals integral over C of (u + iv)(dx + i dy), which equals integral over C of (u dx - v dy) plus i times integral over C of (v dx + u dy).
    C.JPG
    5.5 KB · Views: 479
peripatein said:
But how were these expressions derived? I mean, how could I have obtained them myself?

Your attachment is sorely lacking in detail. How about a picture, labeled, a nice one. All four legs, C1 through C4 starting on the real axis and going counterclockwise. Now, treat it like a fine wine, delicately and in meticulous detail, every square mm of it. Start with the expression:

$$\int_{c1} \frac{e^{iz}}{\cosh(z)}dz$$

what is that? Well, since it's on the real axis, then z=x+iy=x, and we're integrating from -R to R it's simply:

$$\int_{-R}^{R} \frac{e^{ix}}{\cosh(x)}dx$$

let's just leave that one there for now. How about the leg going from (R,0) to (R\pi)? What is that? Well, in terms of the complex variable z=x+iy, that's just z=R+iy but don't forget that dz=idy. Now make the substitution:

$$\int_{C2}\frac{e^{iz}}{\cosh(z)}dz=\int_0^{\pi} \frac{e^{i(R+iy)}}{\cosh(R+iy)}idy $$

Ok, that's #2. How about the other two. Get those and then see what you got. Study it for a little while and see what you can do with it. Maybe things cancel, other things.

You learn what to do by working with them in meticulous detail and then sitting back and studying what you have for a little while, overlooking nothing. Often the answer does not come immediately. It has to simmer a little while in your mind. Don't rush.