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How to show integrals are zero around funny contours.

  1. May 11, 2008 #1
    This is related to my thread on 'integrals over an infinite range (using residue theorem)'. I am trying to understand how certain integrals over semi circles, or the widths of rectangles go to zero under certain functions.

    I want to evaluate the integral [tex]\int\frac{exp(ax)}{coshx}dx[/tex] from -infinity to plus infinity.

    So consider [tex]\oint\frac{exp(az)}{coshz}dz[/tex] around a rectangle. This is equal to

    [tex]\int\frac{exp(ax)}{coshx}dx[/tex] from R to -R

    PLUS

    [tex]\int\frac{exp(a(i*pi+x))}{coshx}dx[/tex] from R to -R (contains one singularity)

    PLUS two integrals corresponding to the sides. They look like this:

    I=[tex]\int\frac{exp(aR+ait)}{cosh(R+it)}idt[/tex] from 0 to pi. Now these integrals apparently EQUAL ZERO when R tends to infinity. Can anyone help me see why this is?

    Perhaps I could use ML inequality but how?
     
  2. jcsd
  3. May 16, 2008 #2
    Let me summarise the problem (hopefully I can get someone to answer now;p)

    I basically want to see why the following integral EQUALS 0:

    I=[tex]\int\frac{exp(aR+ait)}{cosh(R+it)}idt[/tex] from 0 to pi.

    Perhaps I could use ML inequality but how?
     
  4. May 18, 2008 #3
    :(

    still hoping someone can answer; there's so many clever people on here:)
     
  5. May 18, 2008 #4
    Factor out the a inside the exp, and consider what kind of contour R+it makes with t ranging from 0 to pi.

    did that help?
     
  6. May 18, 2008 #5
    thanks for your reply *-<|:-D=<-<

    Hmm a simple parameterisation like z=e^it sweeps out a semi circle on the interval 0<t<pi.

    This integral however is more complicated - how could I work out what contour it makes if I don't know what the integration of the integral actually is?

    Perhaps a better question might be - is it possible to easily evaluate this integral through analytical means?
     
  7. May 18, 2008 #6
    well you get a straight horizontal line with length pi (lets call it [tex]\gamma[/tex]) starting somewhere on the real axis (if R is real), you now have to integrate;

    [tex]\int_\gamma \frac{e^{az}}{\cosh(z)} \mathrm{d}z[/tex]

    does that help you at all? (I have not tested this method, I'm not certain it will get you where you want to be, but i had a feeling it would come through.)
     
  8. May 18, 2008 #7
    yes indeed that was the original integral (I forgot that for a second!). I didn't explain it well, but I wanted to evaluate this integral, with x subsituted for z, from -inifinity to +infinity. To do this we have consider a rectangular contour:

    if you go to the bottom of *page 22* here:

    http://www-thphys.physics.ox.ac.uk/user/JamesBinney/advcalc.pdf

    the problem is illustrated more clearly. Basically the line(s) we're talking about contributes very little to the contour integral when R tends to infinity, and thus can be neglected - it is "less than" 2*pi*exp(-a*r). I don't know how the lecturer came to that conclusion?

    EDIT: Sorry, its the bottom of page 22 and top of page 23
     
    Last edited: May 18, 2008
  9. May 18, 2008 #8
    Okay I've just realised you won't be able to see that link from outside my university network, so I'm attaching it to this post.

    Would greatly appreciate any thoughts:)
     

    Attached Files:

  10. May 19, 2008 #9

    uart

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    Vertices, can you restrict the parameter "a" to |a|<1. That will definitely get you zero on any contour with R going to infinity
     
    Last edited: May 19, 2008
  11. May 19, 2008 #10
    well in that example, |a| IS less than 1.

    So the integrals like:

    I=[tex]\int\frac{exp(aR+ait)}{cosh(R+it)}idt[/tex] from 0 to pi.

    will tend to zero as R tends to infinity? According to my notes, this integral actually is LESS than 2pi*exp(-aR) so yes it does indeed tend to zero as R tends to infinity.

    I don't understand why it is less than 2pi*exp(-ar) though.
     
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