Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    [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:


    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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook