How to show integrals are zero around funny contours.

[vertices]

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 $$\int\frac{exp(ax)}{coshx}dx$$ from -infinity to plus infinity.

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

$$\int\frac{exp(ax)}{coshx}dx$$ from R to -R

PLUS

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

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

I=$$\int\frac{exp(aR+ait)}{cosh(R+it)}idt$$ 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?

 

[vertices]

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=$$\int\frac{exp(aR+ait)}{cosh(R+it)}idt$$ from 0 to pi.

Perhaps I could use ML inequality but how?

 

[vertices]

:(

still hoping someone can answer; there's so many clever people on here:)

 

[Santa1]

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?

 

[vertices]

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?

 

[Santa1]

well you get a straight horizontal line with length pi (lets call it $$\gamma$$) starting somewhere on the real axis (if R is real), you now have to integrate;

$$\int_\gamma \frac{e^{az}}{\cosh(z)} \mathrm{d}z$$

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.)

 

[vertices]

$$\int_\gamma \frac{e^{az}}{\cosh(z)} \mathrm{d}z$$

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

 

[vertices]

Okay I've just realized 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:)

 

[uart]

Vertices, can you restrict the parameter "a" to |a|<1. That will definitely get you zero on any contour with R going to infinity

 

[vertices]

Vertices, can you restrict the parameter "a" to |a|<1. That will definitely get you zero on any contour with R going to infinity

well in that example, |a| IS less than 1.

So the integrals like:

I=$$\int\frac{exp(aR+ait)}{cosh(R+it)}idt$$ 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.