Contour integral- Complex variables

In summary: Again, my mistake. I meant at ##z=ni## I am not great at copying the problem from my work to the computer :P
  • #1
Physgeek64
247
11

Homework Statement


evaluate ##\int \frac{sinh(ax)}{sinh(\pi x)}## where the integral runs from 0 to infinity

Homework Equations

The Attempt at a Solution


consider ##\frac{sinh(az)}{sinh(\pi z)}##
Poles are at ##z= n \pi i##
So I'm considering the contour integral around the closed contour from ##-R## to ##- \epsilon## then around a small semi-circle in the upper half plane of radius ##\epsilon##, then along ##\epsilon## to ##R## then from ##R## to ##R+i##, back along ##R## to ##\epsilon## along x+i, then along another semicircle, enclosing the pole at ##z=n \pi i ## and then similarly on the other side of the integral

I have shown that the two end contributions are zero, and that over the semicircle at ##z=0## the contribution is zero to end up with

##2(1+cosa) \int \frac{sinh(ax)}{sinh( \pi x)}## where this runs from 0 to infinity

and using Cauchy's reside theorem the only enclosed pole is the one at ##z=i## so this will be equal to ##2 \pi i (sinh(ai)) ##

hence I get

that ##\int \frac{sinh(ax)}{sinh( \pi x)}= \frac{- \pi sin(an)}{1+cos(a)}##

but the quoted answer is ##\frac{1}{2}tan(\frac{a}{2})##

I feel like I may have over complicated my contour

Many thanks
 
Last edited:
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  • #2
Physgeek64 said:

Homework Statement


evaluate ##\frac{sinh(ax)}{sinh(\pi x)}##
Many thanks

What is it about ##\sinh(ax)/\sinh(\pi x)## that you want to evaluate? You did not say!
 
  • #3
Ray Vickson said:
What is it about ##\sinh(ax)/\sinh(\pi x)## that you want to evaluate? You did not say!
Oops- My mistake I forgot the integral and to state that it runs from 0 to infinity- I will fix that. Thanks
 
  • #4
Physgeek64 said:
Oops- My mistake I forgot the integral and to state that it runs from 0 to infinity- I will fix that. Thanks

Your function does not have poles at ##z = i n \pi ## because ##\sinh(i n \pi^2 ) \neq 0## for integer ##n##.
 
  • #5
Ray Vickson said:
Your function does not have poles at ##z = i n \pi ## because ##\sinh(i n \pi^2 ) \neq 0## for integer ##n##.
Again, my mistake. I meant at ##z=ni## I am not great at copying the problem from my work to the computer :P
 

Related to Contour integral- Complex variables

1. What is a contour integral?

A contour integral, also known as a line integral, is a type of integral that is evaluated along a specific path in the complex plane. It is used to find the total change of a function between two points along a specific curve or contour.

2. How is a contour integral different from a regular integral?

A contour integral is different from a regular integral in that it is evaluated along a specific path instead of over a specific interval. This path can be any curve or contour in the complex plane, and the value of the integral can vary depending on the path chosen.

3. What are some common applications of contour integrals?

Contour integrals are commonly used in engineering and physics to solve problems involving electric and magnetic fields, fluid flow, and heat transfer. They are also used in mathematical analysis and complex function theory.

4. How do I evaluate a contour integral?

The evaluation of a contour integral involves using the Cauchy integral formula or the residue theorem, which are powerful tools in complex analysis. These methods involve finding the poles and residues of a function and then using them to evaluate the integral.

5. Can contour integrals be used in real-world problems?

Yes, contour integrals have many practical applications in a variety of fields, including engineering, physics, and mathematics. They are often used to solve boundary value problems, calculate work or circulation, and analyze complex physical systems.

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