Integral using residue theorem

In summary, the question is asking to show that using the residue theorem, integrated cosines of zeros of a function at two singularities give the value of pi. However, it does not seem right because the integrand includes both the -i and +i singularities.
  • #1
shebbbbo
17
0
The question asks to show using the residue theorem that

[itex]\int[/itex]cos(x) / (x2 +1)2 dx = [itex]\pi[/itex] / e

(the terminals of the integral are -∞ to ∞ but i didnt know the code to write that)

I found the singularities at -i and +i

so i think we change the function inside the integral to cos(z) / (z2 +1)2

i expanded the cos(z) as cosh(1) - isinh(1)(z-i) -0.5cosh(1)(z-i)2 +...

and i expanded (z2+1)2 as -(1/4)(z-i)2 - i/4(z-i) + 3/16 +...

I did the same for the singularity at x=-i and when i added both the residues together i got

(9/16e + e/16) (this is multiplied by 2[itex]\pi[/itex]i to find residues)

this doesn't seem right? i don't know if what I've done is the right method. please help, I've spent soooo many hours on this one stupid question :(
 
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  • #2
why do you add singularities together? not too sure what you're attempting...

i think the idea is you need to set up a closed contour in complex space and integrate around it. Then teh vlaue of teh integral will be related to singularities contained within the contour.

a good one would be along the x-axis for -R to R, and a semi-circle in the upper half of the complex plane to connect the contour. Then take the limit r->infinity. Hopefully the semi-cirlce contribution tends to zero, then you can relate the x-axis component directly to the original integral
 
  • #3
shebbbbo said:
The question asks to show using the residue theorem that

[itex]\int[/itex]cos(x) / (x2 +1)2 dx = [itex]\pi[/itex] / e

(the terminals of the integral are -∞ to ∞ but i didnt know the code to write that)

I found the singularities at -i and +i

so i think we change the function inside the integral to cos(z) / (z2 +1)2
Wouldn't you have to do that first, in order to find the singularities?

i expanded the cos(z) as cosh(1) - isinh(1)(z-i) -0.5cosh(1)(z-i)2 +...

and i expanded (z2+1)2 as -(1/4)(z-i)2 - i/4(z-i) + 3/16 +...

I did the same for the singularity at x=-i and when i added both the residues together i got

(9/16e + e/16) (this is multiplied by 2[itex]\pi[/itex]i to find residues)

this doesn't seem right? i don't know if what I've done is the right method. please help, I've spent soooo many hours on this one stupid question :(
What path are you integrating over? Does it include both singularities? Obviously to do this real integral as a path integral, you need a closed path that includes the real axis. I don't see how you can do that and get both i and -i inside the path.
 
  • #4
oh yeah... that makes sense...

your right we can't get i and -i in on a path integral across the real axis

maybe i need to build a circle or something? i guess i have less of an idea now?

thanks for your help?

any idea where i should go from here?
 

1. What is the residue theorem and how does it relate to integrals?

The residue theorem is a fundamental concept in complex analysis that allows for the evaluation of certain complex integrals by using the residues (singularities) of a function. It states that the integral of a function around a closed curve is equal to the sum of the residues of the function within the curve. This theorem is useful in evaluating integrals that are otherwise difficult or impossible to solve using traditional methods.

2. How is the residue of a function calculated?

The residue of a function at a singular point is calculated by finding the coefficient of the term with the highest negative power in the Laurent series expansion of the function around that point. It can also be calculated using the Cauchy residue formula, which involves taking the limit of a contour integral around the singular point.

3. What types of functions can be evaluated using the residue theorem?

The residue theorem can be used to evaluate integrals of functions with simple poles, multiple poles, and essential singularities. It cannot be used for functions with branch points or non-isolated singularities.

4. Can the residue theorem be used to evaluate real integrals?

Yes, the residue theorem can be used to evaluate real integrals by using the real line as a contour. In this case, the residues of the function at the singular points on the real line are considered.

5. Are there any limitations to using the residue theorem?

Yes, there are some limitations to using the residue theorem. It can only be used for certain types of functions and integrals, and it may not always provide the most efficient method of evaluation. Additionally, the contour used must be carefully chosen to avoid crossing any branch cuts or other singularities.

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