Integral with singularities at the endpoints

In summary, the integrand in the homework statement is evaluated using contour integration. The branch-cut from -1 to 1 on the real axis is used to define the contour. The argument of the function will change as you go round the loop, which should help you evaluate the integral.
  • #1
HappyEuler2
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Homework Statement



Evaluate the Integral [tex]\int[/tex] dx/((a^2+x^2)*sqrt(1-x^2)) from -1 to 1
Using contour integration

Homework Equations



Residue theorem/Cauchy integral forumula

The Attempt at a Solution



So I know that at the end-points of the interval (abs(z) = 1) that a singularity exists, so a branch-cut from -1 to 1 needs to be made. Additionally, the singularities in the contour used for the residue theorem are located at z = +ia and -ia. After this I am confused. I can't find any resources that describe how to evaluate a contour integral where the endpoints of the interval used to define the contour has branch-point singularities. Any ideas?

I am thinking maybe the contour that I described ( from just above the real-axis counter-clockwise to real 1 and then up into the imaginary plane heading counter-clockwise towards -1 on the real axis enclosing the singularity z= +ia)

Any thoughts?
 
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  • #2
haven't done complex analysis for a while, so take anything i say with a grain of salt...

could you define the branch cut from [-1 , 1] on the real axis then all then integrate all the way around the branch cut, with little loops that shrink to zero around the branch points.

The argument of the function will change as you go round the loop, which should help you evaluate the integral. And the value of the loops should go to zero because of the power of the singularity

if that doesn't work, you could try the branch cuts (-infinty,-1] & [1,infinty) and maybe try a semi circle and a full circle (sort of, but with little loop round the cuts)
 
  • #3
Ok, I've got some new info to add, but I am not sure how to use it.

So I figured out that the contour should have the form of a dogbone/dumbbell encircling the branch-cut from -1 to 1 counterclockwise. But I have no idea how to deal with this. I've gone over branch-cuts only once in class, so working with this problem has kind of been a pain.

Any help would be great
 

FAQ: Integral with singularities at the endpoints

What are singularities at the endpoints in an integral?

Singularities at the endpoints in an integral refer to points at which the function being integrated becomes undefined or has a discontinuity. These points can often cause issues when evaluating the integral.

How do you handle singularities at the endpoints in an integral?

Handling singularities at the endpoints in an integral depends on the type of singularity. For a removable singularity, the integral can be evaluated by taking the limit as the singularity approaches the endpoint. For a non-removable singularity, the integral may not exist or may need to be evaluated using a different method, such as using a Cauchy principal value.

Can singularities at the endpoints affect the value of an integral?

Yes, singularities at the endpoints can significantly affect the value of an integral. Depending on the type and location of the singularity, the integral may be undefined or have a different value than expected. It is important to carefully consider and handle these singularities when evaluating integrals.

How do you identify singularities at the endpoints in an integral?

Singularities at the endpoints in an integral can be identified by analyzing the function being integrated and looking for points where it becomes undefined or has a discontinuity. These points can also be identified by graphing the function and observing any sharp changes or breaks in the graph.

Are there any strategies for avoiding or minimizing singularities at the endpoints in an integral?

In some cases, singularities at the endpoints in an integral can be avoided by choosing a different interval of integration or by using a different method, such as integration by parts or substitution. However, in some cases, singularities may be unavoidable and need to be carefully handled in order to accurately evaluate the integral.

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