Contour Integration with Square Root Function

In summary, the conversation discusses the integral sqrt(x)/(1+x^2) from 0 to infinity and the correct solution, pi/sqrt(2), which was obtained using a keyhole contour. The speaker also mentions their attempt at a solution and a mistake involving the arc on the negative axis. They receive help from another person and realize their mistake, solving previous issues they had encountered.
  • #1
outhsakotad
32
0

Homework Statement


The integral I'm trying to solve is sqrt(x)/(1+x^2) from 0 to infinity.


Homework Equations





The Attempt at a Solution

I've attached my solution. I know it's not right, as I shouldn't get an imaginary solution. The answer is actually pi/sqrt(2) according to my book. The author used a keyhole contour (branch cut along positive real axis and avoiding the branch point at z=0), but I don't see what's wrong with my approach. Any help would be greatly appreciated.
 

Attachments

  • contourintegral_sqrt.pdf
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  • #2
First consider: What is [itex]i^{1/2}[/itex] in cartesian form? (x + i*y). This won't solve the issue, but is just to remind you that to check that i^0.5 isn't actually 1- i (it's not, but it's close).

The real problem is the arc on the negative axis. You set [itex]z = re^{i\pi}[/itex], but forgot an extra factor of e^{i pi} from the differential dz.
 
Last edited:
  • #3
sqrt(i) = e^(i*pi/4)= (1/sqrt(2))*(1+i)... Any idea what is amiss? I still don't see it.
 
  • #4
Whoops, you got in before I edited it. You missed a factor of e^{i pi} that comes from [itex]dz = dr e^{i\pi}[/itex] on your arc on the negative axis. You really have a factor of [itex](1-e^{i 3 \pi/2}) = (1 + i)[/itex].
 
  • #5
Ah, okay. Thanks, so much. That solves some of the issues I was having with earlier problems too I think.
 

FAQ: Contour Integration with Square Root Function

What is a contour integral with the square root function?

A contour integral with the square root function is a mathematical tool used in complex analysis to evaluate integrals along a specified curve or path. It involves taking the integral of a function that contains the square root of a complex variable.

How is a contour integral with the square root function calculated?

A contour integral with the square root function is calculated using the Cauchy integral formula, which states that the integral of a function along a closed path is equal to the sum of its values at all points inside the path. This formula can be applied to evaluate the integral of a function with a square root in its expression.

What are the applications of contour integrals with the square root function?

Contour integrals with the square root function have various applications in mathematics, physics, and engineering. They are used to solve problems involving complex numbers, such as finding the area under a curve in the complex plane, calculating the work done by a force field, and determining the path of a particle under the influence of a complex potential.

What are the main properties of contour integrals with the square root function?

The main properties of contour integrals with the square root function include linearity, additivity, and the Cauchy integral theorem. Linearity means that the integral of a linear combination of functions is equal to the linear combination of their individual integrals. Additivity means that the integral along a closed path is equal to the sum of integrals along its individual segments. The Cauchy integral theorem states that the integral of a function along a closed path is equal to zero if the function is analytic within the path.

Are there any challenges in evaluating contour integrals with the square root function?

Yes, there are challenges in evaluating contour integrals with the square root function, particularly when the function has singularities, such as poles or branch points. In these cases, the integral may not exist or may require special techniques, such as the use of branch cuts or the residue theorem, to evaluate. Additionally, the choice of contour and the orientation of the path can also affect the value of the integral.

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