Understanding Branch Cuts of Ln(x) for Integrating (x^2+1)^2 from 0 to Infinity

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In summary, the author is trying to evaluate the integral \int\frac{ln(x)}{(x^{2}+1)^{2}} from 0 to inf, but is having trouble defining a branch cut. They then try to do the integral from -infty to 0, but find that the contour integral becomes difficult to solve. After trying a few different methods, they arrive at the correct answer.
  • #1
1.I am trying to evaluate the [tex]\int[/tex][tex]\frac{ln(x)}{(x^{2}+1)^{2}}[/tex] from 0 to inf2. I think I am having problems defining what my branch cut should be.

3. Here is what I have tried to no avail. First set the branch cut from -infty to 0. Then changing it to a contour integral gives us ([tex]\int[/tex][tex]\frac{ln(x)}{(x^{2}+1)^{2}}[/tex]) = 2 [tex]\pi[/tex] i (a-1(i)+a-1(-i))=[tex]\frac{i \pi^{2}}{4}[/tex]

I get somewhat lost after this. Please help. I'm studying for finals but i don't really understand this
 
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  • #2
I'd use an upper half-washer with an indentation around the origin. The branch-cut can then be anywhere in the lower half-plane including on the real axis. Then consider the following contour integral:

[tex]\mathop\oint\limits_{W_u}\frac{\text{Log}(z)}{(1+z^2)^2} dz[/tex]

Assuming you can show the integral goes to zero over both large and small semi-circles of the contour, we're left with:

[tex]\mathop\oint\limits_{W_u}\frac{\text{Log}(z)}{(1+z^2)^2} dz=2\int_0^{\infty}\frac{\ln(r)}{(1+r^2)^2}dr+\pi i\int_0^{\infty}\frac{1}{(1+r^2)^2}dr=2\pi i \frac{d}{dz} \frac{\log(z)}{(z+i)^2}\Bigg|_{z=i}[/tex]

where I let [tex]z=r e^{0i}[/tex] on the leg from [tex]0[/tex] to [tex]\infty[/tex] and [tex]z=re^{\pi i}[/tex] on the leg from [tex]-\infty[/tex] to [tex]0[/tex] to arrive at that expression.
 
  • #3
Thanks so much. I worked it out as you have laid it out and received the correct answer. but how did you know to put the contour only in the upper half plane?
 
  • #4
barrybaker said:
Thanks so much. I worked it out as you have laid it out and received the correct answer. but how did you know to put the contour only in the upper half plane?

You could use the lower half-plane and adjust the analysis accordingly. You just have to work a bunch of them to gain experience to analyze it and try things and be willing to fail at it from time to time. Remember, good cooks try again after failing a recipe, bad ones give up. :)
 

What is a branch cut in mathematics?

A branch cut is a discontinuity in a function that is used to define a multi-valued function. It is a line or curve on the complex plane that separates the different branches of the function.

How do you find the branch cuts of ln(x)?

The branch cuts of ln(x) are found by identifying the points on the complex plane where the argument of the logarithm function is negative or zero. These points form a line from the origin to negative infinity.

Why are branch cuts important in integrating functions?

Branch cuts are important in integrating functions because they help us define the domains of integration for multi-valued functions. Without considering branch cuts, we may encounter errors or inaccuracies in our integration results.

How do you integrate (x^2+1)^2 from 0 to infinity using branch cuts?

The integral of (x^2+1)^2 from 0 to infinity can be calculated by breaking the integral into smaller segments along the branch cut. Each segment will have its own branch of the multi-valued function, which can then be integrated using standard integration techniques.

Are there any special considerations when integrating using branch cuts?

Yes, there are a few special considerations when integrating using branch cuts. It is important to correctly identify and define the branch cuts for the function being integrated. Additionally, care must be taken to ensure that the branch cuts do not intersect with the contour of integration, as this can lead to incorrect results.

Suggested for: Understanding Branch Cuts of Ln(x) for Integrating (x^2+1)^2 from 0 to Infinity

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