What Are the Singularities in the Extended Complex Plane for (1/s)Ln(s^2+1)?

[oddiseas]

Homework Statement

(1/s)Ln(s^2+1)

Find and classify all singularities in the "extended" complex plane. Draw but do not evaluate the contour you would use to find the inverse laplace transform.

Homework Equations

(1/s)Ln(s^2+1)

The Attempt at a Solution

Ln(s^2+1)=[Ln(s+i)+Ln(s-i)](1/s)

so the obvious singularities which are branch points are at s=0,i,-i. Now i am having trouble evaluating for singularities at infinity, and in addition drawing this contour.


Letting s=1/t:
t{Ln[(1/t)+i]+Ln[(1/t)-i]

=t{Ln[(i+ti)/ti]+Ln[(i-ti)/ti)]

=t[Ln(i+ti)-Ln(ti)+Ln(i-ti)-Ln(ti)]

=t[Ln(i+ti)+Ln(i-ti)-2Ln(ti)]

now at t=0 we get:

0[Ln(i)+Ln(i)-2Ln(o)]

Now when t=o we get the point i, which is a singularity and log(0) which is undefined, so i am not sure in this example if there is a branch point at positive or negative infinity>


Can anyone show me how to confirm if there is, and how i can draw the branch cuts. Since i already have one along the y-axis from i to negative i.
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[mighty2000]

Thank you for your question. The function Ln(s^2+1) has three branch points at s=0, i, and -i, as you have correctly identified. To determine if there are any singularities at infinity, we can use the following approach:

Let s = 1/t, then the function becomes

Ln(1/t^2+1)/t

As t approaches infinity, 1/t^2 approaches 0, so the function becomes Ln(0+1)/t = Ln(1)/t = 0/t = 0. This means that as s approaches infinity, Ln(s^2+1) approaches 0, which is not a singularity. Therefore, there are no singularities at infinity for this function.

To draw the contour, we can use the following steps:

1. Start at s=0 and draw a branch cut along the positive real axis to infinity.
2. Draw a branch cut along the imaginary axis from i to -i.
3. Connect the two branch cuts with a semi-circle in the upper half of the complex plane.
4. Connect the two branch cuts with a semi-circle in the lower half of the complex plane.
5. Connect the two semi-circles with a straight line along the real axis.
6. This contour encloses all three branch points and will allow us to find the inverse Laplace transform.

I hope this helps. Let me know if you have any further questions. Good luck with your research!