Complex Analysis: Dogbone Contour Example #3

Today, we evaluate the integral from -1 to 1 of sqrt(1-x^2)/(1+x^2). Tried to one take this video (this video is straight from the camera) so it's not as smooth as the others. Might try to do this more often as editing can take a while, and probably not worth it to correct minor mistakes or stumbles. I'll also have more time to film rather than edit, which is always a plus.

[OleJoe]

Thank you for working out the integral I requested.

In the meantime I found this same integral used as an example in two different books. The first was in Ablowitz and Fokas "Complex Variables" pages 250 to 253. The second was Levinson and Redheffer "Complex Variables" pages 238 to 239.

They both are worked out differently.

The shortest method seems to be Levinson and Redheffer. They take the function f(z) = sqrt(1-z^2) and convert it into f(z) = -iz sqrt(1-1/z^2) for the upper edge of the cut.

Not really sure how that works though.

[fordtimelord]

That was awesome. Your clarity of thought is incredible. I learn so much from your videos.

[angelmendez-rivera]

The beautiful part is that sqrt(2) – 1 is the reciprocal of the silver ratio δ = sqrt(2) + 1, hence the final answer can be simplified further as π/δ.

[zakirreshi]

Wow🤩🤩... that was really great. I learnt a lot from it, especially that part of taking care of argument, and residues at infinity. Thanks for being there to spend your precious time for us.

[kummer]

I love the dog bone contour integrals.

[wolfwerewolf]

There's actually a faster way to deal with fractional(or even complex) powers of a complex number. Instead of saying that z^(1/2)=e^(1/2*log(z)) and applying the formula for a logarithm of a complex number, you can just use the exponential for of a complex number: z=|z|*e^(i*arg(z)). This way you get to the result in one step. Essentially, these two methods are exactly the same, because the method you used in a video is just another interpretation of the exponential for of a complex number, when you are defiing the logarithm of a complex number.

[George-ijgm]

Great! 😄😄😄

In the two little gamma integrals, it seems that we cannot simply calculate (1 + i)^(1/2)(1 - i)^(1/2) = [(1 + i)(1 - i)]^1/2 = (1^2 - i^2)^1/2 = [1 - (-1)]^1/2 = 2^1/2, as (1 + i) and (1 - i) have different angle definition ranges. Right?

[andreadimicco]

Why in 22:08 it is possible to split the square root of a fraction in a fraction of square roots, if, in general, it is not allowed?

[mo]

At 4:34 into the video, should not the last line on the screen be * arg(g) instead of * arg(f)?