Complex Analysis: Is this solution way too overkill?

The most amount of arithmetic I've done in a video for quite some time.

[BCQM_BCQM]

20:07
I guess you missed a π there and the final result should be π³/8, right?

[maths_]

Love the new style of using the digital blackboard. Its quite easy to highlight stuff using various colours and that makes it more engaging.
As far as the contour integration is concerned....WHO CARES IF IT'S OVERALL!!!
THIS IS AWESOME!!!!
A mathematician doesn't need efficiency. He needs dopamine from awesome contours!!!
Great job as always bro.

[riadsouissi]

Nice trick with the polynomial. I learnt something new. Thanks.

[Decrupt]

Interesting method, very cool. I like the videos with unique methods involving complex analysis; always nice learning more

[exodus]

Is there a way to know if a public keys private key will be odd or even regardless of how big the range is in secp256k1? Please help

[zinzhao]

I got -pi^3/8 also the graph in desmos shows that the majority of the area is negative so how did we get different answers?

[KhaledRadwan-kubh]

what is your application for this black board plz?

[TheRandomFool]

It's interesting, but honestly, I don't think it's worth the effort. I think it would still be better to change the contour to being across the real axis plus upper half circle.

[nightmareintegral]

I prefer the old way of your videos.
Anyway nice integral 💚

[mo]

At 16:59 of the video I question the validity of replacing (log(z))^2 with P(log(z)). Earlier in the video you showed that P(log(z)) - P(log(z)+2πi) = -6πi(log(z))^2. This means that, in general, P(log(z)) ≠ (log(z))^2.