When a mathematician gets bored episode 3 (feat. proof for the mellin inversion theorem)

[@CM63_France]

Hi, "terribly sorry about that" : 1:10 , 2:39 , 4:35 , 6:16 , 7:01 , 9:37 , 10:52 , 11:17 , "ok, cool" : 1:20 , 2:04 , 4:42 , 5:24 , 8:02 , 10:11 , 12:23 .

[@Calcprof]

The way I typically did this in Integral Transform courses was by change of variables from the Fourier Inversion Theorem, which is essentially what you do here. Great video!

[@maths_505]

Thank you

[@Bituman1293]

@@maths_505 I have a question, please, about how to determine the inverse Laplace-Carson transform of a function F(p)= C+D/(1+ap^-k+p^-h) where p is a complex number, C, D are real constants, and h, k real numbers such as 0

[@maths_505]

@Bituman1293 I'm gonna have to study that a bit more to provide an answer

[@Bituman1293]

@@maths_505 thank you for your time. i appreciate

[@MyOldHandleWasWorse]

The relevance of Mellin transform to the expression in the thumbnail is immediately obvious once you state the domain of analyticity. (That, and the definition of the Gamma function.) Excellent introduction segment!! I totally forgot that part of the definition. Awesome computations, too. I definitely added this to my math playlist.

[@maths_505]

Thanks homie. Wonderful hearing from you after so long. Hope everything's been good at your end.

[@MyOldHandleWasWorse]

@@maths_505 All is going fine, brochacho. I got my MS, like I mentioned last time! Now I teach for a bit before my Holy Grail: a PhD in math.

[@Nottherealbegula4]

Zero idea what I just watched but amazing video as always

[@StevenKrantz-o5m]

this guy gets it

[@MathPhysRadi]

Absolutely beautiful result.

[@OscgrMaths]

Such a good video, I don't think it's possible for you to miss! You know it's good when the integral has pi and e in the answer...

[@maths_505]

@@OscgrMaths thanks mate

[@leroyzack265]

really this was awesome. the reason why I love transforms is because it gives beautiful results and generate very scary integrals.

[@roeelazar]

I love it when Fourier makes an appearance!

[@maths_505]

Oh hell yeah! Interestingly, it's the first time I've ever invoked the transform on the channel.

[@Sugarman96]

@@maths_505 You should keep using it, electrical engineers love it for a reason

[@MrWael1970]

Awesome proof. Thanks.

[@Max-ls2ye]

I self taught myself basic single variable calculus at 15 and all i have to say is: everything gets so much spicier and cooler when you add an integral sign. (Saying this as a physics enthusiast that finds math stale overall)

[@maths_505]

Exactly why I got into the integrals business

[@alfredcognoet]

beautiful result

[@maths_505]

Indeed.

[@Jacob.Peyser]

Awesome result & presentation! I really enjoyed the derivation. Would you be able to make a (semi-rigorous) preliminary video on the Fourier inversion theorem in the future? It's implicitly used everywhere (e.g. deriving the inverse Laplace/Mellin transforms), but all the 'proofs' I've seen use sketchy limits and/or Dirac deltas/Dirichlet kernels along with wild convergence assumptions. Thanks for your awesome content!

[@maths_505]

Will do

[@Jacob.Peyser]

​@@maths_505 You're the best!

[@Decrupt]

There seems to be a missing negative pi that arises from the circle around the pole at zero? If a is zero that is.

[@sh6700]

Everything in the video makes sense, but there’s something I’m confused about: Isn’t this integral just 0? Consider the contour on the complex plane that begins at negative i*infinity, goes up to positive i*infinity, then wraps clockwise back to negative i*infinity in a semicircle. Since the gamma function has no poles inside that contour, the closed contour integral of the gamma function is 0, and by Jordan’s lemma, the circular section is 0, implying the line integral of Gamma(z) over the part of the contour on the imaginary axis (which is equivalent to the integral in this video) is 0. Where’d I go wrong?

[@eu7059]

There is a pole at Γ(0)

[@sh6700]

@ so shouldn’t the integral diverge since it goes right over that pole?

[@eu7059]

@ no just means you can’t apply the residue theorem. If you create a similar countour but with a ball of radius ε going around the pole, and take the limit as ε goes to 0, you would have a full analytic countour and it would probably yield the same result as in the video

[@maths_505]

Exactly, the contour cannot walk over the pole and we would need that semi circle of radius ε. I am still thinking about how to evaluate this explicitly using contour integration.

[@sh6700]

@@eu7059 Oh. I would’ve expected the limit of the ball integral to go to infinity as it approaches a pole? But maybe not since it’s not a simple pole? Idk I haven’t taken complex analysis yet

[@SuperSilver316]

Yeah that’s the nature of integral transforms, for certain functions you have to create a strip of regularity where the function is allowed to remain analytic, otherwise you run into problems with the Inversion. If you wanna see this in action, try the Fourier Transform of 1/(x^4+1), and see what happens to its transform at the origin.

[@maths_505]

Sounds like a plan

[@narkitikx9380]

where do you find these 😭

[@maths_505]

I was looking for a proof of the mellin inversion theorem and Wikipedia just gave an outline. Then I just played around with the result since....well....gamna functions 😂

[@YUNGDADDYDIFFERENTIAL]

This was real

[@maths_505]

You mean complex

[@YUNGDADDYDIFFERENTIAL]

@@maths_505This wasn’t real