Ramanujan's integral is ridiculous!
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- Опубликовано: 14 дек 2024
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Correction: around the 7:15 mark I forgot to switch the limits of integration to cancel out the negative sign. Apologies.
your accent had a singularity at the second s in substitution
I recommend you start working with a script, the calculations already completed beforehand. This way, these mistakes can be avoided.
I agree 100%. It’s very hard to follow when you go down the rabbit hole of sign/algebra errors.@@angelmendez-rivera351
Amazing result. Ramanujan was a genius
But why do people in India only feel the genius of Indian mathematicians ... there were several outside the Indian subcontinent... praise them as well
@@ajaypratap4467 bro this video is related to ramanujan's integral
Moreover we respect every mathematician
@@ajaypratap4467 yes but Ramanujan was one of a kind, his story is unparalleled and his achivements, considering the whole background, are absolutely praiseworthy. And I'm not even indian
@@ajaypratap4467 I do not feel the need to only praise and respect people of my country, I will respect anyone respectable.
He is just a normal mathematician he has used this problems for the research outside india
At 8:52 you can't just split marked integral into two, since both are infinite and you obtain indeterminate form "infinity minus infinity". The problem disappears if you replace the integral from 0 to alpha^2 with lim_{ epsilon -> 0+ } integral from epsilon to alpha^2 .
This guy makes an integral equation sounds like u. Love it!!!!
At minute 8:59, you have to switch the limits of integration to be able to do the next step. So, the first integral should have negative sign. Thank you for your effort.
Yes you're right. I've corrected that in a pinned comment. Thank you professor ☺️
4:04 so smooth
😂😂😂
incredible result...ramanujan was really a genius
Omg this is sooo beautiful!
9:45 When you split integral from 0 to alfa2 in integrals from 0 to 1 and 1 to alfa2 you imply alpha2>1 but this is not per se guaranteed. You told just alpha>0. Is this derivation still valid if alpha
hi maths 505: so are you planning to make a video on proving zeta(4)=pi^4/90? thanks for your time.
Will do
8:59 Can we do that? In the first integral the bounds are the wrong way around. What have I missed?
Bro just switch their places
It should be from a^2 to inf because the - sine du=-a/z^2
@@yoav613oh now I see it!
Thanks bro
5:40 darn that's one isomorphic accent, great vid as always my friend.
Thank you comrad ManStuckInABoxovsky
Sir, you split the limit from 0 to (alpha)^2 into limit from 0 to 1 and 1 to (alpha)^2. But alpha>0 its not necessary that it is greater than 1. Can we actually insert 1 like that between limits? 9:40
❤ nice
Thanks bro
nice indeed
My brain is still spinning!
8:26 - 8:29 No, you cannot do that, because two integrals you are trying to split it into diverge.
9:04 integral from infinity to alpha square plus integral from zero to alpha square= integral from zero to infinity,which law is this sir?
Read the pinned comment😂
int_0^{alpha^2} (1/z) does not make sense
True. Two divergent integrals could sum to anything. Perhaps you should’ve combined the integrals first instead of splitting them first.
Nice
@ t=5:30, you integral representation of the Euler constant is incorrect. Punch that into the calculator.
Wonderfully presented man!!
How you give z 2 identities in one integral
The solution you presented was by ramanujan himself? Where was it published?
Not the accent part.
😂 indeed
@@renerpho but where did you found that ?can you share the link of first published solution by him ,if it's available
that accent escalated quickly
Does this mean genetics is indeterminate after 69 generations?
Really wonderful