Indeed…RIP all lost young geniuses. Ramanujan’s years at Cambridge - as described by his late, great English patron, friend and great mathematician in his own right, G. H. Hardy in ‘A Mathematician’s Apology’ - I found sadder than his eventual passing; for he struggled mightily with the unusual food (so different from that of his native India) and was so frequently found struggling with both the cold climate and a torrent of unfortunate illnesses. Despite this, he achieved so much remarkable Mathematics in his short years with a truly incredible mind for numbers (and arithmetic) - search ‘Hardy’s taxi number’ [edited: I don’t know who tf ‘Harry’ was, or even if either Hardy or Ramanujan knew him] for my favourite!
Of course ,there is a cheap way of looking the integral up in a table of Fouriertransforms . Your explanations however help to understand possible ways of deriving the desired result.
1:10 this looks a little bit similar to the equation hermite orthogonal functions and the fourier transform equation involving them (the one from gamelin's book), I think you can write the function sin(x)/x^s-1 in terms of the orthogonal functions... idk they feel related to me...
Another method to do this - Let, I(t) = integral 0 to ∞ [sin(tx) /x^n+1] dx Taking laplace transform- L{I(t) } = int (0 to∞) [1/(s²+x²)(x^n) ]dx let, x=s tany L{I(t) }=1/s^n+1 int(0 to π/2)[coty]^ndy =1/s^n+1 × π/2cos(nπ/2) Now, taking inverse laplace - I(t) = π/2cos(nπ/2) L^-1{1/s^n+1} = [π/2cos(nπ/2)] [t^n/gamma(n+1)]
Great Video! What I always wonder about, is how long it takes you to come up with those solutions and how you do it? Do you just try out different methods, or are you doing it another way?
17:20 Should be Sin(pi[(1-s)/2]) but the phase shift applies both ways so it’s still your W against my boi Ramanujan🍷🗿
Oh right!
Thank God cosine is an even function....or else Ramanujan's ghost would've put the hit on me😂😂😂
Ramanujan, Abel, Galois, Riemann,... so many brilliant minds lost in their youth.
Indeed…RIP all lost young geniuses. Ramanujan’s years at Cambridge - as described by his late, great English patron, friend and great mathematician in his own right, G. H. Hardy in ‘A Mathematician’s Apology’ - I found sadder than his eventual passing; for he struggled mightily with the unusual food (so different from that of his native India) and was so frequently found struggling with both the cold climate and a torrent of unfortunate illnesses. Despite this, he achieved so much remarkable Mathematics in his short years with a truly incredible mind for numbers (and arithmetic) - search ‘Hardy’s taxi number’ [edited: I don’t know who tf ‘Harry’ was, or even if either Hardy or Ramanujan knew him] for my favourite!
Human beings can take the informations from Akashi(the library of the universe)
Those guys had math{logic} minds so they worked highly on Trigonometry and calculus full logics behind it , we mathematicians appreciate it
Eisenstein too (re Gauss' comment)
RMT has to be my top 5 favourite theorems, but I would really love to see a video on your channel proving this theorem.
Challenging a dead person to a death match is one way to make sure you win...
Considering it's Ramanujan we're talking about, I think this makes the fight about equal, actually.
@@renerpho wrong
Love to finally see RMT on this channel
I wrote 10 pages article on him.
I fascinated by his life.
Prodigy of mathematics lived tiny time in the circle
Of course ,there is a cheap way of looking the integral up in a table of Fouriertransforms . Your explanations however help to understand possible ways
of deriving the desired result.
wonderful. Really creative 😍
1:10 this looks a little bit similar to the equation hermite orthogonal functions and the fourier transform equation involving them (the one from gamelin's book), I think you can write the function sin(x)/x^s-1 in terms of the orthogonal functions... idk they feel related to me...
Sounds worth researching
This is crazy 😮
when my mom walked in i immediately changed to porn cause it was easier to explain
😂😂😂😂😂
She would have seen it as mathematical masturbation!
Another method to do this -
Let, I(t) = integral 0 to ∞ [sin(tx) /x^n+1] dx
Taking laplace transform-
L{I(t) } = int (0 to∞) [1/(s²+x²)(x^n) ]dx
let, x=s tany
L{I(t) }=1/s^n+1 int(0 to π/2)[coty]^ndy
=1/s^n+1 × π/2cos(nπ/2)
Now, taking inverse laplace -
I(t) = π/2cos(nπ/2) L^-1{1/s^n+1}
= [π/2cos(nπ/2)] [t^n/gamma(n+1)]
But integral from 0 to pi/2 (coty)^ndy is diverges for at least integers n, what is the formula next for it? Maybe i don't see something
Thank you!
There is an error in writing the sign of the "s" in the second proof.
Great Video! What I always wonder about, is how long it takes you to come up with those solutions and how you do it? Do you just try out different methods, or are you doing it another way?
Hey kamaal , this can also be done using the maz identity of laplace transform! in just 2 mins :)
@ 16:42 the power of u should be (s-1)/2, not -(s+1)/2.
If you plug in s=0, we know LHS integral is pi/2. but does it hold true in the closed form obtain on the RHS?
That's not surprising....the video is based on the constraint that Re(s) belongs to (-1,0)
12:30 we need x to be positive here (which in our case it is) but watch out
Why do you limit Re(s) to (-1,0)? The integral obviously converges when -1
Can you give the link of master theorem plz?
13:27 int[0,infinite]t^-s should be t^s
13:27 int[0,infinite]t^-s the power of t should be s ,not -s
Where is your accent from? I thought for a second south asian, but then the way you pronounced Ramanujan made me doubt if you were.
Oh I am south asian....specifically I'm Pakistani
@@maths_505🗿🥂
I liked a movie about him
Wow
A very cheap method to do is write sin in exponents and write it directly in terms of gamma function
🎉
atoms and coding ...and AI2
.sin x dx= d(-cos x)
|=|d(-cos x)/(x^2+1)=
- cos x- |(-cos x)(x^2+1)^-2(2x dx)=
=....memang harus varkompleks
You should write the letter "n" properly. Forgive my candor, but tidy up your penmanship.