The generalised Dirichlet integral: integral of (sinx)^n/x^n from zero to infinity
ฝัง
- เผยแพร่เมื่อ 21 ม.ค. 2023
- This is it!!!
The ghost of Dirichlet may now rest in peace!
The approach is kinda strange but absolutely awesome when you put all the pieces together.
You could cancel Gamma function with factorial
Yeah. n!/Γ(n)=n
Ong
Can you please elaborate
@@ShanBojack For any integer n>1, gamma(n) = (n-1)! It's pretty easy to prove by induction
Don't spoil the video
You spoiled the video by saying that the video features gamma function😂
Wow never in my life would I have guessed you could reduce the integrals of (sin(x)/x)^n to a sum of simple integrals of the form ∫ dt/(a^2+t^2)
I am impressed by your generalized solution of this integral. This is the first generalized solution of this integral I have seen so far. That was a good exercise for you. Thanks for sharing.
th-cam.com/users/shortsOadiTfmwjTI
This video is so awesome,i learned so much from you thank you
th-cam.com/users/shortsOadiTfmwjTI
This was feature as Problem 1064 in Mathematics Magazine (in 1979). Two solutions were given.
I see sinx/x, my mind immediately goes to the Fourier transform. The Fourier transform of a window of 1/2 from -1 to 1 is sinw/w, meaning you can use the pattern of the simple convolution to find the nth convolution of said window with itself, at which point the integral just becomes the inverse Fourier transform of a simple, band limited polynomial function.
I thought about the exact same thing, do you know a pattern for the n th convolution of the fonction ?
Yes, it's the central frequency of the fourier transform.
For n = 1 the frequency spectrum is the rect function (one constant function).
For n = 2 the spectrum is a triangle function (two linear functions).
For n = 3 the spectrum is a composite of three quadratic functions.
For n = 4 it's a composite of four cubic functions
For n it's a piece wise composite of n parts which are (n-1)tic functions.
I have to think about how to derive the central frequencies.
@@JobBouwman I also saw that, and calculated the first terms, and understood we just want the value in 0, though we can only do an induction where the whole n-th function is known in the induction hypothesis
You have made my day.
I always wanted someone to do this
Thank you very much
Me too
So I decided why not just take it up myself 😂
@@maths_505 can I ask
How old are you ?
25
@@maths_505 I hope that I will be at that level when I am 25
I mean I understood everything be there is no way I would have thought of doing that
Great work
th-cam.com/users/shortsOadiTfmwjTI
Instead of integrating by parts, you can expand ( sin x ) ^ n as a sum of complex exponentials. This way, you can find a closed formula
integral ( ( sin ( x ) / x ) ^ n , x = 0 , x = infinity ) = n * pi * sum ( ( -1 ) ^ k * ( n - 2 * k ) ^ ( n - 1 ) / ( ( n - k ) ! k ! ) , k = 0 , k = floor ( ( n - 1 ) / 2 ) ) / ( 2 ^ n )
Bruh....
I did in a same way
I generated the exact same to same formula in November 2022...........
I really enjoyed this.. thank you.
I would be very interested to see a plot of I(n) vs n. How does it behave? Is it monotonic? Does it exhibit interesting patterns? How do the even vs odd cases compare?
Great video!
th-cam.com/users/shortsOadiTfmwjTI
First of all, there are papers on this. Nevertheless, I did exactly this as I found it very interesting as well. I evaluated the integrals up until 70 numerically (because my code runs into some problems I dont understand) and plotted them. Maybe you could do this analytically, but I didnt try at all. If you are interested in this, you can read about that in papers, which reduce this all down to one sum.
Anyways, when you plot the integral solutions, its not too interesting at all. Its monoton decreasing, maybe it is converging to some value or to 0 for n->inf, but I dont know. Also, it is decreasing slower than 1/x, so the convergence is rather slow. But numerically there are no real patterns for odd and even
Nvm on the convergence part. For n->inf the integrals converge to 0 as sin(x)/x is bounded by 1, the integral is obviously convergent and therefore, you can interchange the limits. Then the integrand is zero besides at x=0, which does not contribute to the integral
Excellent calculation.
Great video. Thank you
th-cam.com/users/shortsOadiTfmwjTI
Awesome
Excellently laid out, as usual, thanks a lot.
Just a question: I like your path integral solution for n=1, and succeeded applying the same ideas to the case n=2, but I could not solve the case n=3 in this manner. I'd greatly appreciate your comment.
There's a qncubed3 video solving this using contour integration. You should check that out....its quite a nice video
@@maths_505 Thank you for the hint. I am amazed about the high speed with which you use to reply to questions. Great!
This is first class!!!! 🤩
th-cam.com/users/shortsOadiTfmwjTI
I have been able, through some small amount of algebra, to reduce the integral to a sum for even n (odd n is no more difficult). This is:
integral{x=0 to infinity} (sin x/x)^(2n) dx = n pi sum{k=1 to n} (-1)^(n-k) k^(2n-1) / ( (n-k)! (n+k)! ).
This is always a rational multiple of pi. Multiplying and dividing by (2n)! turns the factorials into a binomial coefficient, which shows that the denominator of the rational multiple (when reduced) always divides (2n)!.
In particular, when n=1 (so 2n=2), the sum is the single term (1^1/0!2!) = 1/2 and the integral equals pi/2. When n=2 (so 2n=4), the value works out to 2 pi (-1^3/1!3! + 2^3/0!4!) = pi/3. When n=3 (so 2n=6), the value is 3 pi (1^5/2!4! - 2^5/1!5! + 3^5/0!6!) = 11 pi/40. And so on.
I'm happy to post a PDF with the details.
I'd love to read that PDF
My email is in the about section
th-cam.com/users/shortsOadiTfmwjTI
And I wold be happy to read it
Share it bro share it
YOOOOOOOO WE MADE IT BROOOOO, but we takled the natural final boss, we still need to defeat the real, boss, and of course, dare I say, the c o m p l e x Boss.
It's actually quite intriguing to think about sin^z(z)/z^z, kind of a weird entity, although I think deviating much from the n case, since we're dealing with variables not constants, and that it's value depends on the parameter we choose, also sin^z(x)/x^z is even more intruiging, I'm no expert in choosing contours but I think we can use a rectangular contour? hmm... actually this looks alot like an exponential so I think it can be takled with a semi-circle integral... sorry it takes time for me to brainstorm lol!
but awesome video anyways, it felt like an anniversery since I discovered this haven in the begining of chrismass vacation, and now it's about to end...
btw I starting collecting all the goofy integrals inside a note book of mine called "the Big Book of Integrals" maybe one day we'll collect em all :D
11:54 I spy a missing minus sign.
after watching this video I'm really wondering how Matrix transformation and Diagonalization would help in cases like simplifing reduced formulas like this.... idk Just a weird Idea that came about.
th-cam.com/users/shortsOadiTfmwjTI
Hello sir can you do some jee advanced calculus problems these are some of the toughest undergraduate problems
Take a look at: A direct computation of a certain
family of integrals, by
L. Fornari, E. Laeng and V. Pata
You will find a more general formula, more explicit, and with a simpler proof.
Integral 0 to infinity sin^69(x)/x^69 = 0.260765
Nice
Mans used Wolfram alpha surely
Late comment but could you not reduce the last 2 integrals (for general n) to sums and products using contour integration?
A year or so ago when I first watched bprp’s video about the 3rd Dirichlet integral I decided to try and solve a generalized form of the problem and I managed to come up with a formula involving some finite summations. My solution development was pretty bad and not remotely rigorous (I’d only just started learning multivar within the pasts few months and most of what I knew of it still came from TH-cam) and the end results was still pretty ugly but I was pretty happy with myself considering I’d only just learned Feynman’s technique. I should try it again and see if I can get a bit more rigor involved and then maybe try solving generalized fresnel integrals.
I might try looking back through my math notes from a year ago to find my solution but I’m not sure I’ll be able to because that’s a lot of notes to check through and I didn’t organize them at all
I managed to find it. The answer was:
For odd n:
\frac{\pi n}{2^n} \sum_{k=0}^{\frac{n-1}{2}} \frac{(-1)^k(n-2 k)^{n-1}}{k !(n-k) !}
For even n:
\frac{\pi n}{2^n} \sum_{k=0}^{\frac{n}{2}} \frac{(-1)^k(n-2 k)^{n-1}}{k !(n-k) !}
This wasn’t done at all rigorously so I can leave it to you to actually prove it but it works for all n that I’ve tested. It basically results from doing a power reduction followed by Feynman’s technique going to the n-1 derivative and assuming that the constant after antidifferentiating is always 0 (which is true but I haven’t proved it for the cases where you end up with a cosine in the numerator). Maybe one day I’ll go ahead and do it rigorously but honestly I’m pretty happy to have a solution for the nth Dirichlet integral that only has finite summations.
@@skyethebi th-cam.com/users/shortsDvT_6yRrSOk
I think you should be able to calculate the t-integral for general n by the residue theorem.
I agree, although computing the residues is nasty algebra. I was thinking partial fractions might be easier, since all of the 1/(t^2+n^2) factors integrate nicely to arctans.
@@davidblauyoutube Actually. the residue calculations are fairly easy, since all the zeros in the denominator are simple. In any case, the starting Dirichlet integral can be directly evaluated by the residue theorem. The technique to be used is explained in Ahlfors, Complex Analysis, Section 5.3 (pp. 154-159 in the third edition) Evaluation of Definite Integrals.
Please solve
Integral(-1)^[x]
Correct component
Nice
I got a form which only involves a finite sum,is there a way I can show it to u
I turned the integrals into sums where you just need to substitute
I used a limit so I don't need to change the value of n everywhere in the sum
The sum doesn't work when n is 1 or 2
Maybe if you can simplify the sum it would be the key to a more general formula
This is the sum when n is odd
DLimit((-1)^((t-1)/2)*t*Pi/2*(1/DProduct((2*y+1)^2-1,y,1,(t-1)/2)+t^(t-2)/DProduct((2*y+1)^2-t^2,y,0,(t-3)/2)+DSum((2*x+1)^(t-2)/(DProduct((2*y+1)^2-(2*x+1)^2,y,0,x-1)*DProduct((2*y+1)^2-(2*x+1)^2,y,x+1,(t-1)/2)),x,1,(t-3)/2)),t,n)
And this is the sum when n is even
DLimit((-1)^((t-2)/2)*t*Pi/2*(2^(t-3)/DProduct((2*y)^2-2^2,y,2,t/2)+t^(t-3)/DProduct((2*y)^2-t^2,y,1,(t-2)/2)+DSum((2*x)^(t-3)/(DProduct((2*y)^2-(2*x)^2,y,1,x-1)*DProduct((2*y)^2-(2*x)^2,y,x+1,t/2)),x,2,(t-2)/2)),t,n)
If you need the sums in a different way of writing or photo of them written tell me
I wouldn't have done this without your video and without blackpenredpen's way to do partial fraction (the cover up method)
I wonder if contour integration can deal with this integral.
when you do IBP , there is 1 mistake in the 2nd column 3rd integral. it should be e^-xt/t^2 ..
Good one bhai
What hardware and software do you use to make your videos?
It's the default Samsung notes app.
Wow
This is a NOICE homework?
Swag
interesting
I expect that the value (fraction) gets complicated with increasing n but for n=6 it's simply 11 pi / 40
The partial fractions are mostly arctans so yeah one can expect nice results
th-cam.com/users/shortsOadiTfmwjTI
For the case of n being an odd integer, are you sure that the nominator is also n!? It seems to me it should be n!(n-1) instead.
Is there a closed form answer for this integral?
th-cam.com/users/shortsDvT_6yRrSOk
th-cam.com/users/shortsOadiTfmwjTI
=> sin1 - (sin-1) •dx=0,0349
Homework for n = 69:
I_69 = 998,343,250,657,696,659,388,623,720,040,379,470,133,597,913,727,156,038,875,228,541,953,757,055,024,051,950,731,143,915,115,744,965,383,517,459,741*pi / 12,027,626,526,020,745,490,674,841,999,023,506,972,927,778,751,265,048,277,098,155,425,840,767,196,171,570,579,020,944,634,806,272,000,000,000,000,000
I totally didn't use Mathematica to get this. Pinky promise!
How the fuck you've did it ...???☠️
Nice! However, after unleashing some of my own tricks on this integral I got a closed form solution, or rather two, one for the n=odd case and one for the n=even case. Both are simple finite sums.
ADDED: both cases can be combined to obtain a single sum valid for all positive integer n.
th-cam.com/users/shortsOadiTfmwjTI
@@erictrefeu5041 yes - see my latest comment.
@@TheoH54 oui j'ai vu. Nous sommes d'accord Theo. Cette formule est valable pour m pair ou impair. il m'apparaissait utile de compléter cette vidéo avec une formule générale.
@@erictrefeu5041, bon soir - I would like to ask you, do you have an expression for INT{0;inf} sin(x)^n/x dx ? If yes, maybe we can do something together. I've got a compact expression, apparently not known, with no sums for n=odd.
@@TheoH54 Hello Théo, for (sin(x)/x)^m dx ? , yes of course, follow this link : th-cam.com/users/shortsOadiTfmwjTI
Or for sin(x)^n/x dx ?... je vais réfléchir, je n'ai pas encore cherché (ca doit etre faisable)
Integral 0 to infinity
n=1 , then π/2
n=2 , then π/2
n=3 , then 3π/8
n=4 , then π/3
n=5 , then 115π/384
n=6 , then 11π/40
n=7 , then 5887π/23040
I already solved this for any positive n by using only complex analysis and binomial theorem, and I've my own general formula. You've to only give value of n and then you'll get it
Note: i solved it during my 3rd semester of B.Tech when i was 19 years old.........
that's baddass
@@Antonio-qe2cc
Next time, I'll try to be kickass
th-cam.com/users/shortsOadiTfmwjTI
The solution for the Hw is: 👅
j'ai mieux
I turned the integrals into sums where you just need to substitute
I used a limit so I don't need to change the value of n everywhere in the sum
The sum doesn't work when n is 1 or 2
Maybe if you can simplify the sum it would be the key to a more general formula
This is the sum when n is odd
DLimit((-1)^((t-1)/2)*t*Pi/2*(1/DProduct((2*y+1)^2-1,y,1,(t-1)/2)+t^(t-2)/DProduct((2*y+1)^2-t^2,y,0,(t-3)/2)+DSum((2*x+1)^(t-2)/(DProduct((2*y+1)^2-(2*x+1)^2,y,0,x-1)*DProduct((2*y+1)^2-(2*x+1)^2,y,x+1,(t-1)/2)),x,1,(t-3)/2)),t,n)
And this is the sum when n is even
DLimit((-1)^((t-2)/2)*t*Pi/2*(2^(t-3)/DProduct((2*y)^2-2^2,y,2,t/2)+t^(t-3)/DProduct((2*y)^2-t^2,y,1,(t-2)/2)+DSum((2*x)^(t-3)/(DProduct((2*y)^2-(2*x)^2,y,1,x-1)*DProduct((2*y)^2-(2*x)^2,y,x+1,t/2)),x,2,(t-2)/2)),t,n)
If you need the sums in a different way of writing or photo of them written tell me
I wouldn't have done this without your video and without blackpenredpen's way to do partial fraction (the cover up method)
th-cam.com/users/shortsOadiTfmwjTI