Man, I remember when I watched Peyam solve this integral, you did it on 3x speed, this has been just beautiful, we should call you Magician Kamaal because you're pulling tricks left and right
I couldn't help but notice that alpha=0 limit exsit in that gorgeous answer, so I did it as a homework and got 1/2ln(e^-γ * pi/2) which is a nice bonus result for a special case of first integral at alpha=0
15:09 i just finished calculus I and did not understand most of the things of the video, but i think it is pretty cool watching you solving these integrals
The same denominator of the integrand (x² +2xcosα + 1) appears in the result of the sum ∑ Rⁿcos(nθ), for n ≥ 0. (That is, if we let R=-x and α=θ) I wonder if this has any significance, of if it could be used to solve this integral in another way. edit: nevermind, after 4:00 you used a very similar argument
On a serious note tho, most of your videos actually do in some way or another and it's really enjoyable to watch you layout the solution development for such problems. I really need to sleep now but I will give it another go in the morning am sure.
This is far harder than the coxeter's integral. What makes it insane is that it's linked to geometric series, Fourier series, Laplace transform. Almost all the craziest analysis stuff where required to solve it. It was really a worthy opponent.
Shoutout to one of the few world famous Swedish matematicians. We also have Ivar Fredholm and 50% of Harry Nyquist. The Norwegians of course have Marius Sophus Lie, who's on another level altogether. And the Danes have all the celebrated Physics/QM guys. The Finns have lots of great mathematicians too.
Hello! Is there a "simple" function whose integral has a very "complex" closed form? By a simple function, I mean a function similar to the one in the video, consisting of a few elementary functions, at most rational. By a very complex closed form, I mean expressions that involve various constants, raficals, logarithms, sums and multiplications etc. (ellipitc integrals, hylergeometric, gamma, zeta functions are not allowed, solution must be close). Thanks for any insights!
Me before watching this video: the polinomial in the denominator can be computed as (x+e^itheta)(x-e^itheta). now we define I(s) = int from 1 to infinity of log(x)^s/(x-a)(x-b). applying PFD. a = complement of b a = e^itheta. I(s) = int 1 to infinity of (1/b-a) * (log(x)^s/(x-b)) + (1/a-b) * (log(x)^s/(x-b)) . we split into I1(s) and I2(s). apply u = log(x) e^u du = dx e^u *u^s/(e^u-b) = u^s/(1-b*e^-u) so using some melliny mellin magic I1 evaluates to.. iuhun I'm just going to watch the vid now.
At 10:26-ish, you need to divide the upstairs and downstairs by 2j, not just 2, to get the complex sine definition. Ultimately you're still multiplying by 1, but the point stands 😉
@@maddog5597 the parametrised results has 1/sin(alpha), which would suggest the integral diverges for multiples of alpha that is a whole multiple of pi.
@@Sugarman96 No. As alpha goes to zero, the Gamma functions cancel out leaving you with log(2*pi)^ (alpha/pi). You can bring the alpha/pi out in front of the log to give you a term that looks like alpha/sin(alpha). As alpha approaches zero, this approaches one.
@@maddog5597 why can you just take the limit when the result comes from just plugging in a value of alpha? And how does that work for values of alpha that aren't zero? The gamma functions _don't_ cancel out then
@@Sugarman96 I was just responding to your original post, where you claimed ln(ln(x))/(x+1)^2 diverged. The main problem is that Kamaal should have put limits on alpha. The integral diverges for alpha = pi. It probably diverges for other values of alpha that make the cos go negative.
1) you forgot a - sign at 14:30 in the second sum 2) If you replace cos(alpha) with cos(alpha * x) in the denominator, is this related to Coxeter's integral?
This one was from a patreon subscriber. Sorry I haven't been very responsive to requests, I've been extremely occupied these past few months. Even this integral was recommended to me almost a year ago.
this integral is zeroeth degree murder
that's a horror movie
man I know you do crazy integrals but this is something else bro this is a murder
Finnaly legendery integral!
Next Coxeter integrals! 🖤
Man, I remember when I watched Peyam solve this integral, you did it on 3x speed, this has been just beautiful, we should call you Magician Kamaal because you're pulling tricks left and right
Why am I not surprised that peyam solved it before me😂
Peyam is probably the most underrated RUclipsr in the math space.
I do believe it's time to face your destiny and tackle the Borwein integrals.
“Thank You Math 505” we all said in Unison
I did not survive... Will try again tomorrow.
@@TheDhdk that's the spirit 🔥
I suffered the little death, in the Shakespearean sense.
The important thing is you survived....just as I did
@@maths_505 It means the big O and I'm not talking about notation
same here😅
Coomer series coming in clutch! 🎉
Breathtaking!
@@trelosyiaellinika and you survived!
@@maths_505 Not only. I enjoyed it immensely!
Extremely Insane. Thank you.
Hi,
"ok, cool" : 2:23 , 6:24 , 14:54 , 20:55 , 22:28 , 22:57 ,
"terribly sorry about that" : 11:00 , 11:36 , 14:36 , 18:42 , 26:50 , 27:17 .
I couldn't help but notice that alpha=0 limit exsit in that gorgeous answer, so I did it as a homework and got 1/2ln(e^-γ * pi/2) which is a nice bonus result for a special case of first integral at alpha=0
so early there isn't even a thumbnail
15:09 i just finished calculus I and did not understand most of the things of the video, but i think it is pretty cool watching you solving these integrals
@@felps314 thanks mate
Early in the morning and I'm watching such an integral instead of going to work!!! 😅
Time well spent😂
Is the integral related to the poisson kernel?
You could say that the quadratic was a known kernel
I like Kummer series. Exciting integral!
Como se llama el resultado en 15:36?
The same denominator of the integrand (x² +2xcosα + 1) appears in the result of the sum ∑ Rⁿcos(nθ), for n ≥ 0.
(That is, if we let R=-x and α=θ)
I wonder if this has any significance, of if it could be used to solve this integral in another way.
edit: nevermind, after 4:00 you used a very similar argument
Will come back to this video when my math skills are better
@@euler1 bro this video will literally improve your math skills in one go
On a serious note tho, most of your videos actually do in some way or another and it's really enjoyable to watch you layout the solution development for such problems. I really need to sleep now but I will give it another go in the morning am sure.
Aight bro
Sleep well and keep making those math gains.
This is far harder than the coxeter's integral. What makes it insane is that it's linked to geometric series, Fourier series, Laplace transform. Almost all the craziest analysis stuff where required to solve it. It was really a worthy opponent.
Coxeter is honestly just alot of algebra. This however is an entirely different story.
Wicked awesome.
@@jkid1134 another survivor!
at 10:30 it should be divided by 2i, shouldn't it? ik it doesn't matter but still
Great integral man
Shoutout to one of the few world famous Swedish matematicians. We also have Ivar Fredholm and 50% of Harry Nyquist. The Norwegians of course have Marius Sophus Lie, who's on another level altogether. And the Danes have all the celebrated Physics/QM guys. The Finns have lots of great mathematicians too.
Hello! Is there a "simple" function whose integral has a very "complex" closed form? By a simple function, I mean a function similar to the one in the video, consisting of a few elementary functions, at most rational. By a very complex closed form, I mean expressions that involve various constants, raficals, logarithms, sums and multiplications etc. (ellipitc integrals, hylergeometric, gamma, zeta functions are not allowed, solution must be close). Thanks for any insights!
Shredder! Minor quibble. At ten minutes thirty, divide by 2i to get sine.
Thankfully we are dividing in the numerator and denominator so the constants cancel.
Interestingly, I(a) where a = pi/2 is the result for the Vardi integral, correct?
10:42 part is beautiful
The universe is now speaking to me
I survived, but now I carry with me the weight of the experience.
Wow that was just insane
Me before watching this video:
the polinomial in the denominator can be computed as (x+e^itheta)(x-e^itheta).
now we define I(s) = int from 1 to infinity of log(x)^s/(x-a)(x-b).
applying PFD.
a = complement of b
a = e^itheta.
I(s) = int 1 to infinity of
(1/b-a) * (log(x)^s/(x-b)) + (1/a-b) * (log(x)^s/(x-b)) .
we split into I1(s) and I2(s).
apply u = log(x)
e^u du = dx
e^u *u^s/(e^u-b) = u^s/(1-b*e^-u)
so using some melliny mellin magic I1 evaluates to.. iuhun I'm just going to watch the vid now.
At 10:26-ish, you need to divide the upstairs and downstairs by 2j, not just 2, to get the complex sine definition. Ultimately you're still multiplying by 1, but the point stands 😉
If I were a prof I'd put that in a test just for pranks
Also, a weirdly specific conclusion is that the integral of ln(lnx)/(x±1)^2 over (1,inf) diverges
No! - ln(ln(x)/(x+1)^2 converges
@@maddog5597 the parametrised results has 1/sin(alpha), which would suggest the integral diverges for multiples of alpha that is a whole multiple of pi.
@@Sugarman96 No. As alpha goes to zero, the Gamma functions cancel out leaving you with log(2*pi)^ (alpha/pi). You can bring the alpha/pi out in front of the log to give you a term that looks like alpha/sin(alpha). As alpha approaches zero, this approaches one.
@@maddog5597 why can you just take the limit when the result comes from just plugging in a value of alpha? And how does that work for values of alpha that aren't zero? The gamma functions _don't_ cancel out then
@@Sugarman96 I was just responding to your original post, where you claimed ln(ln(x))/(x+1)^2 diverged. The main problem is that Kamaal should have put limits on alpha. The integral diverges for alpha = pi. It probably diverges for other values of alpha that make the cos go negative.
Can u cover Green's therom and scattering therom
You ruined my lazy Sunday. Will I survive? Greetings from Germany!
1) you forgot a - sign at 14:30 in the second sum
2) If you replace cos(alpha) with cos(alpha * x) in the denominator, is this related to Coxeter's integral?
The negative was factored out. And I don't think this is related to any one of coxeter's integrals.
I think I've survived
Oh wow, didn’t realize this integral would invoke the Laplace transform and the Fourier series… wow
I survived it! Dr Peyam did this integral on RUclips a while ago.
barely survived this one
Where do you find all these integrals? Are they sent by the subscribers? I sent you two in the last few months.
This one was from a patreon subscriber. Sorry I haven't been very responsive to requests, I've been extremely occupied these past few months. Even this integral was recommended to me almost a year ago.
I really enjoyed this. But I have one question I dont understand why you wrote (2*pi)^(alpha/(2*pi)) shouldnt it be just (2*pi)^(alpha/2).
Survived!
13:31 dont mind me, this is just my personal timestamp
15:05 Im... still breathing...
I died, but I turned out fine as well
awesome
My question: WHERE would you encounter such a monstrosity???
Obviously in a near death experience 😂
Love it
Somehow I survived and honestly… I don’t know how
Now it’s time to prove that identity to ln(gamma)
@@alielhajj7769 it's that time alright....
You must devide by 2i to get sin, not 2.
@@willemesterhuyse2547 yes that's 2 errors cancelling out 😂
Lordy. 🏆
Bro you survived too!
@@maths_505 Dazed, but not confused. 👍
I'm okay
Yoooo I’m alive, but my god man
Again poor maskeroni canseling out😂
IDK if it was integral or the notation that killed me. :)
Anything
Sorry Kamal but I died. However upon seeing the raw power of what killed me, Eru intervened and had me reborn
What am I missing: As alpha goes to zero, your answer approaches (1/2)log(2*pi). This is a positive number, but the integral should be negative. Huh?
why do you do this to yourself
Wow.
You survived too!
OK COOL❤
Another survivor!
I didn't know Yngwie had his own integral. (This is a guitarist joke.)
YOU SURVIVED!
HELL YEAH!!!
OK COOL
anything
RIP my boy kamaal(I hope I am writing your name right)
He will be remembered especially for the reverse cowgirl formulation of integral calculus. Oh wait here I am false alarm guys 😂
Unfortunately, I did not survive 😮💨
It's cool, try again tomorrow
i died
Sadly yes...but the good news is you're alive!
Er, I got the wrong channel
😂😂😂
Nice and easy😂😂😂😂😂
😂😂😂
I think it was you who recommended this to me a while back.
@@maths_505 no,but i'm glad someone else did,it is beautiful!
I stopped following you at 9:00 , cause I couldn't follow.