The part where you add zero to the equation is frustratingly sublime. It doesn't seem obvious in retrospect but it makes so much sense that it seems like it *should* seem obvious in retrospect.
that's a really cool integral, i can't wait for the demonstration of that formula, this Is One of the best channels to see integrals and i really love how you use the gamma function
@@maths_505 that's really appreciated, It made me go learn the gamma function as i wanted to follow your videos and i really loved It, i mean, i do love It as i still have to fully understand the beta function, i still have to study multivariable calculus so that part Is a bit unclear but i can now follow what you do and it's honestly so cool and clever that i can't even describe, thank you for this great content
A more generalizable way to get for C is to simply solve for C by absorbing -pi*lnln(2/a) into the integral as -2*lnln(2/a). Then determine which values of a would create an indeterminate form. So 0 and infinity. Then use L'Hopitals with either (I tried with infinity and it worked). If no value of a creates an indeterminate form, or you can't calculate the limit (perhaps because of special functions), then you picked a bad parametrization.
Hi Maths 505, I'm watching your channel and your integrals are awsome. I was wondering is there a book or a course you could suggest since its not clear to me how you select where to put alfa and why. Maybe you could in your videos explain why you select where you put it. In any case excellent work, I'm looking forward to more of your videos.
Like I said in the video it took me 2 days to solve this😭😭 but yeah just keep practicing and you'll get a general idea. I try to place it somewhere to get some cancellations or atleast an easier integral.
Great video. Can someone explain why you’re allowed to use the laplace transform when your original integral bounds are not 0 to infinity? In this case it was 0 to 2pi and we just assert the laplace transform from 0 to infinity?
Does this technique work for only certain types of integrals? Because I feel like it's a massive pain to know when to use u-substitution, integration by parts, or even trigonometric substitution.
Well, I think it will require uniform convergence, but it's mainly useful if an x thing inside some function would cancel nicely with an x thing outside of the function.
Who else used the semi-circle contour of radius=1 and half residue at origin on the function f(z) = (1/z) ln{ln[(z + 1)/2]} 🗿 Believe me you'll get the solution on just half page.
Do you mind explaining how exactly that works? It sounds like a very cool idea, but I’m having a hard time figuring out how you would recover the original integral from this
I did that, but you also ha e singularities at z=1 and z=-1, so the contour has to avoid these two with epsilon semi circle and take epsilon to 0. Not exactly half a page.
Man went through every stage of grief just come up with that save at the end!
Man is stubborn AF 😭😭😭
The intergral's out of this world
Swear to god bro, sometimes it seems like Feynman is carrying all the improper integrals 😂 from his grave
This is such a beautiful integral!!! So many techniques. Great trick with alpha in the end btw!
The part where you add zero to the equation is frustratingly sublime. It doesn't seem obvious in retrospect but it makes so much sense that it seems like it *should* seem obvious in retrospect.
That's what happens when you're too stubborn to quit 😂
that's a really cool integral, i can't wait for the demonstration of that formula, this Is One of the best channels to see integrals and i really love how you use the gamma function
I'm kinda always looking for excuses to apply the gamma function 😂
@@maths_505 that's really appreciated, It made me go learn the gamma function as i wanted to follow your videos and i really loved It, i mean, i do love It as i still have to fully understand the beta function, i still have to study multivariable calculus so that part Is a bit unclear but i can now follow what you do and it's honestly so cool and clever that i can't even describe, thank you for this great content
A more generalizable way to get for C is to simply solve for C by absorbing -pi*lnln(2/a) into the integral as -2*lnln(2/a). Then determine which values of a would create an indeterminate form. So 0 and infinity. Then use L'Hopitals with either (I tried with infinity and it worked).
If no value of a creates an indeterminate form, or you can't calculate the limit (perhaps because of special functions), then you picked a bad parametrization.
Really awesome integral! Keep it up man.
Des intégrales de fou que j’adore !!! Merci beaucoup pour ce que vous proposez …
Thank you kamaal sir u making me to love only the beautiful mathematics , I hope that if everyone who hates maths will love if they watch your vedios
Absolutely the GOAT integral!!
Feynman's technique + Laplace transform + Gamma and beta
in one integral ou ma gwad thats literally the "ok cool" things in one❤ Upper esh
Yeah, was hoping for a digamma too, but my hopes were smashed when those Gamma functions cancelled!
Thank you for this interesting integral. It is a long way to get initial value for I, but you reached successfully.😊
When will ln(2) become a named constant?
really cool don't stop
Hi Maths 505, I'm watching your channel and your integrals are awsome. I was wondering is there a book or a course you could suggest since its not clear to me how you select where to put alfa and why. Maybe you could in your videos explain why you select where you put it. In any case excellent work, I'm looking forward to more of your videos.
Like I said in the video it took me 2 days to solve this😭😭 but yeah just keep practicing and you'll get a general idea. I try to place it somewhere to get some cancellations or atleast an easier integral.
Good, but how can we be sure that the initial integral converges before diving into differentiate it?
Great video. Can someone explain why you’re allowed to use the laplace transform when your original integral bounds are not 0 to infinity? In this case it was 0 to 2pi and we just assert the laplace transform from 0 to infinity?
The integral was from 0 to 2pi. I inserted the Laplace transform as another integral (from 0 to infty).
damn, being stubborn was dope
Does this technique work for only certain types of integrals? Because I feel like it's a massive pain to know when to use u-substitution, integration by parts, or even trigonometric substitution.
Well, I think it will require uniform convergence, but it's mainly useful if an x thing inside some function would cancel nicely with an x thing outside of the function.
i just watched 20min of doing an integral without realising
Bro from where you get all this stuff ?🗿 Genuinely asking
This one's from Micheal Penn's channel so I wanted to take a shot at it.
Not sure if I missed it, but have you done 2024 MIT integration bee?
Yes
Yes I have
"Mr Feynman, how on Earth did you solve it?!"
"It's easy - just differentiate under the integral sign" 😏
😮😮😮
Wow
🐐
Petition to rename ln(ln2) to lnn2
6:20 so many random formulas 🤕
It be like that sometimes 😂
Who else used the semi-circle contour of radius=1 and half residue at origin on the function
f(z) = (1/z) ln{ln[(z + 1)/2]} 🗿
Believe me you'll get the solution on just half page.
Do you mind explaining how exactly that works? It sounds like a very cool idea, but I’m having a hard time figuring out how you would recover the original integral from this
@@GGBOYZ583
Bhai Complex Analysis study karle ek baar acche se !
I did that, but you also ha e singularities at z=1 and z=-1, so the contour has to avoid these two with epsilon semi circle and take epsilon to 0. Not exactly half a page.
Log2 strikes again 😂
And this time comes out on top 😂
Merci Feynam’s trick …
ok, cool
Promo_SM 👌