For the integral J- In the integral I, when you subtituted t=1/x, you just got back to the original integral(J) if you multiply the numerator and denominator by t^2. so, they're equal and you don't need to do the fancy stuff at the end
Everybody knows that good ol' heavy artillery will blow the door open, but it's always more interesting to open it with some completely non-obvious lock picking trick. That artillery may blow yourself up too, you know...
Awesome video - changing of variables which preserves the underlying region of integration is something quite beautiful in calculus. I would highly recommend reading up on some calculus on manifolds if you haven't yet already (start with Spivaks!). For example, he gives you the machinery to prove that if you're integrating a function f over the sphere, if you pre-compose this function with an orthogonal linear transformation L, then the integral is preserved. i.e. integrating the function given by the rule x-> f(L(x)) as oppose to x-> f(x) gives the same result over the sphere!
plugging the substitution of t = 1/x into one of both integrals after factoring respectively a x^(2) or 1/x^(2) out would give the other integral straight away.
Actually, for I, after the substitution, if you bring the 1/t² to the denominator and multiply them out, you'll get J, so it can be shown that I and J are the same without all that manipulation stuff
I = int 0 to infinity (1/t^2)/(t + 1/t)^2 dt = int 0 to infinity 1/(t * (t + 1/t))^2 dt = int 0 to infinity 1/(t^2 + 1)^2 dt = J Would have been faster, but hey, why do it the fast way if you can do it the cool way?
I didn't understand the arctangent part. Arctan(infinity) has infinite solutions π/2 + 2π*m Where m is an integer. I know that that integral has only 1 solution but: How do we know that π/4 is the correct solution? Are the other solutions meaningless?
i tried a solution with using partial differentiation to interchange around ^-1, ^-2, and ^-3, but made a small mistake somewhere in there (i found by doing a simple check, take an equation i had a=b and enter wolfram a-b to see if it gives 0, it didn't.) so i had to scrap it. Then you go along and just presume tangent amidst your mission to avoid trig XD
em eye tee intehgehration bee. Say it, again -- louder. EM EYE TEE INTEHGEHRATION BEE! Chant it!, now! (P.S. I watched the 2006 one and it was baller, all three hours -- thanks for bringing us the spicy intehgaralliboiz, Flammy!)
I am from India and currently preparing for my admission into under graduate college. You really pump me up to take maths and study. It's just motivating and nice to see you tackle these problems. Sir pls keep this going and would it ever possible to make a video just on to guiding people how to become good at maths ik it's shear practice but still anyways you rock
Because after the substitution the integral is with respect to t, you could replace that t with any letter (x, u, Ξ, etc), and so long as it’s also with respect to that letter, it’s exactly the same thing
@@strengthman600 Doesn't it have to do with integral being definite ? I'm not sure, but I think such a trick for indefinite integrals is not allowed. Here is how I understand it. Definite integral is a number, really. Integration variable "lives" only "inside" it, i.e., it has a meaning only within a context of that integral. No matter what letter will you use, you will end up with the same number. The letter (integration variable) is just a tool, which you put aside after integration. Now indefinite integral is a function. This time integration variable also has a meaning outside of integral, being independent variable of said function. If you do a substitution, you change a form of this function, so this time you can only add functions (integrals) of strictly the same variable. Am I right, Papa ? :)
Hello, Papa Flammy, hi, guys! Recently I've faced one interesting definite integral, which I have no idea how to calculate, but I know that it can be expressed in terms of error function: (-oo;+oo) exp(-x^2)dx/(1+x^2). I appreciate any respond!!!
![A Powerful Elementary Integral [ (Almost) Impossible Integrals #1 ]](http://i.ytimg.com/vi/bFeC1Rlajao/mqdefault.jpg)
![WONDERFUL 𝜋ntegral! A Putnam Extravaganza [ Basel Problem Integral Representation ]](http://i.ytimg.com/vi/cOHh2DqZlVQ/mqdefault.jpg)
For the integral J-
In the integral I, when you subtituted t=1/x, you just got back to the original integral(J) if you multiply the numerator and denominator by t^2. so, they're equal and you don't need to do the fancy stuff at the end
Exact same thought after first minute
+
I had to stop watching a flammable math video to watch a flammable math video.
"Certain sophisticated Asian gentlemen"
*Sad BPRP noises*
Just one thing.Complex analysis
Everybody knows that good ol' heavy artillery will blow the door open, but it's always more interesting to open it with some completely non-obvious lock picking trick. That artillery may blow yourself up too, you know...
0:50 I have never seen someone that savage
MIT is my dream college! Didn't expect this one indeed! Cheers for the great work as always, and happy Holidays
@@PapaFlammy69 Cheers man!
Oh my goodness how dare you wage war on christmas! Son of Satan!!
Awesome video - changing of variables which preserves the underlying region of integration is something quite beautiful in calculus. I would highly recommend reading up on some calculus on manifolds if you haven't yet already (start with Spivaks!). For example, he gives you the machinery to prove that if you're integrating a function f over the sphere, if you pre-compose this function with an orthogonal linear transformation L, then the integral is preserved. i.e. integrating the function given by the rule x-> f(L(x)) as oppose to x-> f(x) gives the same result over the sphere!
When you get a headache from glasses and not Extremely Difficult Integrals 💁♂️💁♂️💁♂️💁♂️
Hold on, let me put the glasses on and I will be able to understand this better.
Integaration Biene
Thank you for always posting interesting math videos. Thanks to you, I enjoy every day.
Reduction Formula can be a choice if we don't want to substitute x = tant
Your like ratio is approaching infinity!!
plugging the substitution of t = 1/x into one of both integrals after factoring respectively a x^(2) or 1/x^(2) out would give the other integral straight away.
My guy had a stroke at the start.
This reminds me of that magician who puts a rabbit in the hat and takes out a dove.
Man, I | *love* | this channel.
Actually, for I, after the substitution, if you bring the 1/t² to the denominator and multiply them out, you'll get J, so it can be shown that I and J are the same without all that manipulation stuff
3:41 where did the dx go? Can you please answer this question.
How close is papa to solve the riemann hypothesis ?
" Ihr dummen spasten! Also, please check out my merch! " ... I don't know what to say ^^
I solved the first one quite easily using the residue theorem
3:00 (1/t^2)/(t+1/t)^2 = 1/(t^2+1)^2
so J is equal to I =π/4
The J=I which can be seen slightly faster with your I when it was in terms of t because the 1/t^2 was already there
I = int 0 to infinity (1/t^2)/(t + 1/t)^2 dt
= int 0 to infinity 1/(t * (t + 1/t))^2 dt
= int 0 to infinity 1/(t^2 + 1)^2 dt
= J
Would have been faster, but hey, why do it the fast way if you can do it the cool way?
shit man hope christmas never comes if it means getting more of papaflammysadventcalendar
I didn't understand the arctangent part. Arctan(infinity) has infinite solutions
π/2 + 2π*m
Where m is an integer.
I know that that integral has only 1 solution but:
How do we know that π/4 is the correct solution?
Are the other solutions meaningless?
Where can you get those sick shirts? Theyre so fun xD
Use Glasser's Master Theorum, to end it in one shot.
i tried a solution with using partial differentiation to interchange around ^-1, ^-2, and ^-3, but made a small mistake somewhere in there (i found by doing a simple check, take an equation i had a=b and enter wolfram a-b to see if it gives 0, it didn't.) so i had to scrap it. Then you go along and just presume tangent amidst your mission to avoid trig XD
"boyz and grillz get drilled into unconsciousness by smoking hot maths teacher with glasses", would be a better title imo
em eye tee intehgehration bee. Say it, again -- louder. EM EYE TEE INTEHGEHRATION BEE!
Chant it!, now!
(P.S. I watched the 2006 one and it was baller, all three hours -- thanks for bringing us the spicy intehgaralliboiz, Flammy!)
Today I passed my complex analysis exam, so with rezidue theorem I am done in less than one minute;)❤🙃
How does such an ordinary looking fellow contain such a ginormous amount of charisma ?
*Flammy Khalifa*
Nothing much to see here, just papa flammy destroying yet another elememtary integeral boi 😎
I am from India and currently preparing for my admission into under graduate college. You really pump me up to take maths and study. It's just motivating and nice to see you tackle these problems. Sir pls keep this going and would it ever possible to make a video just on to guiding people how to become good at maths ik it's shear practice but still anyways you rock
If you ever get stumped by an integral, just guess π/4.
if you look closely, after the first substitution, the I integral is already the same as J
Okay, I’m not comfortable with t and x being the same thing, especially when we’ve just set t as a substitution. Could someone elaborate?
Because after the substitution the integral is with respect to t, you could replace that t with any letter (x, u, Ξ, etc), and so long as it’s also with respect to that letter, it’s exactly the same thing
@@strengthman600
Doesn't it have to do with integral being definite ? I'm not sure, but I think such a trick for indefinite integrals is not allowed.
Here is how I understand it. Definite integral is a number, really. Integration variable "lives" only "inside" it, i.e., it has a meaning only within a context of that integral. No matter what letter will you use, you will end up with the same number. The letter (integration variable) is just a tool, which you put aside after integration.
Now indefinite integral is a function. This time integration variable also has a meaning outside of integral, being independent variable of said function. If you do a substitution, you change a form of this function, so this time you can only add functions (integrals) of strictly the same variable.
Am I right, Papa ? :)
I did both using complex integration and residue calculus, I think I'm growing lazy
Hello, Papa Flammy, hi, guys! Recently I've faced one interesting definite integral, which I have no idea how to calculate, but I know that it can be expressed in terms of error function:
(-oo;+oo) exp(-x^2)dx/(1+x^2).
I appreciate any respond!!!
This are the kind if videos I like from YT. No more applied math on your channel pls
you forgot to change the limit of integration before plugging for t.
Nice
Actually, I found that you can immediately say I=J in the second line of your calculation 😂
Well，if we let x=tanu ,then these integrals will be very easy .
This calculations can be simple by putting x=tan@
Why did I find two girls with one cup of chocolate ice cream papa?
I have never pressed “play” button that fast...
Flammable Maths
MIT makes integrals tasty 😍🤤🤤
I love love this video and love you so so so much kisses and huggs
Ahhh yess
x=tan@ was clearly visible and was damnn easy than this tbh
sexy boi now in glasses
Cool
Me 16 trying to understand it 👁️👄👁️
What math class is this lol?
Hi
tan^(-1)(t) instead of arctan(t) :(
Can you plz explain the math behind, vortex from fluid mechanics,
@@PapaFlammy69 thanks a lot
OYAAAAHAHIHIA
she explain better than you. integration by completing the square MIT 18.01SC Single Variable Calculus... you need to start to write good
go and practice to write good
first