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@mohamedmareye3132

Жыл бұрын

Teacher, I follow the lessons you post on KZread I am studying in the college of maths and physics number whats app please tell me you can help me with maths

@user-jb3nr6lm8i

Жыл бұрын

sir from where I will get more videos of definite integration

@mrintegral7348

Жыл бұрын

kzread.info/dron/oLMpMr0JTdLZz4LPdvOf3A.html

@faizurrahmanfr

Жыл бұрын

Lol I don't have much time to go to brilliant when we have enough brilliant content on this channel.

@user-sv6gk8yn4r

Жыл бұрын

Teacher may I know your telegram? I want to ask you something.

this truly is one of the most integral of all time

@keepmehomeplease

Жыл бұрын

Truly. This integral is, in fact, an integral.

@user-lq7lg5jt4k

Жыл бұрын

an integral integral

@fasebingterfe6354

Жыл бұрын

Indeed

@ngoins2010

Жыл бұрын

The most integral?

@knotoftime9680

Жыл бұрын

This integral is indeed an integral

I like this more than the polar coordinates method because it is far easier to understand. No Jacobian, no different kind of coordinate system, just one substitution (really the second one you could do without the substitution, just by inspection).

@holyshit922

Жыл бұрын

There is no Jacobian probably because substitution is done in iterated integral only (he changed only one variable at the time) Another approach is Gamma function with reflection formula

@FleuveAlphee

Жыл бұрын

This is a computationally brilliant method for sure. However, compared to the polar approach the appearance of pi as a result comes across as the outcome of a somewhat artificial-looking substitution process. In contrast, the polar approach relies on the circular symmetry of the "Bell surface" about the z axis, which makes a pi-related value of the integral fairly obvious. Besides, one can relate the square root in the result to the Gaussian curve being a cross-section of that surface, whilst the squaring of the integral in Laplace's approach looks like a mere trick based on nothing but the factorizability of the exponential. For those who are interested, Dr Peyam in his channel does around 20 different derivations of this result!

@purplewine7362

Жыл бұрын

@@FleuveAlphee what's an "artificial" looking substitution process?

@Xoque551

Жыл бұрын

@@holyshit922 That might actually be a line of circular reasoning, since many of the proofs of Gamma function integral rely on the Gaussian integral result!

@holyshit922

Жыл бұрын

@@Xoque551 not necessarily circuloar reasoning Reflection formula can be derived from product representation of Gamma function and Euler's product for sine

I love this integral! Funnily enough in all the physics exams it is always just given 😅

@blackpenredpen

Жыл бұрын

😆

@easondu9236

Жыл бұрын

keep up great work sir

@lechatrelou6393

Жыл бұрын

Because in physic we just use... In math it depends of the subject

@renegadedalek5528

Жыл бұрын

In physics the solution to this integral is an intuitive truth.

@_cran

9 ай бұрын

You're luckyy in engineering my profs made us do it

Never knew you could solve this without using polar coordinates... excellent video!

@abebuckingham8198

Жыл бұрын

I didn't know you could solve it with polar coordinates so I guess we balance the universe out somehow. 😆

@pseudolullus

Жыл бұрын

@@abebuckingham8198 It's very quick to solve with polar coordinates, but there is also a 3rd geometric way to solve this integral.

@azursmile

Жыл бұрын

They're the two proofs outlined in Wikipedia.

@jacoboribilik3253

Жыл бұрын

there are other several ways to prove this remarkable fact.

@azursmile

Жыл бұрын

@@jacoboribilik3253 yes, Dr Peyam presents 12 of them in his video collection here kzread.info/head/PLJb1qAQIrmmCgLyHWMXGZnioRHLqOk2bW

It is great seeing this integral done in cartesian coordinates. All the textbooks I used so far either used the approach over polar coordinates or just used the result. Thank you for this video!

Interestingly, this quite related to polar coordinates but avoids using them :). From the 2D point of view in the first quadrant, we used the coordinates (x,t) where (x,y) = (x,tx). t = y/x is the slope of the line starting from the origin passing through (x,y), in other words t = tan(theta) where theta is polar angle. Cool video :)

@blackpenredpen

Жыл бұрын

Thanks!!

@shuhulmujoo

6 ай бұрын

Wow never thought about this, very interesting thanks!

I remember doing this integral shortly after learning about the jacobian. There is so much joy in doing this integral for the first time, thank you Prof. Steve!

As a stats guy, this is a beautiful detour from the more popular solution. I enjoyed the detailed steps and exploits of the symmetry inherent in the integral. Whilst the polar coordinates approach is easier to explain to anyone who has done trig, this solution is elagant af too.

In 1995 the internet arrived. And this channel is the only good thing worth watching (just a long time waiting :)). Love your style and you teach like someone who understands rather than repeats. Thanks for your hard work!

This is a great video Steve; I've seen this integral solved before, but only with the usual polar coordinates method. Thanks!

As a student who hasn't taken Calc 3 but still uses parts of it, I really appreciate this video and seeing this explanation!

i don’t have the slightest idea about a single thing he said.

@Dergicetea

Ай бұрын

xD

@morpheus6408

12 күн бұрын

How? It’s simple. You just might have to learn the basics of this topic first to understand deeper things

@CalculusIsFun1

Күн бұрын

What part didn’t you understand? It’s okay to admit it. We can help you. the integral is of an even function. This means that it’s symmetrical about the y axis. that’s why he rewrote it in the beginning. From there since it’s a definite integral we can swap out variables so long as the final definite value is the same. So that’s where the integral with respect to y of e^-y^2 came from. Then he just squared it which is the same as multiplying the integrals and got it to 4 times double integral from 0 to inf+ of e^-(x^2 + y^2) Then he substituted for T and rewrite the bounds, then factored out the x^2 and flipped the bounds of integration to make it an integral with respect to x. Then he does a U substitution to compute the inner integral. From there the final integral is a basic trig substitution which yields 2tan^-1(t) from 0 to inf = 2(pi/2) = pi And since this result is the square of the original integral we need to take the square root to get our final answer of root(pi).

Nice route to solving a tricky integral. Great videos ... keep it up!

I have always loved your enthusiasm !! Also, nice way to solve the integral

Franchement très intéressant. Je ne connaissait que la méthode avec passage en polaire. Et j'étais persuadé que c'était la seule méthode possible ! Merci pour cette brillante présentation.

The beauty of the mathematics lies not in the destination but in the elegance of the paths. Thanks for illustrating it.

I didnt know this approach. Thank you for the very clear and instructive presentation.

Thanks for your videos! It's fun to watch your process!

How beautiful the result and the way to solve it , thanks

I had this on my calc 2 exam!! I approximated it using the mclaurin series for e to the x and then integrating that summation!! Cool video

@MohammadIbrahim-sq1xn

Жыл бұрын

I am getting 2 Σ(-x)^(2n+1)/((2n+1)(n!)) [0, inf] isn't this diverging? I know I have done a mistake somewhere but I am not able to spot it

@fooddrive8181

Жыл бұрын

@@MohammadIbrahim-sq1xn i shouldve clarified but my integral was from 0 to x and the intial variable was t. Once you integrate the series you get Σ(-1)^n * x^(2n+1)/((2n+1)(n!)) which is the expansion of sin(x), cheers.

@MohammadIbrahim-sq1xn

Жыл бұрын

@@fooddrive8181 I have a doubt where did you get the (-1)^n from (this might be very silly but I can't seem to figure it out)

@fooddrive8181

Жыл бұрын

@@MohammadIbrahim-sq1xn the (-x^2) is (-1) (x^2) so in my expansion i applied the n to both from (a^m)^n = a^(m*n)

@ginglebaws

Жыл бұрын

I watched 10 minutes of this stuff and we went from a simple formula to a bunch of mumbo jumbo. Can one of you calculus students explain to me what in the world this stuff is useful for? And be specific and have you found a job to where you are actually crunching out formulas to prove they are right or wrong for a living. Can't use teacher or professor either. Some other profession please.

Beautiful :) I did'nt know this trick. BTW when you switch to double integral and polar coordinates, you don't need Jacobian. You just can calculate the volume under the function as sum of volumes of cylinder shells. (Height of shell is e^(-x^2-y^2) = e^-r^2; length is 2 pi r and thickness is dr, so you integrate e^-r^2 . 2 pi r . dr from 0 to infinity.)

@MrPoornakumar

Жыл бұрын

aninob Yes. That is more elegant.

@lawrencejelsma8118

Жыл бұрын

Plus I don't you understand his y = xt where his translation derivatives doesn't equal in dy where dy = x dt + a missing in his formulas t dx term: dy = x dt + t dx I thought? 🤔

@aninob

Жыл бұрын

@@lawrencejelsma8118 In the substitution y = xt in the inner intergral is x in the role of constant. (Like "for given fixed x from the outer integral we shall calculate this inner integral..."). So the substitution y = xt is just recepee for transition from y to t. This is the reason why dy = x.dt.

@lawrencejelsma8118

Жыл бұрын

@@aninob ... Yeah your correct. He should have written it out dy = xdt + tdx but the tdx term equals zero just like t can translate through the integral during summation. I was just use to "two space" partial differentiation mathematics of ut (ut)' = (u')t + u(t'). The recommended solution of real only or even when considering imaginary numbers are always polar solutions of rcosx + jrsinx solutions because trigonometric integral tables solutions are easier to understand.

@aninob

Жыл бұрын

@@MrPoornakumar 3B1B just made a beautiful video about it (plus some mathematical sugar on the top I never knew about). kzread.info/dash/bejne/la1s1JmQg9i6Yaw.html

Nice to see a different way, I really liked the polar method when it was explained in multivariable calc. I also liked the way the Fourier transformation is used to find the improper integral of sin(x)/x. Could be nice to see an alternative way to do that one.

Love the aspect of switching the order of integration, I was wondering how it could be done. But then realised that it is ofcourse summation of product. At the start was confused as to how you could make the function 2* (0 to Infinity) of the f(x), but i had missed that this is a converging function. made math interesting again.. Thanks!

No polar? Something against polar?

@ChollieD

Жыл бұрын

Glad I finally found someone doing this without going to polar coordinates.

@chitlitlah

Жыл бұрын

Polar killed my father.

@davidbrisbane7206

Жыл бұрын

@@chitlitlah Noooo!!!

@lexyeevee

Жыл бұрын

change of variables would be calculus 3 ;)

@imnimbusy2885

Жыл бұрын

Too COLD!

Simple and brilliant, never occurred to me!

Thank you so much. This is the best explanation of this ive ever seen

Well done. You made this very easy to follow. Thanks.

Thank you,I’ve been thinking about a method to do it without polar coordinates cuz I didn’t learn them,great job ❤

Quite impressive in terms of your presentation, well done

This is a cool integral to know in that you can use integrals used by probability distributions to simply rewrite the integral in terms of it's probability distribution and then if they go from -inf to +inf they just become 1

@ruchikarfacts7380

Жыл бұрын

Can you solve this problem? Q. If f[{x + √(1 + x^2)}/x] = x^2. Then find f(x); domain & Range of f(x) =? Video link:- kzread.info/dash/bejne/i4qW0aSNnNC7n5s.html

Beautiful problem, balckpen! Thank you for sharing :)

Great job! Good pace and explanation, an alternative to polar conversion method. Thank You!

3:15 I was like “yeah I get it” 💪

Hi bprp, thank you for the comprehensible and clear derivation, I am now practicing Laplace calculus.

Lovely integral! Thank you.

This has to be one of the most beautiful integrals out there

Hi, I worked on this for years when I was young, until I found the polar solution in a book. But I'm glad to see that there is a method that avoids polar coordinates. Thanks a lot for this 👍

I would love to see you use the Feynman trick with some rigorous explanations (uniform convergence) for the swap between the derivative and the integral! Keep up the good work 😉

@giovanni1946

Жыл бұрын

The Feynman trick has nothing to do with uniform convergence though, you prove it using the dominated convergence theorem - it essentially requires to dominate the integral of the partial derivative

This integral is great. It's amazing how there are several connections between the Exponential function and pi, complex numbers and this at least.

@starpawsy

Жыл бұрын

I'm of the whacky opinion that complex numbers hide (or reveal) a door into other universe, Or universes. No evidence. Just a "gut feel".

@amineaboutalib

Жыл бұрын

@@starpawsy that doesn't mean anything

@starpawsy

Жыл бұрын

@@amineaboutalib Nope. Not a thing.

@holliswilliams8426

Жыл бұрын

@@starpawsy Holy crankometer Batman, it's a kook!

@starpawsy

Жыл бұрын

@@holliswilliams8426 Oh look, I fully recognize the wackiness. That's ok. Im old enough not to care.

Masterpiece . Thanks a lot for your great efforts ,Sir. 💖💖💖

Because of this video, now I understand how upper and lower bounds of integral change due to it's variable change. Thank you so much⭐

13:31 "And there's no +c" CRIES OUTTA HAPPINESS

this is just realy original, congratulations!

Wow! Thank you so much for your videos!

I have seen over 10 different proofs of this result, I don't think I have seen that one before. Great job!

@blackpenredpen

Жыл бұрын

Thank you! Cheers!

@ffggddss

Жыл бұрын

Hey, then you'll now have to make a video about the other 9 methods, to let the rest of us in on them! ;-) Fred

Oh my god you are so amazing. I just loved it. The way you make it so easy for us is commendable. You are incomparable. Thank u so much. Because of you I am able to solve it without remembering polar coordinates typical method.

It is so satisfying to watch you explain the math. (The first thing that catch my eye is the 荼果 doll under the e)

I love this integral and i never saw this approach

Awesome video! Thank you!

amazing sir👍today I had learnt little something,... and understood that, there is a lots of yet to learn ...

Very cool to see someone so passionate about a topic that so many people wrongly think of as boring

OMGGGGGG Thanks you so much, i dont have words 😍😍😍

Very nice , I knew of the Feynman technique , but this is very nice

Excellent presentation 👌

Nicely done! 🙂

I enjoyed watching this. I've taken a few courses in stats and probability, and none of the professors wanted to take the time to show this integral. We just accept it as fact.

beautiful job!!!

That was really nice!

Nice explanation!

The way he swicthes pens is no less than a magician.

Thank you so much You are the greatest teacher in the world🤩🤩🤩

Never mentioned Laplace. Would be interesting to hear just a little about when and how these guys came up with these things. We kinda owe them.

I dont understand calculus one bit, but something about your explanation style just drives me towards your videos

thank you that was very amazing

I feel great, even though I have studied calculus only on my own, not in school yet, I was able to follow along with what's happening. And oh boi, it was beautiful

how good are your tutorials? I passed a calculus 2 course with 70% with little to no help from my professor. keep up the great work man!

Thank you very much... Love your videos...

Wow! This is so good!!

I really doubt polar coordinates since I don't know how it works. With this method, I believe now that the integral of e^-x^2 from -inf to inf is equal to sqrt of pi. Amazing.

@pseudolullus

Жыл бұрын

You can imagine the coordinate and variable switch in the polar coordinate trick as allowing one to sweep from 0 to infinity in a circular way all at once (radially with r from 0 to infinity and circularly with theta from 0 to 2pi). Since the function is actually radially symmetrical from the origin and there is only an infinity (imagine it in 3D), with polar coordinates you do not need to split it up in two (-+inf,0] halves.

@carultch

8 ай бұрын

Start with the original integral: Integral e^(-x^2) dx Square it: (Integral e^(-x^2) dx)^2 = double integral e^(-x^2) * e^(-x^2) dx dx Change one of our variables of integration to y: double integral e^(-x^2) * e^(-y^2) dx dy Using properties of exponents, the product of the two exponential functions becomes e^(-x^2 - y^2): double integral e^(-x^2 - y^2) dx dy In polar coordinates, r^2 = x^2 + y^2.. The differential area dx*dy is equivalent to r dr dtheta. Thus: double integral e^(-r^2) r dr dtheta The limits on this integral are the full domain of x and y. In polar coordinates, it is 0 to infinity for r, and 0 to 2*pi for theta. This generates the derivative of the inside function, so we can use u-substitution. Work with the inner integral first, and theta isn't involved. integral r*e^(-r^2) dr Let u = -r^2. Thus du = -2*r dr, and dr = du/(-2), and the integral becomes: -1/2*integral e^u du, which evaluates to -1/2*e^u + C. In the r-world, it becomes: -1/2*e^(-r^2) + C We'd like to evaluate this from r=0 to r=infinity: (-1/2*e^(-inf^2)) - (-1/2*e^0) = 1/2 The outer integral is trivial, it's just integral 1 dtheta, which is theta + C. Evaluate from 0 to 2*pi, which is 2*pi. Multiply with the r-integral result, which gives us the result: [integral e^(-x^2) dx from 0 to infinity]^2 = pi Since we originally squared the integral, take the square root to get the original integral we want: integral e^(-x^2) dx from 0 to infinity= sqrt(pi)

Very nice solution dear teacher. 👍👍👍👍👍👍👍

You could also try x=rcosa and y=rsina for solving double integral

when you define y=xt therefore t=y/x, how can you say that 1+t^2 is constant in the x world? t=y/x so it varies with x, it doesn't seem constant to me. What am I getting wrong? thank you!

@abddibani

9 ай бұрын

√(1+t) the correct translation

@abddibani

9 ай бұрын

When you change the variable X to U

@loweffortdev

8 ай бұрын

Yes, can someone please explain the process

I remember learning to do this one!

This was quite easy to follow - which is weird, I'm very bad at integrals. Honestly the only point I tripped up was in the end where 2*int[0,inf] (1/1+t^2) * dt became 2*tan^-1 (t) [0,inf]. But that is probably because I don't remember formulas for anti-derivative of 1/(1+x^2) and didn't know derivative of tan^-1 (x). My big fail in trigonometry is remembering all the formulas that can be used. I only remember that sin^2+cos^2 = 1. Very entertaining and informative video - thanks!

#blackpenredpen I was wondering how you come up with the idea y=xt and why it works? thanks in advance.

I remember solving the indefinite integral version in my calc 2 class by using the Taylor series expansion of e^x

It would be nice to have a video where you solve this integral using complex analysis (residue theorem). It's a bit longer but it is a very fun calculation.

I am so happy to live in a world where bprp exists! greetings from Brazil!!

@blackpenredpen

Жыл бұрын

Thank you

Waiting for 100 integral part 2 😌

I´m calculus 2 student from Argentina and I understand it so well, I´ll believe in Chen Lu

Thank you so much sir

If you substitute t= y/x then you have substituted the tan of the angle for polar coordinates. Also the substitution for u is minus the square of the radius in polar coordinates. So you have used polar coordinates, it is just disguised.

Dhanyawad bhaiyaa 🙏🏻🙏🏻. Love from BHARAT 🇮🇳

Awesome resolution ❤

It's cute, but you cant hide from the trig on this one. You end up with the arctan anyway. Thats kinda obvious because the answer has pi in it, but still fun to watch you sweep it under the rug for as long as possible. I like different approaches, but I still like the move to polar better at the end of the day.

BPRP, can you integrate √(x+√x) with respect to x? Yes, there is an elementary answer :D

y = xt , when y goes to 0, the value of t does not have to be always 0. because in y = xt , the value of y also depends on x. in the integral we can see that x varies from 0 to infiniti. For example if x = 0, then t can have many values while xt remains 0. Can you explain why we assume when y = 0 that t = 0.

@marshallsweatherhiking1820

Жыл бұрын

The value at some isolated point never effects the integral as a whole. The substitution is okay if still have convergence approaching zero. Polar coordinates has the same problem as theta can be anything at the origin.

Very clever, I love it! I have a little problem with it though. I could be wrong, but it seems to have a dimensional mismatch between the two sides of the equation. The left-hand side, I, represents an area under the curve, which is 2-d, a number to the power of 2. The right-hand side is the square root of a double integral, which represents a volume-a number to the third power. Taking the sqr. root of that number produces a power of 3/2, which is not 2. Am I seeing it wrong?

@duurc

Жыл бұрын

Your definition of dimension is bizzare and certainly doesn't apply to integrals that flexibly. How many dimenisons cylinder have, knowing that it's integral is defined by DOUBLE integral in 2D polar coordinates multiplied basically by a constant? How many dimensions integral of sphere has, knowing that it's integral is TRIPLE integral? Inconsistent.

@duurc

Жыл бұрын

Also double integral came not from a 'natural observation' of a geometrical object, but was a product of multiplication of two usual integrals, which would rather double the 'dimensionality' of I becoming I^2 rather than adding 1 to it

@lih3391

Жыл бұрын

Nothing says the dimensions have to match up, I think it was the squaring of an area and then labeling that extra area with a new variable that caused your confusion

@xinpingdonohoe3978

Жыл бұрын

∫ydx=∫∫dydx What changed, really?

Yayy, part 2. Let's goooo

Apart from this awesome video supplied by ur easy explanation . Actually, Like the Gamma function, since there is no explicit integral to f(X) = e^-x² , another special function can easily be defined to be the distribution function F(X) = the integral of f(X) between -∞ and x. Obviously it is positive , strictly increasing and limited. and actually , since ex is equal to the Taylor series that has the terms 1/n! .X^n . Then e^-x² is equal has the Taylor series with terms : 1/n! . (-x²)^n . Thus F(X) can easily be obtained by integrating of that series. I think This can be helpful in numerical analysis as the you can have an approximation to F(X) by studying a polynomial of sufficiently large degree and dropping the rest of the power series.

I love calculus Nice video:)

this method is so much nicer than using polar coordinates and the equation which has a difficult to understand derivative

Excellent! It is fun even if (or maybe because) it is more complicated than going through polar coordinates. I watched it because I was curious to see how the π number would appear without that. I remember it was a wonder to me when I found out that trigonometric functions can be built by just integrating functions defined using only the 4 basic arithmetic operations plus the square root, the latter not even always necessary, like here. More generally, it is fascinating how π can materialize where it is not expected, like for instance in the sum of reciprocal squares.

I’m a masters in ML but I love math and calculus, love these videos❤

@paigeturnah927

Жыл бұрын

さとみくんの｢まず｣って言い方好きなの分かりますか？！🥺 youtubemn.com/watch?v=zZt0708hbPO お母さんに言われた時は天国へのカウントダウンしよっかなって思っちゃったけど

@janami-dharmam

Жыл бұрын

nice, but what is ML??

@zannyrt

Жыл бұрын

@@janami-dharmam Machine Learning

@janami-dharmam

Жыл бұрын

@@zannyrt I see; these are rather infant sciences. Math is well established.

Nice, thanks a lot

You are outstanding!

how are you able to treat the "t" as constant in the first integral when it depends on x and the integral is with respect to x?

Question: If you are defining t=y/x, then x cannot be equal to zero. But you are letting that x be the same as the x in the other integral which is requires x=0 on the integral limits. How is this reconciled?

@marshallsweatherhiking1820

Жыл бұрын

Yea. I think the function can be broken up when there are only a finite number of discontinuities and you can prove the absolute value after substitution still converges approaching the now asymptomatic discontinuity. It’s probably one of those theorems you prove using real analysis techniques. Not that I remember much of that.

You are better than 99% of calc teachers!