how Laplace solved the Gaussian integral

[@blackpenredpen]

Learn more calculus from Brilliant: 👉 brilliant.org/blackpenredpen/ (20% off with this link!)

[@mohamedmareye3132]

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

[@user-jb3nr6lm8i]

sir from where I will get more videos of definite integration

[@mrintegral7348]

kzbin.info/door/oLMpMr0JTdLZz4LPdvOf3A

[@faizurrahmanfr]

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

[@សុខសុគន្ធបញ្ញា]

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

[@jonasdesmedt3965]

this truly is one of the most integral of all time

[@keepmehomeplease]

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

[@user-lq7lg5jt4k]

an integral integral

[@fasebingterfe6354]

Indeed

[@ngoins2010]

The most integral?

[@knotoftime9680]

This integral is indeed an integral

[@ShaunakDesaiPiano]

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]

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]

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]

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

[@Xoque551]

@@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]

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

[@zhelyo_physics]

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

[@blackpenredpen]

😆

[@easondu9236]

keep up great work sir

[@lechatrelou6393]

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

[@renegadedalek5528]

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

[@_cran]

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

[@AlexandreRibeiroXRV7]

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

[@abebuckingham8198]

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

[@pseudolullus]

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

[@azursmile]

They're the two proofs outlined in Wikipedia.

[@jacoboribilik3253]

there are other several ways to prove this remarkable fact.

[@azursmile]

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

[@qm_turtle]

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!

[@davidalexander4505]

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]

Thanks!!

[@shuhulmujoo]

Wow never thought about this, very interesting thanks!

[@wayneqwele8847]

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.

[@fooddrive8181]

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]

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]

@@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]

@@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]

@@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]

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.

[@carterwoodson8818]

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!

[@aura-audio]

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

[@aninob]

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]

aninob Yes. That is more elegant.

[@lawrencejelsma8118]

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]

@@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]

@@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]

@@MrPoornakumar 3B1B just made a beautiful video about it (plus some mathematical sugar on the top I never knew about). kzbin.info/www/bejne/maqbo2qNiNqHZ6s

[@jul8803]

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

[@Syntax753]

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!

[@ericguillet8986]

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.

[@CliffSedge-nu5fv]

Google translates _la méthode avec passage en polaire_ as "the fleece method."

[@anastasiosefthimiadis6831]

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

[@blackpenredpen]

Thank you! Cheers!

[@ffggddss]

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

[@fatitankeris6327]

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]

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]

@@starpawsy that doesn't mean anything

[@starpawsy]

@@amineaboutalib Nope. Not a thing.

[@holliswilliams8426]

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

[@starpawsy]

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

[@redvine1105]

3:15 I was like “yeah I get it” 💪

[@a3hindawi]

Thanks

[@blackpenredpen]

Thank you!

[@davidbrisbane7206]

No polar? Something against polar?

[@ChollieD]

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

[@chitlitlah]

Polar killed my father.

[@davidbrisbane7206]

@@chitlitlah Noooo!!!

[@lexyeevee]

change of variables would be calculus 3 ;)

[@imnimbusy2885]

Too COLD!

[@gal-zki]

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

[@blackpenredpen]

Thank you

[@BulaienHate]

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]

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:- kzbin.info/www/bejne/j4fFoHWKodKIpZo

[@MadhukarMoogala]

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

[@venniethompson8473]

My own technique also has value: create an infinite alternating series through iterative integration by parts. After creating the first few terms the series rule can be inferred and condensed into series summation nomenclature ( Σ). It gives rise to a very nice little "sum i" with multiples of e^-2x^2 in the numerator and a fun little concatenation "π" product in the denominator.

[@boredgamesph4872]

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]

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]

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)

[@CM63_France]

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 👍

[@ans72411]

You are one of the best maths teacher on internet

[@johnanderson4638]

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

[@kushagramishra1729]

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

[@davidm9442]

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

[@aurelian3268]

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!

[@Jack_Callcott_AU]

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

[@9146rsn]

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!

[@chungus816]

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

[@francoocampo5286]

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

[@johnredberg]

6:02 Slopping over possible divergence at x = 0. 8:10 Circular argument.

[@FX9426]

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)

[@zannyrt]

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

[@paigeturnah927]

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]

nice, but what is ML??

[@zannyrt]

@@janami-dharmam Machine Learning

[@janami-dharmam]

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

[@dottemar6597]

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.

[@Vinaya999]

This is amazing. Now I won't forget integral of e^(-x²) from 0 to ∞. I just have to learn the equation but knowing the proof just makes it easy to remember.

[@andresmillanmillan5155]

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

[@acelm8437]

When he was doing the first integral I wondered how the π would show up. Then he wrote down 1/2(1+t^2) and it hit me!

[@ysfhanikai995]

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

[@darinhitchings7104]

Quite impressive in terms of your presentation, well done

[@sebastians7346]

Can someone explain why you can just "put the e^-x^2" inside the red integral? (4:00)

[@ezxd5192]

Because it's considered a constant according to the red integral

[@sarcastikluffy4820]

Sir , at 6:32 , i didnt understood the part where u divided y by x saying it is a positive number , so ♾️ /+ve = ♾️, but the value of x ranges from 0 to ♾️ and then we would get two sets of different integral forms ie ♾️/0 and ♾️/♾️ , can you pls explain

[@Inglesemente]

You are better than 99% of calc teachers!

[@pratyushdahal3667]

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

[@OdedSpectralDrori]

Simple and brilliant, never occurred to me!

[@jatag100]

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

[@mohamadabdo5521]

When you changed variables around 10:00, why is it assumed that t is independent of x (constant) even though t=y/x. I would understand if it was written that y=tx for whatever x in 0 to inf

[@maximegr3992]

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

[@pedrodeoliveiracamargo2413]

this is just realy original, congratulations!

[@luigibeccali2840]

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]

√(1+t) the correct translation

[@abddibani]

When you change the variable X to U

[@loweffortdev]

Yes, can someone please explain the process

[@Grecks75]

Good point. It wasn't a constant at first in the dt-integral right after substitution. But it does become a constant for the dx-integral after application of the Fubini-Tonelli theorem that changes the order of integrations. Now the inner integral runs over x values (x ist the integration variable) and t is a constant here. Only when evalutating the outer dt-integral, t is the integration variable and cannot be considered constant.

[@rotemlv]

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.

[@jannegrey]

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!

[@AshokKumar-ii4ny]

Very neat and direct. For those interested in a simple way, using polar coords, ( and simple to understand) I suggest you refer to Prime Newtons' video- it is nice for beginners.😊

[@charlesalmeida3268]

His face in the end saying "i love this one so much" is priceless. Greetings from Brazil

[@saregeh8083]

No more words need to be said. Just... Elegant.

[@larsybarz]

Question at 7:36 : why is (1+t^2) constant? If we’re integrating over all x and all y, and t is defined as y/x, and the function in question isn’t linear, how could t be a constant?

[@abebuckingham8198]

It's constant with respect to the inner integral, not in general. Once you make the change of variables you no longer have to deal for these complications since the dependency has been accounted for. You can treat it as a proper variable from then on.

[@mehrdadmatinfar5470]

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

[@1862anthony]

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

[@Dergicetea]

xD

[@morpheus6408]

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

[@CalculusIsFun1]

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).

[@Pratikt2828]

​@@CalculusIsFun1 your efforts will be remembered bro

[@CliffSedge-nu5fv]

It was multiplication followed by addition. And then there was a square root at the end.

[@benjamingross3384]

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.

[@rafibot4276]

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

[@pemfiri]

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]

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.

[@Grecks75]

That's a good question. @marshallsweatherhiking1820 already gave you the reason why the the substitution is still valid/correct. If you change the integrand of the Riemann integral at only a finite number of points, the integral stays the same. So the value of the integrand at x=0 doesn't matter and you can ignore it. You can also think of this as taking the integral on the *open* interval (0, +inf) so that x is always positive. Note that you're actually doing something similar on the upper limit of the integral: inf is not a real number that you can divide by x. Here, too, the upper limit itself is not actually included in the computation of the integral; a limiting process is used instead.

[@Ninja20704]

Really still. Love how pi just comes out nowhere when we are doing something related to exponentials

[@abebuckingham8198]

There is an intimate connection between the circle and the complex exponential so this isn't too surprising. Honestly think of the complex numbers as the circle numbers for this reason.

[@AdrienLegendre]

Substitute s=x^2 and you get the gamma function for 1/2. Gamma(1/2)*Gamma(1-1/2)= pi/sin(pi/2) (Gamma reflection formula). Reason is that zeros of sine(pi*x)/pi function match the poles of this product of gamma functions , think of factorials. This is where pi comes from.

[@fantiscious]

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

[@aminebenmessauod5228]

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

[@hexcadecimaldhcp1105]

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⭐

[@sdquinlan]

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]

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.

[@sahibpreetkaur7917]

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.

[@muhammadamr5073]

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

[@user-gr8fc3de9w]

if y ranges from 0-infinity how can t do the same, ie if y is infinity then surely t wont be infinity aswell? 6:30

[@okbro2198]

This is one of the youtube videos of all time

[@rajendramisir3530]

Is this the method Laplace used to solve the Gaussian Integral? Was the Laplace Transform or Inverse Laplace Transform used in this method?

[@honourabledoctoredwinmoria3126]

He did hit on this method, and he later defines the Laplace Transform in the same book. So maybe it was the other way around, that thinking about this and also generating functions led him to transforms? Someone would have to go through his unpublished papers and see what he was working on, because he doesn't tell you how he came to something useful, just the thing he came to. (Usually after saying it can be easily seen....)

[@rajendramisir3530]

@@honourabledoctoredwinmoria3126 Thank you Sir for your prompt and informative reply. I was complimented by Professor Leif for how well I learnt Laplace and Inverse Laplace Transform to solve differential equations. I was a student in her ODE class at KCC(CUNY).

[@JamesJoyce12]

when you finally realize that dudes born centuries before you are infinitely smarter than you will ever be

[@acdude5266]

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

[@diegoramirez7901]

Nice vid! Question: Geometrically, what does mean to square the integral?

[@mustafasaracaloglu13]

probability? wrt y

[@lih3391]

e^(-x^2)->e^(-x^2-y^2) r=sqrt(x^2+y^2) tan(theta)=y/x ->e^(-r^2) So imagine spinning the e^(-x^2)=y graph around the y axis through your screen in the 3d dimension

[@vogelvogeltje]

Oh no, you watch Whatifalthist? 😂😂😂 The guy that says the scientific method is sh!t, and scientists are wrong?

[@CM63_France]

Nothing at the first place. The gemetrical meaning only comes when you transform it into double integral.

[@draftymamchak]

I don’t even know basic functions, except for linear functions, yet I still watch his videos as If I understand something.

[@YoutubeUser-yl9ys]

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

[@nilsmarione4768]

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]

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

[@Frogieder]

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

[@muhammedfuadpt5137]

Waiting for 100 integral part 2 😌

[@qazar7906]

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

[@procerpat9223]

Geometrically speaking, the area under 2/(1+x^2) from 0 to infinity is exactly the square of the area under e^-x^2 as is the case with pi*x*e^-x^2 from -infinity to infinity

[@kono152]

I love this integral and i never saw this approach

[@prollysine]

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

[@Grisostomo06]

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

[@Österreich-f6m]

Not calculus but high school student. And yeah, your explanation is excellent

[@zahari20]

This is possibly the best way to solve this integral without polar coordinates.

[@tzebengng9722]

Not apparent that use of iterated improper integrals as double improper integral and the interchange of integration in the iterated improper integrals behave this way. An explanation with improper integrals as limiting values would be lengthy. But with I(t) = (integral of e^(-x^2)) from 0 to t)^2 and g(t) =integral from 0 to 1 of e^(-t^2(x^2+1))/(1+x^2) with respect to x, we can show that I(t)+g(t)= is a constant = g(0) = integral of 1/(1+x^2) from 0 to 1 and is Pi/4. We can use differentiation under the integral sign here. Letting t tends to infinity you get your answer since g(t) tends to 0.

[@guylocation9823]

Thank u !! I returned 20 years to my college days

[@saggycoconuts6104]

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

[@258thHiGuy]

You write the best infinity symbols I've ever seen

[@Grecks75]

It's an application of the Fubini-Tonelli Theorem.

[@dubbidude777]

Amazing video you explained it very well

[@Johnny-tw5pr]

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

[@prakhars962]

Even though in real engineering work we never use these analytical techniques I love to spend my dinner time watching your video.

[@YassFuentes]

Beautiful problem, balckpen! Thank you for sharing :)

[@TomJones-tx7pb]

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.