There is some "complex" stuff happening here...

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A nice integral related to the Gaussian integral.
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Опубликовано:

18 дек 2021

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Комментарии : 114

@tomeknowak4200 2 года назад

10:30 The fact that we can integrate over half-line from 0 to infinity, instead of half-line from 0 to alpha times infinity is a consequence of Cauchy integral theorem. In my opinion doing it the same way, as we would in real analysis course , without any explanation is a certain abuse.

@pedrosso0 2 года назад

thank you

@AustinNoLimits 2 года назад

thinking the same during watching and thought he might revisit this later. Thanks for proving this context.

@clementcoine1581 2 года назад

At 9:17 this really requires some justifications using Cauchy's formula. But well, the result is correct :)

@matthiasbergner8911 2 года назад

10:28 : There is something I do not understand: You say that u is just a real variable. But x is already a real number and u = alpha * x where alpha is a complex number with nonzero imaginary part, so u cannot be a real number. Technically, the integral in the u-variable is not along the positive real axis but along some straight ray in the complex plane. I am not sure whether this integral has the same value as the wellknown Gaussian integral.

@randomname7918 2 года назад

I had the same question so I'm leaving this comment here to get a notification

@vinvic1578 2 года назад

@@randomname7918 same I would love to understand

@cernejr 2 года назад

Yes, I have the same concern.

@matiasnovosad7330 2 года назад

Yeah... i woud like to know how does that work too

@charlottedarroch 2 года назад

If instead of taking the Gaussian integral along the ray from 0 along the positive axis, you instead pick any ray with argument on [-π/4,π/4], then you get exactly the same value for the integral. If you instead pick a ray with argument on [-π,-3π/4]U[3π/4,π], then you get the negation of the usual Gaussian integral. Taking the integral along any other ray from zero results in a divergent integral. So if α had been in a different sector of the complex plane, the integral might not have converged. So definitely this should have been given closer inspection and justification.

@shanestrickland9557 2 года назад

15:57 : The moment I realized you were going to use the 1/2 formula that way, set the calculation god knows how many steps b4, and busted out laughing like a madman. I trade stocks and options for a living now adays... My God I miss solving these problems! Some days, watching your channel is what gets me through day of life in a cold dark world that has abandoned logic, math, and physics. Seriously, thank for your dedication!

@BikeArea Год назад

😊

@nathanisbored 2 года назад

9:31 im not sure why we're allowed to say that u tends to infinity here, since its skewed in the complex plane... certainly if alpha were negative, we'd have to say u tends to -infinity instead of regular infinity... so in this case, dont we need to say that u tends to complex infinity or something? and doesnt that affect how we can apply the gaussian integral? something feels fishy but i cant figure out why it doesnt affect the solution...

@rocky171986 2 года назад

Because in this case the complex function is holomorphic, so the limit tends towards infinity in a well-defined manner

@nathanisbored 2 года назад

@@rocky171986 which complex arguments (angles in the complex plane) can you have that still make it to infinity? since -infinity and +infinity are distinguished, why arent other infinities?

@xCorvus7x 2 года назад

Seems like it should be precisely α times infinity.

@tharunsankar4926 2 года назад

The frickin integral screamed: RESIDUE AND JORDAN’S LEMMA to me when I saw this monster lol

@The1RandomFool 2 года назад

I used Feynman's technique with the extra parameter in the sine function. Differentiating allowed me to eliminate x^2 in the denominator. I then complexified the cosine and turned it into a contour integral in the complex plane. My substitution is beta=sqrt(1+a*i), where a is my extra parameter. Michael Penn's substitution is sketchy here. He makes a leap to a real-valued integral without showing why that's the case. With the proper application of complex analysis, it can be shown it does. Much of my work is the same as his here, but at the end, I have an additional integral to evaluate.

@SuperSilver316 2 года назад

I think I did the same thing, once you integrate that Gaussian you have to find the real part of the square root of a complex variable and then integrate that one more time, which can be done using some hyperbolic trig subs and identities and the integrand reduces really nicely!

@The1RandomFool 2 года назад

@@SuperSilver316 In the final integral I actually ended up using a tangent substitution, then doing a secant^2 substitution in place of secant, which made it trivial. In total, I think this was the same amount of work as the video.

@SuperSilver316 2 года назад

Yeah I think that works to, it’s interesting that hyperbolic and tangent substitutions can be used so interchangeably.

@user-yt198 2 года назад

Somebody pointed at the back of our professor with laser. Euler maybe? 😛

@nikolaynikolov4086 2 года назад

Me: ugh my life is so complicated... Me after watching this video: ok, nevermind, I take it back!

@SuperSilver316 2 года назад

It’s also at least formally the Gaussian Transform of sin(x^2)/x^2, taking into account the appropriate normalizations and constants which can uniquely reduce down to other Transforms given the appropriate substitutions.

@manucitomx 2 года назад

Very nice. Thank you, professor!

@cernejr 2 года назад

Wolfram Alpha: sqrt(pi/sqrt(2)*(1-sqrt(2)/2.0)) , approx. 0.81, about 90% of the value without the sin(x^2)/x^2 which is sqrt(pi)/2 as mentioned in the video.

@Alex_Deam 2 года назад

I keep thinking there's a red laser target on your back

@riadsouissi 2 года назад

After substitution, the integral should be from 0 to inf+inf*i where i is the complex number. Then we can split the "interval" into an integral from 0 to R (this one can be evaluated in a classical manner when R tends to inf) and an integral from R to R+R*i (this one tends to 0 when R tends to inf because of the e^(-R) term). He just got lucky his substitution worked because the complex interval is zero but often it is not the case...

@vladimirriabchun1872 2 года назад

Thank you for such a beautiful solution. I think, you should have said something about another value of alpha - there are two square roots of i + 1. The prove that we get the same values in the end or we took the value we needed should be there. And another tricky thing happened when you squared (sin - cos) and then rooted. If the value was negative, you would have got the wrong answer. Of course, it’s quite obvious that it’s a positive number, but if there was a more complicated expression this fact should have been noticed.

@admink8662 2 года назад

9:36 i have some doubt with this

@robertapsimon3171 2 года назад

It seems to me that the final solution is slightly more elegant to write as sqrt(pi/2*(sqrt(2) - 1)). But a surprisingly straightforward and elegant solution to this difficult integral!

@VinayakRaina1610 2 года назад

Extremely lucid explanation. Thanks!

@yoantardy8062 2 года назад

Great video ! I want to say that I really enjoy watching these videos that are really original and kind of "high level" compare to other maths videos. I have an alternative proposition for the step of the integration by part : maybe putting a parameter t in the exponential and calling the all fonction phi (t), see it as an parameter integral and derive it with respect to t. You have a simplification beween the square of x that divide the expression and the square of x that arrive because of the derivation and you have directly a calcuable expression of the derivative of phi (t) which is calculable by the same way you are doing in the end of the video.

@PhysicalMath 2 года назад

Awesome video!

@vaibhavcm7503 2 года назад

I think you can easily solve it, by the substitution x²=t, and then use the complex definition of sinx and then some Laplace transform.

@SuperSilver316 2 года назад

Having looked at two other methods to solve this problems, this one is definitely the easiest and gives the answer rather quickly. One remark is that integral itself is just Gaussian transform of sin(x^2)/x^2, evaluating the new function at 0 (and making some substitutions to acknowledge normalizations and certain constants). It’s interesting to see this also reduce into a Laplace Transform, I wonder if there exists an equivalent result for a Fourier Transform.

@hudson11235 2 года назад

Here is one idea. After the first simplification, one needs to calculate the integrals of sin(x^2)e^(x^2) and cos(x^2)e^(x^2), which are the imaginary and real parts of the integral of e^{ix^2}e^{x^2} = exp((i-1)x^2). This is a real function with values in R^2 (or C) and we integrate each coordinate individually. Next, define F(y) = \int exp((iy-1)x^2) dx, for real values of y. If we can differentiate inside the integral, F'(y) is F(y) divided by a complex like -i(1-yi). The solution of this complex ODE will be a F(0) times a term with the sqrt of (1-iy). After that, one just calculates F(1). The advantage of this method is that one doesn't need residuals nor limits on C. Also, it is not hard to see we can really differentiate inside the integral using the exponencial decay and uniform convergence (a la Weiestrass). Hopefully we get the same answer.

@davidbrisbane7206 2 года назад

@9:35 surely if u = 𝛼x, where 𝛼 = √(1 + i) i.e. 𝛼 is complex then as x → ∞, then u *does not* → ∞, as 𝛼 is complex number.

@a52productions 2 года назад

Representing sin and cos as real and imaginary parts of a complex exponential is a nice little trick -- I always expand them out to their full form as a sum of complex exponentials and it ends up being a real mess for a little bit before I can collapse it back down.

@SuperSilver316 2 года назад

I always like coming to these videos after having tried the problem just to see if Mike does it a different way. But yeah, I think like Someone else I introduced a t in the argument of the sine, took a derivative which gets rid of the x^2, and then it turns into the integral of the Real Part of exp(-x^2(1+it)). Finding the real part of of this result is nontrivial and so is the integration of that real part but it can be done and reduces very nicely with a few hyperbolic trig subs.

@goodplacetostop2973 2 года назад

17:09

@vinvic1578 2 года назад

Thanks you're putting in the good work

@MathSolvingChannel 2 года назад

Hey!

@goodplacetostop2973 2 года назад

@@vinvic1578 👍

@goodplacetostop2973 2 года назад

@@MathSolvingChannel Hi 👋

@DarkArc_ 2 года назад

So. Much. Numbers. And. Letters.

@polychromaa 2 года назад

Integrals are my favorite!

@claudiosilvestri6865 2 года назад

Awesome integral

@thierrychenevier3508 2 года назад

Nice integral, thanks !! Two small comments : first, as we are talking about integration among real numbers, the "trick" to use complex notations might be valid, and might not. As already legitimately noted by a few other comments, this is not as obvious as "goes to infinity so everything is ok"... and a bit of an additionnal justification might be useful. Second, when you say that cos (pi/8) - sin (pi/8) is equal to the square root of its square, you still need to check whether this number os positive, since the square root of a square is the absolute value of the number, not always the number.

@wynautvideos4263 2 года назад

True i didnt like how he simplified cos(pi/8) - sin(pi/8) and he couldve done it much easier

@DeanCalhoun 2 года назад

Well we know pi/8 is less than pi/4 and cos is greater than sin on the interval [0,pi/4) so it must be positive

@tomholroyd7519 2 года назад

8:34 - shirt change

@mihaipuiu6231 2 года назад

Very interesting integral and,...of course very nice proof.

@konraddapper7764 2 года назад

It is alway great to see differnt Solutions to the same Problem. It would never crossed my mind to solve the Problem in that way. Because to me it screams to used cauchys interal therm In oder to transform it to the regular gaussian integral.

@nuranichandra2177 2 года назад

We owe it to both the Integral suggester and the integral slayer.

@markuswalden9186 2 года назад

Way to go 👍

@viktoryehorov4314 2 месяца назад

btw, it could be solved using Ramanujan's master theorem

@lincolnyoung549 2 года назад

No one going to mention that at 14:18 he literally forgets an “i” and messes up the whole thing and doesn’t even notice :/

@schweinmachtbree1013 2 года назад

he doesn't forget an i - at 12:56 he calculates 1/alpha on the side because he needs its real and imaginary parts, which he plugs in at 14:18

@_skyslayer Год назад

I don't agree that at 9:17 the boundaries of integration do not change. integral_0^oo is a completely different thing from integral_0^(oo*i). In this example I believe you should end up with an integral_0^(oo + oo*i), which you should split into a sum of integral_0^oo + integral_0^(oo*i). I bet the second part of this sum tends to 0, but you didn't show it!

@MathSolvingChannel 2 года назад

Exactly, integral by part, same result here :)

@2ntwins 2 года назад

You said a while back that you were going to do videos on complex analysis right? Similar to your number theory videos? Is that still going to happpen? It was a course I never got to take in undergrad unfortunately.

@snehasismaiti342 2 года назад

Very nice integration

@mrminer071166 2 года назад

Improper at x = 0. Basically you're looking at sin x approximated by x, divided by x^2, so it blows up at x =0

@stmmniko7836 2 года назад

Nice

@immolator6666 2 года назад

Can someone explain to me why we can assume that u is a real number (10:20), although it comprises of alpha which is a complex number?

@Edsonrsmtm 2 года назад

a-b is not [(a-b)^2]^{1/2} if a-b is negative.

@victoriakurt441 2 года назад

Why does he write the infinity symbol like that?

@bravobessa3684 2 года назад

A big shortcut... but it is correct

@Ahmed-Youcef1959 2 года назад

Can we solve this integral without using complexe number ?

@swapnilshinde6368 2 года назад

Quiet a complex

@RozarSmacco 2 года назад

Prof. Penn, could you use Frullani’s theorem and Ramanujan’s Master theorem to evaluate this?

@SuperSilver316 2 года назад

Just from a cursory glance I think you will run into a problem not because integrand, but because the formula will require you to find Gamma(-1), using the complex exponential does give you the complex power series that you need, but you will have to evaluate gamma at a negative integer because of the x^(-2) term.

@SuperSilver316 2 года назад

Not sure if an immediate way to tackle this with the frullani integral either, but I’d be curious to see it tried.

@SuperSilver316 2 года назад

I’m dumb Ramujan’s Theorem does work, just make the u=x^2, and then turn then turn the integral into the imaginary part of 1/2*int(u^(-3/2)*exp(-u+iu)du from 0 to inf You can expand the complex exponential into a nice alternating power series, find that the coefficient that you want is a(k) = (1-I)^k. Than the integral is just the imaginary part of this. Rly neat result. 1/2*Gamma(-1/2)*a(1/2)

@yvindmathiassen7617 2 года назад

I´m definitly square now

@TheGroundskeeper 2 года назад

Michael, groundskeeper here. Appreciate your studies. I have a prime number problem I’d love to share some day 👍

@gerryiles3925 2 года назад

Doesn't that simplify down to sqrt( pi/2 * [ sqrt(2) - 1 ] ) ?

@GirishManjunathMusic 2 года назад

‌ √[(π/√2)(1 - (√2/2))] = √[π(2 - √2)/2√2] = √[4π(√2 - 1)/8] = √[(π/2)(√2 - 1)] So yes, it could. I don't like leaving irrationals in the denominator either, lol.

@antoninnguyen1617 2 года назад

Sorry I don't understand at 10:00 how can Re(exp(-u^2))/α is equal to Re(α)× exp(-u^2)

@xCorvus7x 2 года назад

In the video he's claiming that u is a real variable and the integral of exp(-u^2) du as such a real integral. Completely real terms can indeed be factored out of the Real/Imaginary Part functions. The bigger problem seems to be that original claim because u=αx where α is a complex number with non-zero imaginary part. He omitted a demonstration of the integral over u being in fact equal to the regular Gaussian integral over a real variable. As other comments pointed out, though, this equality is the case.

@jplikesmaths 2 года назад

Just cancel out the x^2 😛

@tomholroyd7519 2 года назад

Shadow of mic in picture. Do a backflip and nuke that mic nice parts, tho (integration by partitions?)

@MichaelPennMath 2 года назад

In the new year I am going to figure out some lighting that removes the shadow of the mic. I am also going to add some more sound treatment.

@db-os9kb 2 года назад

Be careful with using l'Hopital's Rule for sinx/x since the property that lim sin x/x = 0 is required to find the derivative of sin x. So (to be pedantic (apologies!!)) you cannot lean on l'Hopital to verify it.

@user-vk8fs6jb7i 2 года назад

What 😶

@morten3536 2 года назад

fuckin hell man

@ssarkar2996 2 года назад

I don't get why we can assume u is real.

@AhmedDZUSA 2 года назад

20 like

@bobbwc7011 2 года назад

Seriously, the notation is not great: What is sin x²? Is it sin(x²) or is it (sin x)² = sin²x? I understand that it must be sin(x²) in this context due to the chain rule for finding du, but nevertheless: parantheses never hurt. If it is sin(x²) it should be written as that, and not as sin x².