A nice integral.

Рет қаралды 44,609

3 жыл бұрын

We solve a nice integral using Feynman's trick and Fubini's theorem.
Playlist: • Interesting Integrals
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Пікірлер: 226

@DeanCalhoun 3 жыл бұрын

came for a specific solution, left with a general one: now that’s good math!

@ariels.claudino7045 Жыл бұрын

I have just thought of that!!!!! That is what math is all about!

@noahtaul 3 жыл бұрын

4:48 That antiderivative shouldn’t have a negative sign. But at 9:21, even though you mentioned needing to change the sign, you quietly didn’t change the sign, in order to fix the earlier antiderivative error. Pretty sneaky!

@giladu.6551 3 жыл бұрын

my thoughts exactly! i even tried to work it out myself to see if the mistake was mine, but when i saw the mistake-cancelling-mistake, i finally understood.

@Meilo0110 3 жыл бұрын

I was also wondering why he didn't switch the signs at 9:21. But that was a creative way to fix a mistake

@elaadt 3 жыл бұрын

I caught the first mistake and was waiting for a correction at some point. I missed the slight of hand at 9:21 and decided to check the comments to get the puzzle resolved. Thanks.

@MSCCA 3 жыл бұрын

Thanks for clearing that up. It had me confused

@yoavshati 3 жыл бұрын

He probably has a rough solution printed out, so he copied the correct step at 9:21 He has these double sign mistakes in a few videos

@angelogandolfo4174 3 жыл бұрын

2:25 - “Ok. Great.” Indeed!!! I just LOVE your channel; you have such a great variety of problems, of all types, and from literally, ALL around the globe. I just love your solutions, they’re SO satisfying! Thank you! From London, UK.

@frozenmoon998 3 жыл бұрын

Assuming 21:15 is continuous, we can see that under the limit it converges to "ok, that's a good place to stop".

@charlottelockett820 3 жыл бұрын

Your channel has gone a long way to igniting a newfound love for math. Thank you for sharing!

@EduardoBatCountry 3 жыл бұрын

The way you solve this is pretty beautiful, even making it look simple. Love your channel!

@chrstfer2452 Жыл бұрын

Listen, ive studied math for like 8-10 years now and have gone through the whole calculus sequence--some of them a few times, because credits didnt transfer--and none of the diff eq or pde work i did made the usefulness of diff eq's for solving problems (rather than modelling) quite so clear as the few moments you spent solving I''(t)=4I(t). Like, i knew they were powerful, but nothing i encountered before was explained so concisely or intuitively. You truly are a giant among men in terms of explaining mathematical problem solving on this platform. And so prolific too, multiple videos a week. Just wow, professor penn, i know ive been in the comments giving you praise before like its nbd, but man you are amazing.

@veenasv954 3 жыл бұрын

We can solve this integral using Contour integration, Laplace transform(integrand is an even function, so can be written as 2*Integral from 0 to infinity) and Fourier transform

@veenasv954 3 жыл бұрын

Also using Feynman's technique

@drsonaligupta75 3 жыл бұрын

And cauchy residue theorem

@laxminarayanbhandari855 2 жыл бұрын

@@drsonaligupta75 that is contour integration 😐

@laxminarayanbhandari855 2 жыл бұрын

@@veenasv954 I know two of the four methods. 😀

@bendaniels7346 2 жыл бұрын

@@veenasv954 that was the first one I thought of, glad to see someone else thought of it too

@a_llama 3 жыл бұрын

fascinating how feynman's trick gives a sort of general solution!

@angelr.5123 3 жыл бұрын

Leibniz Regel*

@a_llama 3 жыл бұрын

@@angelr.5123 differentiating under the integral was popularised by feynman

@angelr.5123 3 жыл бұрын

@@a_llama It is unrelevant.

@reelandry 3 жыл бұрын

LOL the salt

@a_llama 3 жыл бұрын

@@angelr.5123 Unrelevant is not a word. But that is irrelevant, as is this thread.

@user-wu8yq1rb9t 2 жыл бұрын

It was just amazing, I just enjoying. Thank you

@ion5964 Жыл бұрын

Fantastic video 👏🏻

@MarcoMate87 3 жыл бұрын

Fantastic video. It's useful to observe that f(x) = sin(x)/x is an incredible example of a function which has a finite Riemann improper integral but is not Lebesgue-integrable, because it can be shown that |sin(x)/x| has a divergent integral.

@fhlbadenhorst 3 жыл бұрын

For a detailed treatment of "Feynman's Favorite Trick", see Chapter 3 of Paul J Nahin's book "Inside Interesting Integrals" link.springer.com/chapter/10.1007%2F978-1-4939-1277-3_3

@CTJ2619 Жыл бұрын

Another good video - thanks

@wiwaxiasilver827 3 жыл бұрын

The general formula I just came up with is (1/b)e^(-b*abs(t)) for any value of t and b^2 added to the denominator in the place of 4.

@SuperSilver316 3 жыл бұрын

I think I got the same thing with a contour integral, but I like this way as well!

@andyb6177 2 жыл бұрын

That was pretty wild how he turned a integral problem into an ODE. Very slick...

@harshraj460 3 жыл бұрын

Awesome techniques ❤❤❤

@lori2364 Жыл бұрын

Good one!

@holyshit922 3 жыл бұрын

3:34 This integral i prefer to calculate via Laplace transform (In fact this is Leibniz's differentiaton under an integral sign but in one place calculation of L(sinx/x) is needed for method which i use)

@jayantimajumder4151 2 жыл бұрын

When you came up with the double integration and change the order of those.You should use Laplace transform of sinx in term of y, but that's a another great method to evaluate the integal❤️❤️❤️👍👍

@wiwaxiasilver827 3 жыл бұрын

I think that an absolute value on the t for the (pi/2)exp(-2t) at the end is important for a more general value of t, since the entire steps of integral operations that led to the solution was based on the symmetry of the integral, and I’m wondering if we can come up with a more general formula for when a different number is added to the x^2 in the denominator.

@yonatanrosmarin4135 3 жыл бұрын

There is something spooky here. Notice that if we try to evaluate I(t) for negative numbers we shilud get the same values as for positive ones, due to symmetry. But thefunal result does not show that, and the "true" I(t) is not differentiable at 0.

@pbj4184 3 жыл бұрын

That is because we don't consider the -ve t. I have seen an alternative approach to the final step which answers your question. After solving the D.E, we get I(t)=C1*e^2t+C2*e^-2t Now we know |cos(tx)|

@pbj4184 3 жыл бұрын

In Michael's method, taking (-t) instead of t causes I'(0) to be different which changes the value of C1 and C2. I haven't done the arithmetic but I'm pretty sure when you introduce (-t) in the definition of I(t) instead of t, it not only flips the sign of t in the solution of the D.E but also switches C1 and C2 which altogether which leads to the same result. It seems like the answer breaks down but it actually doesn't. That was a great question. Keep up that inquisitivity 👍

@reynaldopanji2066 3 жыл бұрын

It is so hard for me to solve this problem, because there are so many things come in my mind when solving this But after watching this video, I came to understand what should I do to solve the problem Thank you WolframAlpha 👍

@douglasmagowan2709 2 жыл бұрын

Of course, this integral is a snap to evaluate using complex analysis. Nice to see it done "the hard way" though.

@YoutubeModeratorsSuckMyBalls 2 жыл бұрын

Nice, definetly like

@user-xp9rv7iq7t 3 жыл бұрын

wonderful

@CM63_France 3 жыл бұрын

This method of "secondary integral" is great, you teach me. For fun: 0 "and so on and so forth", 3 "I'll go ahead and", including 1 "I'll go ahead and do that" 1 "let's go ahead and", 5 "let's may be go ahead and", including 1 "let's may be go ahead and do that", 3 "great", including 2 "ok, great", and the "and that's a good place to stop" at the end.

@demenion3521 3 жыл бұрын

"in other words" would be interesting as well :D

@CM63_France 3 жыл бұрын

@@demenion3521 Yes, but I am afraid there may be too many of them :)

@carl13579 3 жыл бұрын

Another approach is to write the cosine as a sum of complex exponentials and then use Fourier transform results involving two-sided exponentials (the latter of which are easily derived).

@veenasv954 3 жыл бұрын

Or you can simply calculate Fourier cosine transform since the function is even

@giuseppemalaguti435 3 жыл бұрын

Molto elegante la soluzione dell' integrale.....complimenti

@meiwinspoi5080 3 жыл бұрын

brilliant

@fhlbadenhorst 3 жыл бұрын

A short article on Feynman's trick over at Medium: medium.com/cantors-paradise/richard-feynmans-integral-trick-e7afae85e25c

@kqp1998gyy 3 жыл бұрын

🌷you make my day. Thank you for what you are doing that we see and especially thank you for what it takes for you to do it that we don't know.

@winky32174 3 жыл бұрын

That was a doozie!

@yoav613 3 жыл бұрын

you can solve this integral easily, using contour integral

@thephysicistcuber175 3 жыл бұрын

+1. Complex analysis

@rishikperugu9381 3 жыл бұрын

yuss!!!

@hopegarden7636 3 жыл бұрын

Where's the fun in that?!😉

@arvindsrinivasan424 3 жыл бұрын

Residue theorem for the win

@RozarSmacco 3 жыл бұрын

Cauchy residue theorem is amazingly powerful.

@BatmanPooping 3 жыл бұрын

@Michael Penn you cannot use Fubini for the second tool, as sin x/x is not absolutely integrable.

@chris12dec 3 жыл бұрын

Such an unexpected answer!

@BatmanPooping 3 жыл бұрын

What am I missing? Why does he change integration order when we dont have integrability in absolute value?

@tordjarv3802 2 жыл бұрын

A very interesting solution. I evaluated it with the residue theorem and Jordan's lemma, which was easy, resulted in the same result but was not as interesting as your solution.

@benardolivier6624 Жыл бұрын

16:47 Something's not right here. If I replace t with 0 the second part is zero and the first part is -pi. But if I replace t with 0 in the original version of the integral there is also sin(t*x) which is the only place there is a t variable, so the value should ALSO be zero. In my world, I'(0) cannot be simultaneously equal to 0 and -pi. My only explanation here is that since we calculate I'(t) for t=0 we cannot use the substitution x=u/t because that would imply a division by zero.

@killermakd2015 3 жыл бұрын

where did you use the tan inverse integral tool? Nice video!

@perimetros314 2 жыл бұрын

We can calulate that just only with residue theorem. ∫[-∞,∞]cos(nxi)/(x²+4)dx = ∫[-∞,∞]exp(nxi)/(x²+4)dx ( ∵ imag. pat is odd ) = 2πi res( exp(nxi)/(x²+4), x=2i ) = 2πi exp(nxi)/(x+2i) ← x=2i = π/2 exp(-2n)

@edwardjcoad 3 жыл бұрын

Isnt the integral of e(-xy) got a superfluous negative in it? Int e(-xy) = -e(xy)/x so isnt the negative outside that integrate for tool #2 incorrect?

@rezabidar1042 2 жыл бұрын

Hello. Do you have any videos showing how to evaluate improper real integrals using Residue Theorem in Complex Calculus? That'd be also a great video topic.

@slenderman4788 2 жыл бұрын

Ngl, this is another one of many problems that make me love complex analysis. Solved this one in 5 mins with it.

@krisbrandenberger544 3 жыл бұрын

There was no need for the minus sign before the integral in which 1/x was expressed. The partial derivative of -e^(-x*y)/x with respect to y is e^(-x*y) and not -e^(-x*y).

@danielinfinito6304 2 жыл бұрын

You can't apply Fubini's theorem to the integral of exp(-xy)sin(x) because this integrand is not absolutely integrable in [0,oo)^2 because the integral of |sin(x)|/x in [0,oo) diverges to infinity

@yannld9524 2 жыл бұрын

It seems that nobody notice this comment, but this is probably the biggest mistake of the video. Most of people don't care what are the conditions to apply Fubini's theorem...

@Steindium 3 жыл бұрын

Im assuming the thumbnail is supposed to be cos(3x)?

@AlexandreRibeiroXRV7 3 жыл бұрын

I love his videos but I sometimes wonder if he is actually paying attention to the thumbnails or the equations he's trying to solve... Not to mention when he fails to cut the mistakes he made on video on the jumpcuts.

@divyanshaggarwal6243 3 жыл бұрын

@@AlexandreRibeiroXRV7 actually you can solve the thumbnail integral by the same approach. so it might not be a mistake

@djvalentedochp 3 жыл бұрын

@@AlexandreRibeiroXRV7 ele deve ser distraído mesmo

@malawigw 3 жыл бұрын

@@angelmendez-rivera351 He did say he is. I know he is kinda busy ATM with other stuff

@hybmnzz2658 3 жыл бұрын

On the bright side we got a general solution that solves the thumbnail and video.

@davidbrisbane7206 3 жыл бұрын

Michael could have used Feynman's integration technique to evaluate the Dirichlet integral by introducting e^(-tx) into the integrand and evaluating (sinx/x)*e^(-tx) rather than sinx/x.

@dieguenec 3 жыл бұрын

Wait the - sign while developing the second tool isn’t supposed to be there since the integral of e^(-xy) is -e^(-xy)/x itself, so if we put the minus sign we change the minus in our result to a plus sign?

@Bayerwaldler 3 жыл бұрын

At 9:00 he should have switched the signs - but then the result should be negative Pi/2. I'm a little confused now. Edit: Aaah - a twofold sign-error - those are the best! 😉

@Walczyk 2 жыл бұрын

this looks like a classic contour integral problem

@yohanguy8216 2 жыл бұрын

Would it work if you instead try to do the integral over the complex plane & use the Residue Theorem & Jordan’s Lemma to get the value of the integral?

@knvcsg1839 3 жыл бұрын

I have faced this question a few months ago and I hated myself for not being able to solve it. Because, the integral looked easy but still, I wasn't able to do it. And then after watching your video, I hated myself again for not knowing that the integral of sinx/x over inf to 0 is pi/2. 😥😥

@xl000 3 жыл бұрын

Can you solve it to infinity and beyond ?

@holyshit922 Жыл бұрын

Laplace transform would be a little simpler I(t) = Int(cos(tx)/(x^2+4),x=0..infinity) then calculate L(I(t))

@tomatrix7525 3 жыл бұрын

Classical, have to say I think integrals are my favourite in maths

@M-F-H 3 жыл бұрын

That's weird (around 17'), if you set t=0 in the first expression of I'(t) then you get 0 right away and that seems totally legit (anyway that's how I'(t) is "defined" in the first place, if it weren't correct for all values of t including t=0 there would be a problem here already and it would invalidate all later steps in the case t=0) ; but then you doodle around and finally get -pi + 0, where the 0 is nearly the same integral of the form f(x)sin(tx) with f -> 0 (x -> oo) in both cases!! ... hmmm ?

@noelani976 3 жыл бұрын

sin(0) =0. Therefore the entire integrand becomes zero.

@M-F-H 3 жыл бұрын

@@noelani976 That's what I'm saying. Yet he finds that it's equal to -π.

@noelani976 3 жыл бұрын

@@M-F-H I got your point. But it doesn't work just that way. First you must realize that that integral is one of the LAPLACE integrals in mathematics, which's solvable given the condition that t>=0. Therefore, putting zero too early defeats the purpose of reaching the GOAL.

@megauser8512 3 жыл бұрын

I know why the first integral of I'(0) is NOT equal to 0: there is an x in the numerator, so as x --> ∞ the integrand x * sin(0*x) / (x^2 + 4) --> (∞ * 0) / ∞, which is an indeterminate form.

@M-F-H 3 жыл бұрын

@@megauser8512 no that's not the point, you can divide numerator and denominator by x to get sin(...) / (x + 4/x) which clearly goes to zero as x→∞. If sin were a positive constant one could argue that the integral diverges as integral (1/x) ~ ln x, but in the case t=0 the sin and thus the whole integrand is zero.

@Jaeghead 3 жыл бұрын

16:53 But if you plug in t = 0 directly into the first integral representation of I'(t) we get 0 because the integrand is always 0.... How do you justify the substitution x = u/t if you want to evaluate I' at 0? I'm confused because in the end the result is correct again....

@Jaeghead 3 жыл бұрын

@@angelmendez-rivera351 True but then still I'(0) = 0 and only the limit of I'(t) as t goes to 0 is -pi, so the differential equation at the end has to be set up differently.

@PyarMatKaro 3 жыл бұрын

I'd like to see an explanation too. I guess we have to take a limit but it is not explained. I think we are taking t>0 so we need to take the limit

@leif1075 3 жыл бұрын

And at 5:03 if you plug infitntiy into y you get zero which when multiplied by 1 /x gives you zero NOT 1 /x so it doesnt work.. please clarify

@leif1075 3 жыл бұрын

@@PyarMatKaro and at around 6:30 since when does order of integration matter for an integral?? You get thebdsme result no matter which variable you integrate first because the other variable is held constant while you integrate one.

@kingthanatos6093 3 жыл бұрын

@@leif1075 The order of integration does matter sometimes - en.wikipedia.org/wiki/Fubini%27s_theorem#Counterexamples

@CommanderdMtllca 3 жыл бұрын

So I did not follow the +4-4 step at all. Why were you able to just add 4 to x^2 in the numerator? If you’re adding 4, shouldn’t it go at the very end instead being sandwiched in the middle?

@CraigNull 3 жыл бұрын

At 15:30, it's very strange the integral of sin(tx)/x over the interval is the same as the integral of sin(u)/u over the same interval. Horizontally scaling that sine function does *nothing*? The same would be true for any integrand of the form f(tx)/x when the interval is over [0, infty). Feels like this fact merits dissection on its own

@Noam_.Menashe 2 жыл бұрын

It does do something if t is negative, and I'm not sure about how complex t would change things.

@coderdemo9169 3 жыл бұрын

I like sir u from India always 💯👌

@tretyakov3112 3 жыл бұрын

17:07 but if we put t = 0 in the second line we’ll have I’ (0) = 0. Why is that?

@awrRoman25 3 жыл бұрын

The general solution I(t) = pi/2*exp(-abs(2*t)). We can't take derivative at t=0. But we can take limit I'(t) when t->+0

@JSSTyger 3 жыл бұрын

And thats a good place to stop... ...because my brain just got wrecked.

@user-A168 3 жыл бұрын

Good

@Zaxx70 Жыл бұрын

17:05 Why does the second part disappear when we set t = 0, while the same two integrals having a sin(tx) just above don't. I'm totally confused here. I feel like the negative pi comes from nowhere, as if we set t=0 immediately from taking the derivative of I(t) we get zero... Why is the integral of sin(tx)/(x*(x²+4)) zero while the integral of sin(tx)/x is not zero?

@juniorstephanovictoriolope1675 3 жыл бұрын

NICE

@josephhajj1570 3 жыл бұрын

You can just use complex analysis

@manfredwitzany2233 3 жыл бұрын

A nice method, but extremly complex. The solution is much less complex, if you aplly complex analysis, as I did to solve the integral in about 2 minutes: 1. Using the rsidue theorem we have to make the function complex analytic, which is done by replacing the cosine function by an exponential function and multiplying the argument by i. f(x)=cos(3x)/(x²+4)=exp(3ix)/(x²+4) 2. Find poles of the denominator of this function by solving the quardatic equation x²+4=0, which are +/- 2i. Only the positive pole is used. 3. The integral along a semicircle comprising the x-axis with a radius greater than 2 will contain the residuum 2i only. According to the residue theorem this integral can be calculated as: I=2 pi i Res(exp(3ix/(x²+4); 2i)=2 pi i exp(-6) /(4 i) For the limit R of the semicircle going to infinity the integral along the arc will tend to zero, as the denominator is quadratic in x. Therefore this integral will be equal to the integral along the real axis from -infinity to +infinity, which is the goal integral. 4. The factor 2i in the numerator and denominator cancel out leaving the result integral from -inf to +inf dx cos(3x)/(x²+4) = pi/(2 exp(6))

@saadmansakib1549 3 жыл бұрын

How did you replace cos(3x) with the exponential function?

@manfredwitzany2233 3 жыл бұрын

@@saadmansakib1549 cos(3x)->exp(3ix) The real part of exp(3ix) equals cos(3x) . Therefore exp(3ix) is a complex expansion of cos(3x). As the exp function is analytic, the residue theorem works with it. By this replacement I changed the complex part of the original function, which dose not matter, as we are only interested in the real part of the same.

@saadmansakib1549 3 жыл бұрын

@@manfredwitzany2233 thank you very much.

@mukulgupta4008 3 жыл бұрын

Wonderul❤❤

@KuroboshiHadar 3 жыл бұрын

Hello Michael! I miss your overkill series, do you have anything planned for that in the near future?

@MichaelPennMath 3 жыл бұрын

I am kind-of at a creative loss for overkill ideas at the moment... Maybe I could get some inspiration?

@KuroboshiHadar 3 жыл бұрын

@@MichaelPennMath maybe a non-traditional proof of pythagorean theorem? There are plenty out there. If I recall correctly, there is even a proof through differential equations. (Although I don't know how it goes). Also, proofs that √2 is irrational are pretty common, but I bet there are some outlandish ones out there. Anyways, thanks for the videos, much love from Brazil :)

@malawigw 3 жыл бұрын

@@MichaelPennMath Perhaps som evaluating some "standard" integrals using Feynman trick?

@TwilightBrawl59 3 жыл бұрын

Michael Penn n choose k = n choose n - k ?

@yaasdpalala4492 3 жыл бұрын

@Adam Romanov Great suggestion!

@ArifSolvesIt 2 жыл бұрын

two minus mistakes but in the end these two errors cancel out.

@thephysicistcuber175 3 жыл бұрын

Thumbnail doesn't match problem. I think this has been happening for a while. Why is that?

@a_llama 3 жыл бұрын

true but technically the general soln found applies to the integral in the thumbnail as well

@ivoh2384 3 жыл бұрын

It's in purpose? To make people play the video....

@Rahul-ky8mm 3 жыл бұрын

Suggest me a proper Olympiad math Books (higher mathematics)

@omarsamraxyz 3 жыл бұрын

Michael, please explain to me what happened at 14:18, repeated that part several times and didn't get the idea!! How did you insert 4-4 into a multiplication? I know that's like adding 0 but didn't get the idea pretty good!! Anyone explain please ❤️

@CrazyPianoMan1 3 жыл бұрын

I had to stop and think about this as well. Here's the step by step as I see it (just for the numerator): (x^2)*sin(tx) (x^2 + 4 - 4)*sin(tx) (x^2 + 4)*sin(tx) - 4*sin(tx)

@omarsamraxyz 3 жыл бұрын

@@CrazyPianoMan1 thanks broo, I figured it out on the same day I asked in this comment. But thank you bro I appreciate it so much ❤️👍.

@giuseppemalaguti435 3 жыл бұрын

Integrale Sinx/x così non l'avevo mai visto

@peterdriscoll4070 3 жыл бұрын

Very nice derivation. But making a differential equation of it seems like a step backwards to go forwards. I long for a more direct approach for that last step. Just me.

@maelhostettler1004 3 жыл бұрын

The integral on the thumbnail isnt the sâme... IT is cos(2x) on the thumbnail and cos(3x) in the vidéo

@malawigw 3 жыл бұрын

Well then the answer to the thumbnail is pi/(2*e^4) ;)

@M-F-H 3 жыл бұрын

Anyway I wonder why he didn't leave "t" arbitrary.

@maelhostettler1004 3 жыл бұрын

@@M-F-H cuz it doesnt give à "beautifull" Finite result

@noahzuniga 3 жыл бұрын

sinx/x can also be solved with derivative under the integral

@harshraj460 3 жыл бұрын

This can be simply done using laplace transform 😎😎

@arandomcube3540 3 жыл бұрын

A nice title.

@M-F-H 3 жыл бұрын

yes, so specific and complete, self-contained explanation of the contents! :-D !

@edmundmcthundercock2796 3 жыл бұрын

Well fml, i just learnt the power rule

@noelani976 3 жыл бұрын

Another LAPLACE INTEGRAL, but this time with sine function in the integrand. integral(-inf,inf) of (xsin(tx)/(k^2 + x^2))dx where k>0, t>0

@harish6787 3 жыл бұрын

Heavily impressed sir I love u sir

@whyyat3470 2 жыл бұрын

Sorry, but around 19:00, why did you write two functions and add them when one would suffice for I(t)?

@Noam_.Menashe 2 жыл бұрын

Because you want the most general solution. After that, you can find coefficients and they could be zero.

@The1RandomFool 3 жыл бұрын

The video thumbnail says cos(2x), but the goal in the video says cos(3x).

@Meilo0110 3 жыл бұрын

Simply a spectacular video, u and bprp taught me 90% of all I know and I do know a lot BTW I'm only 18

@thecodingpheonix2073 3 жыл бұрын

Who is bprp?

@demenion3521 3 жыл бұрын

@@thecodingpheonix2073 black pen red pen

@JoseFernandes-js7ep 3 жыл бұрын

👌

@niczoom 3 жыл бұрын

Coming from someone whos knows very little regarding these integrals, what's the real world application of the above?

@demenion3521 3 жыл бұрын

i can't tell you that much about "real world" applications of this kind of integrals. but this is a kind that very often appears in physics, either in quantum theory (for propagators) or in the theory of stochastic processes (for characteristic/response functions) or in the theory of (soft) condensed matter (for response functions/elastic moduli). in general, you get this often when solving differential equations using fourier transform

@noelani976 3 жыл бұрын

@@demenion3521 PDEs to be exact.

@demenion3521 3 жыл бұрын

@@noelani976 PDEs are the most common types that you run across, but you can solve ODEs just as well with fourier transform ^^

@noelani976 3 жыл бұрын

@@demenion3521 sure of what you're saying? LAPLACE transforms only apply to ODEs. FOURIER transforms come in when you are talking of BOUNDARY VALUE PROBLEMS (BVPs). You can very this claim yourself.

@demenion3521 3 жыл бұрын

@@noelani976 the thing is: laplace transforms and fourier transforms both can be applied to either ODE or PDE. it is just more common for one type of problem to use one or the other transform. if your functions are sufficiently well-behaved, there shouldn't be a large difference anyway as the fourier and laplace transform can then be thought of as analytic continuations of each other. of course, this correspondence does not hold in general, but in most cases which are relevant for physics. i tend to use fourier rather than laplace just because i know the transforms of some standard functions better

@keshavb3128 3 жыл бұрын

use contour integral. You don't need to use so much work.

@MichaelPennMath 3 жыл бұрын

You are not wrong... so far I have stayed away from complex analytic approaches for some reason.

@leif1075 3 жыл бұрын

@@MichaelPennMath If you don't know those two tools is it possible to solve? And why wiuld anyone think of rewriting the sin x/x integral as e raised to the xy like that?? Hope you can please respond.

@sasharichter 3 жыл бұрын

contour integration would result in a 3 minute video and not a 21 minute one, not to mention it wouldn't allow one to showcase a set of esoteric techniques reserved solely for integrals that can't be solved by simpler or more standard methods such as complex integration.

@sirlight-ljij 3 жыл бұрын

Can't this be done by residuals of e^2iz/(z^2+4)?

@sasharichter 3 жыл бұрын

that's how it should be done - by the method of residues (not residuals :-) - instead of using esoteric techniques that are not easily generalizable and hardly applicable in most cases

@goodplacetostop2973 3 жыл бұрын

21:15 Also, completely unrelated to this video but here’s an exercise for you. 100 children took part in a Mathematics competition consisting of four problems. 90 children solved the first problem 85 solved the second 80 solved the third 75 solved the last problem What is the minimum number of children who could have answered all the questions correctly?

@donaldbiden7927 3 жыл бұрын

So fast

@xcarnage8632 3 жыл бұрын

This is a simple question of set theory

@aaronwelson 3 жыл бұрын

@R0M4ur0 3 жыл бұрын

Out of the 75 who correctly answer number 4, 10 could be the ones who did not answer the 1st (but did 2nd and 3rd), 15 those who did not answer the 2nd (but did 1st and 3rd) and 20 those who did not answer the 3rd (but did the 1st and 2nd). The 30 left necessarily answered all questions

@chandravo501 3 жыл бұрын

@@xcarnage8632 it is not simple, it took me 20 min to solve

@bertrandviollet8293 3 жыл бұрын

What is the level of michael Penn among other mathematicians?is he a great mathematician or average university prof like many others in the world?

@erfanmohagheghian707 2 жыл бұрын

One line solution: 2pi*i*Res(exp(3iz)/(z^2+4) @ z=2i) lol

@oder4876 3 жыл бұрын

you Forget a dx """"" 12:16 """" a lot of time i get noticed by profs after the Exam

@charlesbromberick4247 3 жыл бұрын

I like the way you do math, but think someone has to be pretty darn good just to follow - sometimes I can´t. (I enjoy the way you congratulate yourself with all the "good"´s.)

@neur303 3 жыл бұрын

To me the "good"'s means something like "we have reached a result"/"this subsection is finished"

@charlesbromberick4247 3 жыл бұрын

@@neur303 agreed

@luizarnoldchavezburgos3638 3 жыл бұрын

I(0) = π i think

@federicopagano6590 Жыл бұрын

13:16 I'(0)=0 then 16:50 I'(0)=pi no way

@maddog5597 Жыл бұрын

I agree 100%. Something fishy is going on here. Michael Penn - care to e plain?

@maddog5597 Жыл бұрын

e plain = explain

@Ch1pp007 3 жыл бұрын

Is "quantity squared" a thing now? I did a maths degree back in the good old days and never heard the term, we just said "squared".

@Ch1pp007 3 жыл бұрын

@@angelmendez-rivera351 or you say "x over ... two squared" rather than "x over two ... squared". All you need is a correctly placed pause and there's no confusion. I can't remember any ambiguities from before people said "quantity squared".

@Ch1pp007 3 жыл бұрын

@@angelmendez-rivera351 Warwick, UK within the last decade so fairly up to date. Maybe it is an exclusively American thing as I have never heard it outside of this guy's videos. Also, things are almost always being written on a board or paper at the same time so there isn't really any ambiguity for the language to correct.

@zygoloid 3 жыл бұрын

This seems to be at least a UK/US thing. In the UK I never heard "quantity squared"; we'd use a pause or say "all squared". But in US mathematics it seems pretty common.

@Ch1pp007 3 жыл бұрын

@@zygoloid Yeah, now you mention it I think I remember "all squared". Certainly rolls off the tongue better than "quantity squared".

@noelani976 3 жыл бұрын

The GOAL integral presented by Michael Penn was LAPLACE Integral. Did anyone identify it as Laplace Integral? If that's the case, you can use Fourier Cosine Integral formula to solve it too.