A nice integral.

[@angelogandolfo4174]

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.

[@noahtaul]

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]

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]

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]

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]

Thanks for clearing that up. It had me confused

[@yoavshati]

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

[@frozenmoon998]

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

[@dingo_dude]

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

[@ariels.claudino7045]

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

[@chrstfer2452]

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.

[@EduardoBatCountry]

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

[@veenasv954]

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]

Also using Feynman's technique

[@drsonaligupta75]

And cauchy residue theorem

[@laxminarayanbhandari855]

@@drsonaligupta75 that is contour integration 😐

[@laxminarayanbhandari855]

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

[@bendaniels7346]

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

[@charlottelockett820]

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

[@MarcoMate87]

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.

[@Jaeghead]

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]

@@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]

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]

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]

@@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]

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

[@a_llama]

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

[@angelr.5123]

Leibniz Regel*

[@a_llama]

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

[@angelr.5123]

@@a_llama It is unrelevant.

[@reelandry]

LOL the salt

[@a_llama]

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

[@holyshit922]

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)

[@andyb6177]

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

[@yonatanrosmarin4135]

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]

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]

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 👍

[@benardolivier6624]

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.

[@ion5964]

Fantastic video 👏🏻

[@user-wu8yq1rb9t]

It was just amazing, I just enjoying. Thank you

[@wiwaxiasilver827]

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.

[@reynaldopanji2066]

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 👍

[@CTJ2619]

Another good video - thanks

[@M-F-H]

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]

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

[@M-F-H]

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

[@noelani976]

@@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]

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]

@@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.

[@SuperSilver316]

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

[@CM63_France]

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]

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

[@CM63_France]

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

[@carl13579]

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]

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

[@tretyakov3112]

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

[@wiwaxiasilver827]

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.

[@Steindium]

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

[@AlexandreRibeiroXRV7]

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]

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

[@djvalentedochp]

@@AlexandreRibeiroXRV7 ele deve ser distraído mesmo

[@malawigw]

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

[@CraigNull]

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]

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

[@goodplacetostop2973]

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]

So fast

[@xcarnage8632]

This is a simple question of set theory

[@aaronwelson]

25

[@R0M4ur0]

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]

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

[@rezabidar1042]

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]

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

[@danielinfinito6304]

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]

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...

[@Zaxx70]

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?

[@harshraj460]

Awesome techniques ❤❤❤

[@edwardjcoad]

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?

[@yoav613]

you can solve this integral easily, using contour integral

[@thephysicistcuber175]

+1. Complex analysis

[@rishikperugu9381]

yuss!!!

[@hopegarden7636]

Where's the fun in that?!😉

[@arvindsrinivasan424]

Residue theorem for the win

[@Swybryd-Nation]

Cauchy residue theorem is amazingly powerful.

[@jayantimajumder4151]

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❤️❤️❤️👍👍

[@krisbrandenberger544]

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).

[@douglasmagowan2709]

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

[@dieguenec]

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]

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! 😉

[@perimetros314]

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)

[@holyshit922]

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

[@CommanderdMtllca]

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?

[@ArifSolvesIt]

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

[@tordjarv3802]

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.

[@KuroboshiHadar]

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

[@MichaelPennMath]

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

[@KuroboshiHadar]

@@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]

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

[@TwilightBrawl59]

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

[@yaasdpalala4492]

@Adam Romanov Great suggestion!

[@omarsamraxyz]

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]

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]

@@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 ❤️👍.

[@lori2364]

Good one!

[@yohanguy8216]

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?

[@giuseppemalaguti435]

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

[@maelhostettler1004]

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

[@malawigw]

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

[@M-F-H]

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

[@maelhostettler1004]

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

[@kqp1998gyy]

🌷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.

[@편대범-p9k]

wonderful

[@davidbrisbane7206]

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.

[@chris12dec]

Such an unexpected answer!

[@knvcsg1839]

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. 😥😥

[@meiwinspoi5080]

brilliant

[@JSSTyger]

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

[@killermakd2015]

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

[@manfredwitzany2233]

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]

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

[@manfredwitzany2233]

@@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]

@@manfredwitzany2233 thank you very much.

[@winky32174]

That was a doozie!

[@whyyat3470]

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

[@Noam_.Menashe]

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

[@Walczyk]

this looks like a classic contour integral problem

[@coderdemo9169]

I like sir u from India always 💯👌

[@niczoom]

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

[@demenion3521]

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]

@@demenion3521 PDEs to be exact.

[@demenion3521]

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

[@noelani976]

@@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]

@@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

[@tomatrix7525]

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

[@federicopagano6590]

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

[@maddog5597]

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

[@maddog5597]

e plain = explain

[@peterdriscoll4070]

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.

[@thephysicistcuber175]

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

[@a_llama]

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

[@ivoh2384]

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

[@arandomcube3540]

A nice title.

[@M-F-H]

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

[@xl000]

Can you solve it to infinity and beyond ?

[@user-A168]

Good

[@juniorstephanovictoriolope1675]

NICE

[@josephhajj1570]

You can just use complex analysis

[@edmundmcthundercock2796]

Well fml, i just learnt the power rule

[@sirlight-ljij]

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

[@sasharichter]

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

[@Rahul-ky8mm]

Suggest me a proper Olympiad math Books (higher mathematics)

[@fhlbadenhorst]

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

[@noahzuniga]

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

[@giuseppemalaguti435]

Integrale Sinx/x così non l'avevo mai visto

[@The1RandomFool]

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

[@mukulgupta4008]

Wonderul❤❤

[@harshraj460]

This can be simply done using laplace transform 😎😎

[@keshavb3128]

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

[@MichaelPennMath]

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

[@leif1075]

@@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]

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.

[@Meilo0110]

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]

Who is bprp?

[@demenion3521]

@@thecodingpheonix2073 black pen red pen

[@bertrandviollet8293]

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?

[@oder4876]

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

[@noelani976]

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]

Heavily impressed sir I love u sir

[@Ch1pp007]

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]

@@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]

@@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]

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]

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

[@fhlbadenhorst]

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

[@charlesbromberick4247]

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]

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

[@charlesbromberick4247]

@@neur303 agreed

[@JoseFernandes-js7ep]

👌

[@luizarnoldchavezburgos3638]

I(0) = π i think

[@erfanmohagheghian707]

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

[@noelani976]

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.

[@KingGisInDaHouse]

Is it just me or does he skip over too many steps