check out the twist at the end of this integral.

[@sweetcornwhiskey]

Nice video! There was one step that was left out - when you perform integration by parts here, the limit as y -> infinity of sqrt(2y-1)*arctan(1/y) is zero, but there's more steps that are required to show this. This is because the limit is of type infinity * 0. It's not too difficult to show that this is zero using L'Hopital's rule though

[@clementcoine1581]

Or simply use that arctan(t) ~ t in 0

[@DrR0BERT]

I agree. It isn't obvious that this would be 0.

[@Kapomafioso]

Yeah that was a little weird how quickly he went through that bit. It's not that straightforward that the term goes to 0 at oo, because the square root bit grows to oo, too, but slower than arctan(1/y) goes to 0.

[@gregsarnecki7581]

11:53 what a superb semicircle he drew! Totally nailed all three points. That's a lot harder than it seems. Kudos Michael!

[@laurensiusfabianussteven6518]

that real number extension is simply beautiful

[@Hipeter1987]

"Spicy contour integral" - quite the teaser! 😂

[@byronwatkins2565]

At 20:35, sqrt(2y-1) approaches infinity making that an indeterminate form and making it NOT obvious that the limit is zero.

[@The1RandomFool]

At 14:12 he's missing a factor of e^(i*theta) in the parameterization of the half circle, but the conclusion is the same.

[@StanleyDevastating]

but in polar co-ordinates dz = r d\theta . Isn't he just using that?

[@The1RandomFool]

@@StanleyDevastating If z = R*e^(i*theta), then dz = R*i*e^(i*theta) * dtheta. He moved R*i outside the integral, but missed the factor of e^(i*theta) in the integrand.

[@Zaxx70]

9:05 When t tends towards infinity, there is an indetermined form infinity times zero, should have done a bit more work to show that the limit is indeed zero.

[@Ravi-ng3ee]

Make a playlist for integrals!!!

[@thierrytitou3709]

At 9:08 : sqrt(2y-1)×arctan(1/y) = sqrt(2)×[y^(1/2)]×sqrt(1 - 1/2y)× [1/y + O((1/y^3))] and therefore is equivalent to sqrt(2)×y^(-1/2) as y -> +oo thus the limit is 0

[@Calculus58]

I found it easier to differentiate both sides of the t substitution at the start, then substitute for cos and sin in terms of t and rearrange. Saves having to remember the derivative of inverse cos.

[@juniorcyans2988]

Amazing!

[@SEJEONGLEE-h1g]

Dear Dr. Penn, thank you for your nice work, excellent board schemes and well-prepared explanations. It is a real joy to watch and learn from your videos. Can I suggest that, in case you tackle integrals, it might be good to show how the functions (integrand) graphically look like. Thank you! P.M.H.

[@Noam_.Menashe]

Don't you need to isolate x=1/2 from the contour? It's where a branch cut happens.

[@xizar0rg]

I believe that happens in his argument explaining why we only need the Real part of the contour integral.

[@nameless3556]

I agree that this should at least have been brought up. I'm very rusty on contours and complex square root functions, though.

[@SuperSilver316]

How he chose to do this avoids this particular hiccup, but I would in general I would have opted for a keyhole contour around with a branch cut taken from (-inf,1/2). That makes for a harder problem overall since you need to estimate the circle around the branch point also, but you should ultimately arrive at the same answer. You could also avoid it by make the variable change 2t-1 = z^2, and contour integrating that rational expression, but I need to look at that a little deeper just to make sure, this just gets rid of the square root and also the branch point by proxy.

[@SuperSilver316]

Yup both methods I suggested yield the same answer, if you go the route of branch point you have to be very careful with how you compute residues for, as your choice branc will determine what those square root functions will yield.

[@pawel_maslanka]

that was indeed a really surprising twist, i did not expect that

[@erfanmohagheghian707]

I was surprised that you didn't even mention that the square root is multivalued and you need either a branch cut or semicircle around z=1/2 It's true that the integral will turn out zero, but leaving it out was bad.

[@holyshit922]

My approach t = tan(x) substitution Integration by parts with u = arctan(2/(1+t^2)) dv = dt After my propositions we should get integral of rational function 2Int(2t^2/(t^4+2t^2+5),t=0..infinity) 2(Int((t^2+sqrt(5))/(t^4+2t^2+5),t=0..infinity) + Int((t^2 - sqrt(5))/(t^4+2t^2+5),t=0..infinity)) 2(Int((1+sqrt(5)/t^2)/(t^2+2+5/t^2),t=0..infinity) + Int((1 - sqrt(5)/t^2)/(t^2+2+5/t^2),t=0..infinity)) Int((1+sqrt(5)/t^2)/(t^2+2+5/t^2),t=0..infinity) u = t-sqrt(5)/t Int((1 - sqrt(5)/t^2)/(t^2+2+5/t^2),t=0..infinity) v = t + sqrt(5)/t Or we can do partial fraction decomposition

[@richardheiville937]

I have proceeded in the same way!

[@SuperSilver316]

Interestingly enough, a function as crazy as this, does admit to an anti derivative that has a closed form Solution, it’s defined rather wildly with ar tangents and imaginary units but it’s a pretty cool result if you play around with it.

[@holyshit922]

In fact one of these substitutions for integral of rational function will be problematic on this interval so partial fraction decomposition seems to be better option Indefinite integral also can be quite easily calculated

[@Alan-zf2tt]

lol @ ~ 13:00 Is it Re as in Real or is it Re as in R times exp( # )? Nice and subtle for sure 🙂

[@bot24032]

9:08 is it not 0*infinity?..

[@emanuellandeholm5657]

the inverse tangent "wins". Try L'H or a truncated Taylor series

[@bot24032]

@@emanuellandeholm5657 ok. still believe this should have been said though

[@emanuellandeholm5657]

@@bot24032 sure

[@reubenmanzo2054]

Actually, the numerator in the y substitution can be simplified. z=arctan(1/y) tan(z)=1/y y=1/tan(z)=cot(z) z=arccot(y) Therefore, arctan(1/y)=arccot(y)

[@TheEternalVortex42]

How is that helpful though?

[@leif1075]

Can't younsolve WITHOUT the second substitution and just skip to integrationnby parts?

[@reubenmanzo2054]

@@leif1075 It looks cleaner, which 90% of the time, means the process will be cleaner as well.

[@leif1075]

@reubenmanzo2054 but you can still.solve if you skip the second substitution and go into integration by parts directly as i wouldhave done, can't you? Thanks for answering. Not sure how someone would think it would look cleaner though unless you've seen this type of problem before.

[@reubenmanzo2054]

@@leif1075 From experience, you're usually better of with a stand alone variable, than dealing with a reciprocal. This is especially the case when working with differentiation and integration because you won't have to worry about the chain rule.

[@gp-ht7ug]

Was it necessary to pass through a contour integral?

[@davode76166]

@maths_505 can you solve it by Feynman technic? For example with d/da arctan(a.(cosx)^2)/(cosx)^2)?

[@rainerzufall42]

It doesn't change anything of the argument, but how do you get i R in front of the integral? z = R e^(i \theta) => dz = i R e^(i \theta) d\theta [ = i z d\theta ]. Okay, I admit, that explains i R, but where's the missing e^(i \theta) gone? It is lost! But as I said, it doesn't change the argument for R -> \infinity...

[@rainerzufall42]

@The1RandomFool concluded this in a shorter comment... ;-)

[@davidcroft95]

Just a classic Micheal Penn mistake

[@rainerzufall42]

@@davidcroft95 I believe, he does it intentionally! He wants critical thinking from the viewers!

[@rainerzufall42]

Or it's just laziness! ;-)

[@Happy_Abe]

Why is there an iR before the integral at 13:45

[@inigovera-fajardousategui3246]

When differentiating the parametrization z=Re^itheta, you get dz=iRe^itheta and then you pull iR out of the integral

[@Happy_Abe]

@@inigovera-fajardousategui3246 why do we differentiate the parameterization? Isn’t there a missing e^itheta then in the integral?

[@SuperSilver316]

He might have dropped the term, but we do need the differential of our parametrization

[@Happy_Abe]

@@SuperSilver316 why’s that? I’m not familiar with the concept

[@neilgerace355]

​@Happy_Abe Yes but at 14:30 he shows the integral goes to 0 anyway.

[@monikaherath7505]

Question: how do people come up with or discover such crazy integrals? Do people just stumble upon it while solving crazy integrals? Or are there Ramanujan-like mathematicians out there who are able to just "see" things like this or "sense" that such integrals will have special results?

[@ianbennett2443]

im not super sure but i'd imagine a lot of the time people start with a result and work backwards. like for indefine integrals, it's easy to come up with a weird function, differentiate it, then put the derrivative under an integral, and then give that integral to someone. it also might be pattern recognition, where when people solve a *lot* of integrals, they start to notice that when you have "this thing" in the integrand, then the result will look like "this other thing", and then the process would be combining different "things" you think are interesting/funny into one intergral in a new way, and seeing if you can solve the intergral you just made.

[@MrMctastics]

The same question could be asked of all further mathematics

[@TomFarrell-p9z]

Mathematicians get to pick their integrals. Engineers and physicists have to solve (or not) the ones they are given! 🙂

[@КонстантинАлександрович-г5э]

@ianbennett2443 Is it much different to what Ramanujan did?

[@CM63_France]

One method is to perform a "zero's order integration" as Michael call them and explained us a while ago. You can start with a function as crazy as you want.

[@gilperalta4068]

In real life, from a layman's perspective, what hypothesis does the original equation try to show or represent?

[@leif1075]

Jesus Christ WHY IN GODS NAME WOULD ANYONE substitue t for 1/y thats random and out of nowbere..why not solve witbout thst substitution..like uae integration by larts at that point or substitue the whole sqrt 2t -t^2 ewuals y would make so much more sense..come on now.

[@d-nize]

0 is a Pole in This case. Improper Integrals are easier to handle. Thats why you substitute t=1/y.