The solution is an important constant.

[@danv8718]

Michael, you should write a "Book of Gnarly Integrals" with tons of examples like this full of amazing tricks. I'd definitely buy it (even pre-order it!)

[@orphanfrequencyyeah1742]

Try Cornel Valean's "Almost Impossible Integrals", or the slightly less exotic "Inside interesting integrals" by Paul Nahin. If you want an old-times treasure trove, try the two volumes of Joseph Edwards' "Treatise on the Integral Calculus" For anything related to mathematical constants, nothing can beat Steven Finch's "Mathematical Constants" volumes 1 and 2. These are not exactly easy books btw.

[@danv8718]

@@orphanfrequencyyeah1742 Thanks for the references! I've got a copy of Nahin and Edwards but I'd never heard of Finch's Mathematical Constants.

[@TypoKnig]

When you convert 1/(1-z) to an infinite sum, is there a hidden assumption that abs(z) < 1, so the infinite sum converges?

[@norbi275275]

There is no hidden assumption. 0 < z < e^2y but y < 0 , because of the bounds of integration so e^2y < 1 therefore z is guaranteed to be below 1

[@TypoKnig]

@@norbi275275 Thanks! That's what I get for posting a comment before I've had caffeine.

[@8jhjhjh]

@@TypoKnig haha don’t worry bro i had the same thought

[@forcelifeforce]

@@norbi275275 You mean e^(2y)

[@josephmartos]

@@norbi275275 thanks for this part, wasnt so dificult but still sometimes we cant notice it very quickly

[@PelosiStockPortfolio]

Follow along for 7 minutes, then fall asleep. When you wake up the answer will be on the chalk board. Its good to know this trick still works 20 years after taking calculus

[@Tentin.Quarantino]

And that’s a good place to wake up.

[@yoav613]

Short solution: u=sinx then dx=du/cosx and we get integral from 0 to 1 of 0.5ln(u)ln(1-u^2)/u du then by taylor series we know ln(1-u^2) = the sum -1( 1/n u^(2n)) n from 1 to inf and by simple ibtegration by parts we get 1/8 the sum from n=1 to inf of 1/(n^3)

[@GGBOYZ583]

Doesn’t work, the denominator in the question is tan(x). 0.5ln(u)ln(1-u^2) gives ln(sinx)ln(cosx) like you suggest, but the question also divides this by tanx, not sinx like you did here

[@albertoferis8250]

@@GGBOYZ583 well 1/tanx = cosx/sinx, then let u=sinx and du=cosxdx ---> dx/tanx=du/u soo, what he said does work

[@yoav613]

@@GGBOYZ583 it works well and i solved it in my paper very shortly.try to solve it too in paper before you saying it dose not work

[@GGBOYZ583]

Oh crap you’re right, my bad. I completely forgot about the dx part lol

[@sushildevkota350]

Its the correct way but not for other integrals. I also used the expansion of logarithm and applying the integral by parts of variable u.e^u,u.e^2u......., we get the exact answer. But this way is aliter method i.e. analytical method, but the way michael penn did is the actual method and quite provide the better way of solution for other types of complex integral. So converting or adding subtracting summation to and from integral respectively, very sum can be expressed in very ideal and unique way.

[@OleJoe]

I would have never figured that out on my own. Great job. Going to add this integral to my notebook.

[@goodplacetostop2973]

13:37 Sadly, the answer is not the Jen’s Constant 😔

[@all462]

Nice

[@ianrobinson8518]

Interestingly Zeta(3)/8 is the sum of the even terms of the zeta function.

[@vivientane3060]

Hello can you explain please ?

[@darkmask4767]

interesting to see an integral involving Apéry's constant

[@Reuleaux99]

You spoiled me >:(

[@damascus21]

Thanks for spoiling it EDDIE

[@gregsarnecki7581]

I assume you've seen the one where it's a triple integral over the unit cube...beautifully symmetric.

[@haibai1766]

That's the next recommended video XD

[@emanuellandeholm5657]

My first thought was an ln(sinusoid) subst and my second thought, after that ingenious double integral transformation that I did not anticipate, was that y/(1 - z) looks like a geometric series. I'm ever so slowly learning from this channel!

[@Kirill.Dubovitskii]

Using series expansion for log would do the job directly. No need to introduce integral over z. But cool way to find Taylor series of log

[@sathvik1007]

Yes, totally agree since none of them attend 1 or -1 due to its neighbour.

[@user-nb6zu3rk4f]

When I read about Apery’s constant in the comments I immediately thought about Taylor because the series expansion of log is pretty simple and there won’t be any mess with integration.

[@PowersOfTwo]

Damn. That was some black magic! 👏🏼👏🏼

[@nasim09021975]

The substituted integral could've been solved much sooner if ln(1-e^(2y)) was written as an infinite series (since 0< e^(2y)

[@CTJ2619]

I love seeing “gnarly” integrals

[@esorminihaz8269]

I thought this was gonna be one of those symmetric u sub integral lol

[@anshumanagrawal346]

Same, I did at first too

[@Spaokobb]

Gnarly Sine Waves.

[@michaelempeigne3519]

zeta ( 3 ) = apery's constant

[@kevinmartin7760]

Hand-waving alert at 12:00. The limit for the lower bound is of the form 0 times infinity and so L'Hopital's rule should be applied (or at least mentioned) to decide what the final result is.

[@bluestrawberry679]

I disagree I think it's well known enough, that in a limit like that, the exponential Term is the only thing which really matters in deciding the limit

[@jacksonstarky8288]

I found this channel early in the new year thanks to KZbin finally figuring out that math videos are interesting. 🙂 Every time I see a video title like this one, I hope for something whose solution is the Euler-Mascheroni constant... but this one is just as interesting, and where the two meet is particularly interesting; I've been looking for a visual explanation of the role of gamma in the zeta function for a few years now, ever since finding 3Blue1Brown's original video on the zeta function... which I think is five years old now(!).

[@jerrysstories711]

9:13 You can do that integral more easily by converting the integrand to (1/2)*d(e^(2my))/dm, then moving the d/dm outside the integral, then solving the simpler integral. I love that method so much I use it sometimes when it wasn't even necessary. :-)

[@theloganator13]

Or a quick substitution to make it into Gamma(2)

[@monikaherath7505]

@@theloganator13 woah what substitution is that?!? explain pleaseQ!

[@monikaherath7505]

could you explain why that's permitted? thanks EDIT: It's probably beyond the scope of a comment, so does that method have a name? I can see why it is probably true, as it is applying it to independent variables, but I was wondering if there was a rigorous proof or a name for it. Thanks!

[@jerrysstories711]

@@monikaherath7505 Sure. It's a little hard to type integrals here, so I hope you can read this. The key thing is that there's an integral over y of a derivative wrt m, but you can convert it to a derivative wrt m of an integral over y. Integral(-Inf,0, y * e^(2my), dy) = Integral(-Inf,0, 1/2 * d(e^2my)/dm, dy) = 1/2 * d(Integral(-Inf,0, e^2my, dy))/dm = 1/2 * d(1/2m)/dm = -1/(4m^2)

[@theloganator13]

​@@monikaherath7505 We're looking at ∫_-∞^0 y exp(2my) dy. Make the substitution 2my = -u, so dy = -du/2m, and the integral becomes -1/4m^2 ∫_0^∞ u exp(-u) dy = -1/4m^2 Gamma(2) = -1/4m^2 Same result as in the video, but without the pesky integration by parts. Just need to know the special function Gamma(n) = (n-1)! for natural n.

[@davannaleah]

All the steps you did were quite straight forward but led to an amazing result!

[@NeverTalkToCops1]

I don't think Elon Musk solved hundreds of integration puzzles (but I bet he could explain integration to a five year old). Neither should anyone. Only solve integrals that matter in your field. There is an infinity of integral puzzles, so what?

[@user-wu8yq1rb9t]

One the greatest video, you recently released. I just enjoyed and learned, thank you so much teacher.

[@TheOnlyAndreySotnikov]

Wolfram Mathematica evaluates this integral after several seconds of thinking.

[@MTrivial]

"Sometimes is a cheat" ? Why not prove -1/2=+infinity with some magic sum manipulation ?

[@CM63_France]

Hi, For fun: 0:00 : good time to shave, 4:14 : "great", 13:24 : "and so on and so forth".

[@KarthikMalisetty]

Couldnt we have expanded the logarithm directly instead of writing it as an integral, expanding and integrating it back. basically same thing

[@usptact]

I would've never managed to solve by myself! Glad that I was able to follow!

[@raffellabaldassare9718]

Is the most.important ask always . ... why ???????????????????

[@egillandersson1780]

The Apéry's constant ! Great video ! I have revised some integration trick and learned others. Thank you !

[@stephanet8294]

I dont understand the beginning int ( -1/ 1-z) 0 ; e2y. How do you that ?

[@holyshit922]

4:23 There would have been a lot easier if he had used integration by parts

[@reubenmanzo2054]

4:35 Why not just use integration by parts?

[@alexestefan7521]

Zeta(3) is also called apery's constant

[@ericsmith1801]

Where did the expression -1/(1-z) come from. It is the geometric series which can be expressed as an infinite sum...but how did it end up here in the equations ?

[@tgx3529]

I used in this example more substitutios than Michael, substitution for tgx=a for integral 1/2* ln(cos^2 x)*1/2 ln(1- cos^2 x)/ tg x. Then I got integral[ (1/4) * ln(1/(1+a^2))*ln(1-1/(1+a^2))]/a(1+a^2)) da where a (0; infinity). After substitution v=1/(1+a^2) I got integral [ (-1/8)* ln v*ln(1-v)]/(v-1) dv where v in (0;1) ( we can see zeta(3)) Finally substitution y=ln(1-v) there Is integral (-1/8)*y*ln(1-exp y)dy where y in (0;- Infinity) For ln(1-exp y) I used series (-exp(ny)/n) where n=1....., I think, there Is uniform convergention by M test f_n(y)=-y*exp(ny)/n, supremum Is for y=-1/n. If I change suma and integral I have suma[(1/8)* integral(exp ( ny)/n^2) dy where y in (-infinity;0) n=1......There Is the same result As Michael.

[@ytang3]

Nice shout out to BPRP!

[@r6e686]

Couldn't you just turn ln(1-e^(2y)) into a summation? Since -infinity

[@함함-b9t]

I always click your youtube always expecting to hear"here we gonna look into a nice problem~"or other forms which end with "a problem~" ㅋㅋㅋㅋ

[@alnitaka]

I see you put up a problem involving Zeta(3) = Apery's constant = 1.2021 to 4 decimal places, today, in the year 2021. Sometimes I call this year the Apery Year.

[@schrodingerbracat2927]

zeta(3) = Apéry's constant en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant

[@angeloluisrocattojunior3425]

Now I got why the Numberphile video about Apéry's constant is in the related videos.

[@divyakumar8147]

thanks

[@laurentveysseire7705]

Interesting, using another method, I got the following expression, with successive derivatives of Euler's Gamma function at1: 1/8*(3/2*Γ'(1)*Γ''(1)-Γ'(1)^3-1/2*Γ'''(1)). Your expression looks nicer.

[@davidsweeney111]

Some people pronounce ln as lon is this correct or should I just say natural log?

[@noelani976]

Of course I do watch *blackpenredpen* and the D-I method works just perfectly fine.

[@takumimatsuzawa1774]

This was so much fun! Could you make a video about when you could swap the summation and the integral?

[@holyshit922]

I would be able to calculate this integral myself I would probably try to get series Answer can be presented in terms of Zeta function

[@pavlopanasiuk7297]

One may expand logarithm instead, he will get same summation over same integral, but way shorter

[@pacojacomemaura2129]

Why can we exchange the order of summation in 6:35 ?

[@kafianan6586]

Absolute convergence

[@aakksshhaayy]

pretty tough integral... nice summation trick

[@koenth2359]

Wow, very interesting and (to me) a somewhat surprising result!

[@cicik57]

hey in the middle i started to hope that you will calculate Zeta(3) using this trick...

[@synaestheziac]

@5:27 what was the point of “slamming” the factor of y into the integral dz?

[@mathunt1130]

As he's done series in the final instance, then he should have done the expansion after the y substitution. This would have shortened the calculation considerably.

[@The1RandomFool]

I tried this before watching the video, and the substitution I took was u = sin x. I used cos^2 x = 1 - sin^2 x in the other logarithm, then expanded it out after substitution using logarithm rules. I then saw it was in a good form for integration by parts using the polylogarithm function and got the same result. I didn't consider the substitution in the video, though.

[@MultiCarlio]

How about solving some hard limits? :-)

[@motekkz9997]

Try integral e^x Tan(x) dx plz ..

[@SplashEazy]

Uhhh so I'm talking with a buddy and I'm confused about the part where he pulls the sum over m out of the integrand. Is that allowed in general or is this some special case where that is allowed?

[@aakksshhaayy]

It is allowed in general... think about expanding the summation out, the terms with m are constant with respect to each integral so you can pull them out then put it back in sigma form again

[@राजनगोंगल]

👏👏👏

[@camtalkdaily5541]

Thank you !😍

[@giorgibliadze1151]

DI)

[@robinjones1]

Nice

[@berzerksharma]

How do you know that 1/1-z can be expanded Without assuming z is between (-1,1)

[@m4riel]

remember how y is bounded between -infinity and 0? that means e^2y is between 0 and 1, and z is bounded between 0 and e^2y

[@berzerksharma]

@@m4riel well thanks, I never noticed it.

[@forcelifeforce]

OP -- Grouping symbols are needed: 1/(1 - z)

[@forcelifeforce]

@@m4riel Grouping symbols are needed: e^(2y)

[@hamdiel-sissi7760]

Pretty stupid problem with loose sloppy solution!!