The Glaisher Integral

We look at a nice view suggested integral.
Playlist: • Interesting Integrals
Please Subscribe: www.youtube.co...
Merch: teespring.com/...
Personal Website: www.michael-pen...
Randolph College Math: www.randolphcol...
Randolph College Math and Science on Facebook: / randolph.science
Research Gate profile: www.researchga...
Google Scholar profile: scholar.google...
If you are going to use an ad-blocker, considering using brave and tipping me BAT!
brave.com/sdp793
Buy textbooks here and help me out: amzn.to/31Bj9ye
Buy an amazon gift card and help me out: amzn.to/2PComAf
Books I like:
Sacred Mathematics: Japanese Temple Geometry: amzn.to/2ZIadH9
Electricity and Magnetism for Mathematicians: amzn.to/2H8ePzL
Abstract Algebra:
Judson(online): abstract.ups.edu/
Judson(print): amzn.to/2Xg92wD
Dummit and Foote: amzn.to/2zYOrok
Gallian: amzn.to/2zg4YEo
Artin: amzn.to/2LQ8l7C
Differential Forms:
Bachman: amzn.to/2z9wljH
Number Theory:
Crisman(online): math.gordon.edu...
Strayer: amzn.to/3bXwLah
Andrews: amzn.to/2zWlOZ0
Analysis:
Abbot: amzn.to/3cwYtuF
How to think about Analysis: amzn.to/2AIhwVm
Calculus:
OpenStax(online): openstax.org/s...
OpenStax Vol 1: amzn.to/2zlreN8
OpenStax Vol 2: amzn.to/2TtwoxH
OpenStax Vol 3: amzn.to/3bPJ3Bn
My Filming Equipment:
Camera: amzn.to/3kx2JzE
Lense: amzn.to/2PFxPXA
Audio Recorder: amzn.to/2XLzkaZ
Microphones: amzn.to/3fJED0T
Lights: amzn.to/2XHxRT0
White Chalk: amzn.to/3ipu3Oh
Color Chalk: amzn.to/2XL6eIJ

[@AFastidiousCuber]

The fact that you did all that work just to get a result of 0 is giving me complex analysis flashbacks.

[@johnunverzagt9387]

Cauchy-Goursat.....

[@RozarSmacco]

Well I sent the problem to prof. Penn and I got this out of the Edwards book and IT HAS the Goal integral = pi^2/8

[@shailynaagar5048]

@@johnunverzagt9387 yes

[@johannesh7610]

There is a mistake at 10:40 that results from the second tool being wrongly written down. The tool should have the fraction as its multiplicative inverse: x^2+1 = 2 [1+t²] / (1+t)². (as proven in the video). Before the numerator is 1/(1-t^2) (which already results from a wrong cancellation) and after the equals sign it is just (1-t^2).

[@tonyhaddad1394]

100% you re right

[@theloganator13]

As always, the correct answer doesn't mean no mistakes, just an even number of mistakes 😂

[@tonyhaddad1394]

@@theloganator13 "even number of mistake " good describe i loved it

[@s4623]

The correct lines are at 3:49 but the left side of the board was never adjusted for it.

[@WriteRightMathNation]

I saw a mistake at 10:21. I've not made it to 10:40 yet. Continuing now to watch...

[@UrkTheKeg]

The green box "fact" flipped the t polynomials from the proof ( 3:54 has both onscreen ). The blue cancelling at 9:25 is wrong with what's on the board but right if the green box error is fixed. The 2 errors cancelled :)

[@Catilu]

Actually, if you make an even number of mistakes, it's the same as if you didn't do any! :)

[@AKSatMusic]

@@Catilu if it's the exact same kind of error. If you make an even number of errors that don't have anything to do with each other you just have a bunch of errors

[@sergiokorochinsky49]

@@Catilu She cheated and I forgave her, that was my first error. Then I marry her, and that was the second. Of course she is still cheating. Even number of errors, twice as fucked. And eventually she will get half of my money. I wouldn't call that an error cancellation...

[@starshipx1282]

3:50. The derivation and 2nd result are slightly mismatched.

[@carlosgiovanardi8197]

as usual. no surprise.

[@starshipx1282]

@@carlosgiovanardi8197 what do u mean ‘as usual’?

[@leickrobinson5186]

@@starshipx1282 He’s referring to the fact that many, if not most, of Mike’s videos contain errors. It really makes me sad that he doesn’t take a little more care, because otherwise his videos would be awesome, but I fear all of the errors can only serve to confuse students who are new to this stuff and come to this channel to learn something. However, this video is far more egregious than most, because he derives the correct answer but then *ignores* it, and then later when he uses the incorrect result, he just blindly follows the script and performs some absolutely heinous false algebraic manipulations.

[@carlosgiovanardi8197]

@@starshipx1282 sometimes some errors appear, like here kzbin.info/www/bejne/iaLMhZWMntyaeck pity there are no edits.

[@starshipx1282]

@@leickrobinson5186 Right. I did watch some videos before and didn’t notice much. Thanks for the heads up. If he read more comments it would help him.

[@demenion3521]

somehow it feels like there should be a way to show that the value of the integral is 0 by a more straighforward substitution. especially it is easy to see that the integrand changes sign only at x=sqrt(2)-1, so that you could split the integral there and try to work towards two integrals over the same region of integration. but when trying that i got stuck with messy integrands that don't look alike at all :D

[@yoav613]

you can use x=tanu.then the integral will be arctan(3(tanu+1)/(1-tanu-tanu(tanu+1)))=arctan(3/((1-tanu)/(tanu+1)-tanu)) from 0 to pi/4. we have the formula tan(a+b)=(tana+tanb)/(1-tanatanb) so we get arctan(3/(tan(pi/4-u)-tanu)).using z=pi/4-u we get arctan(3/(tanz-tan(pi/4-z)). we know that arctan(-x)=-arctan(x) so we see that our integral I=-I which means it is0

[@pbj4184]

@@yoav613 Yes I have generalized this a bit more in my comment and shown the motivation behind the x=1-t/1+t sub which otherwise looks magical

[@sudhanshumishra6482]

Loved the number theory series michael

[@curtmcd]

The integrand is discontinuous at -1-sqrt(2) and -1+sqrt(2), and looks rather asymmetrical. The first discontinuity consists of a sign change near 0.4, yet the integral from 0 to 1 is properly zero. Moreover, integrating from -inf to +inf also gives zero, in three weird segments! Exploration here on desmos: www.desmos.com/calculator/8pksunhhml

[@RozarSmacco]

The Edwards book on Integral Calculus Vol.2 p.964 gives the goal integral’s value = pi^2/8

[@curtmcd]

@@RozarSmacco That is amazing! How did you find it tucked away there? Unfortunately, Edwards just left it as an exercise. But looking at the function graph, it seems pretty clear Edwards was wrong and Mr. Penn is right. Unless there is an impulse at the discontinuity.

[@abdallahal-dalleh6453]

Wait! 9:25 when substituting x^2 + 1 the fraction is in a denominator flips out, so (1 + t)^2 stays in the integral's denominator and does not cancel with the (1 + t)^2 of the dx substitutions.

[@Catilu]

He did that in order to correct his previous mistake, in which he had flipped the fraction corresponding to x²+1

[@abdallahal-dalleh6453]

@@Catilu Just saw it at 3:52 . Thanks!

[@thanasleraj9596]

I think you got the second fact wrong

[@ikarienator]

The cancellation at 9:25 is wrong.

[@lisandro73]

Dr. Penn, I think you have a mistake in the second formula, 3:55

[@bobbobson6867]

The curve is discontinuous at x=sqrt(2)-1. Does this matter?

[@xCorvus7x]

So the lesson is that every integral for which there exists a substitution, i. e. transformation to a different vector space, that turns the integral into itself but negative is equal to zero?

[@thephysicistcuber175]

9:28 excuse me?

[@amirb715]

he messed up big with the 1+x^2 but in the end the answer is right.

[@tonyhaddad1394]

Mst Micheal youre mistake is intentinally , you like to see some comment like my comment 😉

[@alejandrojimenez108]

Omg can everyone shut up about the mistake, like just stop saying the same crap, we all saw, you aren’t smart because you point it out, please stop.

[@pbj4184]

Ok.

[@TrainingCuber]

Hi

[@leickrobinson5186]

It’s regrettable that many, if not most, of Mike’s videos contain errors. However, this video is more egregious than most, because he derives the correct answer (3:52) but then *ignores* it, and then later when he uses the incorrect result, he just blindly follows the script and performs some absolutely heinous false algebraic manipulations (9:26, 10:19) to force the correct answer. It really makes me sad that Mike doesn’t take a little more care, because otherwise his videos are awesome, but I fear all of the errors can only serve to confuse students who are new to this stuff and come to this channel to learn something. :-(

[@rwjoyner]

Mr. Robinson - I do hope you manage to make a full recovery from what appears to be a deep depressive episode.

[@pbj4184]

I hope you understand that he doesn't solve these questions live but instead solves them the pen-and-paper way off-camera. I think you're being a conspiracy peddler here by suggesting he's just copying answers off of somewhere else. When he does something like that, he mentions it clearly in the beginning (like in the (-1)^n/(2n+1)^3 sum video) He already knows the solution due to having solved this before which is why he usually writes the correct thing at the end even after making mistakes

[@wesleydeng71]

This one is a simple oversight, not a logical mistake. I'd give him some slack since he produces these videos at such a high pace that it is bound to have some errors. But yes, I agree that care should be taken.

[@pbj4184]

@@琥珀-u3o Saying Michael blindly follows a script _is_ saying that he copies his answers off of somewhere else instead of actually knowing the solution

[@hybmnzz2658]

@@pbj4184 sorry for my rudeness.

[@TedHopp]

It would have been a good video if not for the second "tool" having numerator and denominator reversed. Then the result is saved by the ridiculously faulty cancellation of fractions at around 10:00 which just happens to exactly compensate for the earlier error. Definitely not up to your usual fine standards.

[@CTJ2619]

en.wikipedia.org/wiki/James_Whitbread_Lee_Glaisher

[@sayantanbiswas6038]

I think, it should be In the hint: x²+1 = 2(1+t²)/(1+t)²

[@orenfivel6247]

nice cancelation in 9:30 and correction in 10:20

[@sayantanbiswas6038]

@@orenfivel6247 nicely done

[@Metal_dead]

Oh. thank you explained where the real error is. I thought I went crazy :)

[@Fun_maths]

Thanks for the correction, I thought something went terribly wrong

[@minhphatpham7263]

Yeah, the same to me @.@

[@pbj4184]

Since I don't think anyone has talked about the origins of the sub here, I will. Let the original integral be I=Int(0,1) f(x)/1+x^2 dx. When we sub x=tan u, the bounds (0,1) change to (0,π/4) and the 1+x^2 of the denominator becomes sec^2(u) which cancels the extra sec^2(u) created by the sub. So we're left with I= Int(0,π/4) f(tan(u)) du Now we apply the famous v=b+a-u sub which transforms the integral again to Int(0,π/4) f(tan(π/4-v) dv. We're now ready to add the secret sauce which is tan(π/4-v)=1-tan(v)/1+tan(v) So I=Int(0,π/4) f(1-tan(v)/1+tan(v)) dv Now we finally resub y=tan(v) to end this neatly I=Int(0,1) f(1-y/1+y) dy Therefore in general Int(0,1) f(t)/1+t^2 dt = Int(0,1) f(1-t/1+t)/1+t^2 dt

[@pbj4184]

At a glance/ Essence: The two essential ingredients required for this sub are the (0,1) bound and the 1+t^2 in the denominator. The (0,1) sub is transformed into (0,π/4) which allows us to use the tan(π/4-u) formula and the 1+t^2 in the denominator cancels the extra sec^2(u) which allows us to have only tan(u)s in the integral. These two are essential in the condensed form sub too (obviously) but for slightly different reasons (atleast on the surface level) Note that this expanded form and the condensed form are both equally correct and this is not exactly the "derivation" of the sub but more so, the motivation behind it

[@sinecurve9999]

9:30 (1+t)^2*(1+t)^2 != 1... An Error!😣

[@skylardeslypere9909]

It's because the second fact was wrong. The fraction in the big denominator should be flipped, which results in the same answer. He just made 2 mistakes that 'canceled out'

[@tomatrix7525]

@@skylardeslypere9909 yea, I noticed his first error but didn’t notice that one with the complex fraction. I was wondering how he was ending up with the right answer still. I guess two wrongs really do make a right afterall....

[@RozarSmacco]

As to the origin of this integral it first appears in Glaisher’s 1877 paper “Transformations of some definite integrals” in Vol. VI of Messenger of Mathematics. The paper uses the integral in question to further illustrate an infinite integral theorem originally found by Liouville. Glaisher transforms Liouville’s result into this Integral from 0 to inf of f(1+x)*dx/(1+x^2) == Integral from -1 to 1 of f(2/(1+x))*dx/(1+x^2)

[@reshmikuntichandra4535]

Hello Mr. Penn, I want to let you know about a mistake u did in the computation of the integral. At 3:53, u deduce that x^2 + 1 = 2(1+t^2)/(1+t)^2 ..... But in the sidenote the formula you wrote was wrong.. It was 2(1+t)^2/(1+t^2).... So your second formula in the sidenote was wrong. You go on to incorporate this formula into the computation of the integral at 9:03... But, as told, that formula was wrong, and thus the entire computation of the integral is incorrect. A petty mistake, but it's ok, we all make mistakes!Nevertheless, keep on the great work!

[@willsawyerrrr]

You are correct in saying that he made a silly mistake by writing and substituting the second fact incorrectly. However, you should notice that his simplifications after the subsitution do not rely on the way he has written it on the board, but instead on how he knows the solution should continue. There are two mistakes regarding the complex fraction created by this substitution which "cancel each other out", for want of a better word. I recommend you read some other top-level comments who say a similar thing to what I have. Also, maybe don't be so quick to conclude that a minor mistake renders an entire solution null and void, particularly during videos on this channel. Mr Penn regularly makes silly mistakes in the form of writing something different to what he says, but you should not let that take away from the fact that he is a brilliant mathematician.

[@reshmikuntichandra4535]

@@willsawyerrrr Hey dude, I never meant to disrespect him, infact I wrote this comment in the most polite way possible. He is indeed a great problem solver, and the method of the solution is correct as well, but that one mistake matters largely.... Remember, At the end of the day, It's Mathematics... You get it either correct or incorrect. Not a hate comment in any way.

[@reshmikuntichandra4535]

Also, I love how he simply deals with the mistake instead of freaking out and corrects it without anyone noticing at 9:27

[@yoav613]

you can use x=tanu.then the integral will be arctan(3(tanu+1)/(1-tanu-tanu(tanu+1)))=arctan(3/((1-tanu)/(tanu+1)-tanu)) from 0 to pi/4. we have the formula tan(a+b)=(tana+tanb)/(1-tanatanb) so we get arctan(3/(tan(pi/4-u)-tanu)).using z=pi/4-u we get arctan(3/(tanz-tan(pi/4-z)). we know that arctan(-x)=-arctan(x) so we see that our integral I=-I which means it is0

[@goodplacetostop2973]

12:54

[@pauloedmachado9137]

Thanks

[@cepatwaras]

thanks!

[@goodplacetostop2973]

@@pauloedmachado9137 My pleasure 👍

[@goodplacetostop2973]

@@cepatwaras No problem 👍

[@alainrogez8485]

At first glance, I was crying when I saw this integral.

[@RozarSmacco]

This integral appears on page 964 of Joseph Edwards’ A Treatise on the Integral Calculus Vol. 2 and it’s value is given as pi^2/8. John Whitbread Lee Glaisher, FRS FRSE FRAS ( he was a fellow of the Royal Society, the Royal Society of Edinburgh and Royal Society of Astronomical Sciences) originally evaluated it and he actually anticipated Ramanujan in many results. When Glaisher evaluated an integral he was usually right but even Desmos says the value is Zero. What Gives Prof. Penn??

[@mathunt1130]

These types of integrals are done with t=tan(x/2).

[@Czeckie]

yep, and this is basically what happens here. I'm a bit rusty, so there might be an easier way to see it. You start with a substitution x=tan^2(a/2) ..and then you chain it with substitution t=cos(a), this yields x=(1-t)/(1+t)

[@peterburbery2341]

Does this integral involve Weierstrass substitution?

[@wirajithuthpala6355]

too many mistakes = no mistakes 😍😅

[@yoav613]

you can use x= tanu.then you left with integral from 0 to pi/4 with arctan(3(tanu+1)/(1-tanu^2-2tanu)) =arctan(3(tanu+1)/(1-tanu-tanu(tanu+1)))=arctan(3/((1-tanu)/(1+tanu)-tanu)).we have the formula tan(a+b)=(tana +tanb)/(1-tanatanb)so we get integral of arctan(3/(tan(pi/4-u)-tanu))from 0 to pi/4.we use z=pi/4 -u and by using the identity arctan(-x)=-arctan(x) we get integral from 0 to pi/4 of arctan(3/(tanz- tan(pi/4-z)))= - arctan(3/(tan(pi/4-z)-tanz)) .now we can see that that our integral I=-I which means it is 0

[@yoav613]

z=pi/4-u and the integral is then arctan(3/(tanz-tan(pi/4-z)))=-arctan(3/(tan(pi/4-z)-tanz))

[@djsmeguk]

green box is upside down, but you end up cancelling it as it should have been, rather than as it was written..

[@moltorne]

I actually like the content a lot but it is simply way too many commercials on these videos.

[@jackrubin6303]

Dear Michael, please correct your video.

[@chaparral82]

I know a similar Integral Trick, however with an easier looking integral: Integrate: 1/x * sin (x - 1/x) dx from 1/1000 to 1000 ;-) Hint: could also be from 1/a to a

[@pbj4184]

Haha true but it's almost obvious here, especially with the bounds and the knowledge that the sub the bounds insinuate gives dx= -dy/y^2

[@chaparral82]

@@pbj4184 just substitute u=1/x and you will get dx/x = - du/u

[@pbj4184]

@@chaparral82 That's certainly a more aesthetic way of writing it

[@djvalentedochp]

that's cool

[@leponpon6935]

cool!

[@swingardium706]

10:22 I don't understand how this logic works. To me it appears that the assumption that the original integral is equal to the final integral implies that x=t, but we know from the definition of t that this isn't the case. Can someone with a better grasp on this sort of thing explain this?

[@violintegral]

Well clearly x≠t because the original integral is equal to the negative integral. And as he said, t is just a dummy variable in this case. A change of variable doesn't change the value of a definite integral.

[@ethanjensen7967]

Isn't glaisher the guy who found a bijective proof that the number of odd partitions equals the number of distinct partitions?

[@GaneshKumar-vh6ts]

Wow superbbb sir I thought the answer would be very difficult but it ends with very simple steps. But I have an doubt that what is that arc.

[@WriteRightMathNation]

Mistake at 10:21: should be -\int_0^1[(1+t^2)arctan([3+3t]/[1-2t-2t^2])]dt

[@user-A168]

good

[@brettaspivey]

boy, a lot of errors, you really needed to check the video

[@tomatrix7525]

Such a nice one. Integration is really epic...

[@douglasmagowan2709]

7:30 you are missing the exponent on the t^2 term.

[@sudhanshumishra6482]

I think there's a typo in the x^2 + 1 part. Please check.

[@EnteiFire4]

That's what was weird! I couldn't understand at the end why the (1+t)^2 cancelled.

[@JankkPL]

There is, but then he flipped the fraction again when substituting so the overall answer is correct.

[@sudhanshumishra6482]

@@JankkPL yeah surprisingly becoz he flipped twice that mistake did not progress 😂

[@sudhanshumishra6482]

Honestly this brought back some JEE memories when two mistakes somehow cancel each other and you get the correct answer

[@kasun1998lk]

Mistake at canceling (1+t)2

[@monikaherath7505]

Does anyone have a graph of what the function would look like? Thanks!

[@karolakkolo123]

I'm on the phone, so I can't really provide a link, but check on Desmos. The function is discontinuous on [0,1] which makes me question whether what he did in the video is valid

[@jeffreycloete852]

Hi Prof Penn..thanks for another great video.. i appreciate all the competition maths..but when are u going to do another algebra stream..like Lie Algebras. .or questions and answers on Vertex Operator Algebras. .

[@Ty-sc2ri]

Integrals aren’t competition maths

[@jeffreycloete852]

@@Ty-sc2ri ..Thanks for the comment. .you are CORRECT. .was still thinking of the previous Putnam question. .just looking forward to more Abstract Algebra

[@red0guy]

What I like the most about this channel are the mistakes .... They are fun.

[@Superman37891]

I think Professor Bruce H. Edwards had an antiderivative for sin(x^2) dx in his Great Courses Plus Calculus 1 episode about approximation techniques. Can someone with an active membership verify this? Thanks

[@vikashdhukat4249]

Please solve for x and y y x^3+x y^3=(10(x+y)^3)/9 And y x^4+xy^4=(2(x+y)^3)/3

[@i.am.jihoonk]

x=0,y=0

[@antonryzhov]

Oh, wait! Should denominator of denominator go to the numerator? If you divide arctan(..) to (1+t)^2 / (1+t^2) and then cancel (1+t)^2, it is still 1/(1+t^2), not 1+t^2.

[@abdallahal-dalleh6453]

Using fraction division, the denominator is flipped and moves up so the (1 + t)^2 stays in the denominator of the integral hence not cancelling with the (1+ t)^2 of the dx.