This is my favorite kind of monster integral. Everything you need to solve it would be things taught in a high school AP Calc class, or taught in a college freshman semester of Calc 2. But this integral is far more intricate than anything you would see in a textbook and needs to be processed through multiple layers of manipulation, including non-obvious substitutions you would not likely see in an average elementary Calc class.
Just saw this video, and I was totally blown away by how easy this integral is once I let x^2 = tan(t), and this whole thing turned into a nice beta function (√2/16)*int_(0 to pi/2) 2*(sin(2t))^(1/2) (cos(2t))^(-1/2) d(2t) = (√2/16)*B(3/4, 1/4) = pi/8. It only took like 2 minutes.
@@MichaelPennMath You're doing great, Stephanie! I love seeing Dr. Penn get a team to further the channel, it's hard to believe it used to be just him and his chalkboard. You guys have been rocking it lately!
You can also do this as a contour integral using a quarter circle. Taking care to account for a sign change due to the branch cuts, we get the contour integral, as the radius of the circle -> oo and the paths around the branch cuts approach the branch cuts, has value 2*I + 2i*I, where I is the value of our original integral. But by the Cauchy residue theorem and noting the residue at 1/sqrt(2)+i/sqrt(2) is 1/8-i/8, this is 2*pi*i*(1/8-i/8) = pi/4 + i*pi/4. It follows I = pi/8.
It turns out you can do this with a single substitution: u=sqrt(1-x^4)/x. (Hard to think of or motivate, but then so were the substitutions that Michael used!) It is then routine to check that the original integral is equal to the integral of 1/(u^4+4) from u=0 to infinity, and it is a challenging but classic partial fractions problems to show this equals pi/8.
u can solve it also with the substitution: u=sqrt(tanx) , this will lead to the integral of sqrt(tanx) from 0 to pi/2 which we already know how to solve this integral.
Let f(x)=x^2/((1+x^4)*sqrt(1-x^4)) and Int(f(x),x=0..1)=I. It is then not difficult to check that the complex valued integral Int(f(x),x=-infinity+i0..infinity+i0)=2I. In fact you can close the complex integral in the upper half plane since f(R)*R goes to zero in absolute terms as R->infinity. The resulting closed contour integral encloses a pole at x=e^{i*Pi/4} and at x=e^{i*3Pi/4}. Furthermore it encloses a cut from x=i to x=i*infinity. Picking up the residues gives a value of Pi/2 and the closed contour can be written as encircling solely the line (i,i*infinity) counterclockwise plus the Pi/2. Rotating this contour clockwise by 90° with the substitution x=i*u (picking up the phases for the square-root und substituting u=1/t), it is found 2I=Pi/2-2I which gives I=Pi/8.
the crazy thing is this solution means that this indefinite integral is actually doable, it becomes 1/16*log(4x^2/(1-x^4)-4sqrt(x^2/(1-x^4))+2)-1/16*log(4x^2/(1-x^4)+4sqrt(x^2/(1-x^4))+2)-1/8*arctan(1-2sqrt(x^2/(1-x^4)))+1/8*arctan(1+2sqrt(x^2/(1-x^4))) EDIT: it can also be expressed more compactly as 1/8*(artanh(2x*sqrt(1-x^4)/(x^4-2x^2-1))-arctan(2x*sqrt(1-x^4)/(x^4+2x^2-1))), though getting the right value for the definite integral from this involves choosing the correct branches of the multi-valued inverse trig/hyper functions
Yes. But if you watch it again, then you get another 1/8th of a pie and so on. So if you watch it 8 times, you'll have a full pie. -Stephanie MP Editor
Michael uses the Sophie Germain's identity, but he makes a little mistake when he writes it on the blackboard. Anyway, thank you very much for such an awesome exercise.
An amazing elementary method of evaluation! I would have rewritten it as a hypergeometric function and simplified using its special relations, but that feels more like cheating.
I don't think that you are cheating by using hypergeometric functions; in fact it could be awesome do this "elementary" integral using more sophisticated methods.
He explains the first one: take a part and see what the derivative is, then construct a 1 out of it by multiplying it with it's inverse. Half of what you have multiplied in automatically substitutes to du. Nice method - I wonder if that strategy has an unofficial name. XD The second one is more straight-forward, you get a feeling for these after solving a couple similar ones. :)
if it was truly a gnarly integral, the video would be 35 minutes long and you'd already have the boards filled with pre-requisite identities that you have to prove first... plus there would be a smattering of phi's, tree (3) exponents, and congruences equal to 5 (mod pi)
Call me pedantic, but maybe you could have said a few words about the singularity in the integrand at x=1? I played with this briefly in my Computer Algebra System (Scientific Workplace, now sadly defunct), but not enough. It appears that if you expand the integral as an infinite series you may be getting one of those weird counterexamples to the necessity of Leibnitz' Alternating Series Test. (Don't bet the farm, though, I was not able to work out the pattern in the time that I invested. I suspect things would become clearer by changing the variable to (1 - x). ) Really appreciate your work!
This was the perfect closure to my work week - calculus just has an unmatched natural beauty to me.
Рік тому+1
isnt there a mistake? he workt out that 1-x²= 2/(u+1) but actually he needed x²-1 wich ist -2/(u+1) for the hole process it doesnt matter because it gets squared so the negative cancel out 14:00
I think in the main part (right side of the board) it was supposed to be (1-x^2)^2, which would be more consistent with the rest of it. But either way it clearly doesn't affect the result.
While watching video, I've found another but similar way to solve the integral. I = ∫[0, 1] (x^2 dx /{(1+x^4)sqrt(1-x^4)}) First, divide numerator & denominator by x^2 → ∫[0, 1] (1 /{(1/x^2 + x^2)sqrt(1/x^2 - x^2)})dx/x Next, let "x^2=y, dx/x=dy/2y" to reduce order of x → ∫[0, 1] (1 /{(1/y + y)sqrt(1/y - y)})dy/2y = ∫[0, 1] ((1/y^2 +1)dy /{2(1/y + y)^2 sqrt(1/y - y)}) Let "1/y - y=z, -(1/y^2 +1)dy=dz" so that "(1/y + y)^2=z^2+4" → ∫[∞, 0] (-dz /{2(z^2+4) sqrt(z)}) = ∫[0, ∞] (dz /{2(z^2+4) sqrt(z)}) Then let "z=u^2, dz/(2sqrt(z))=du" to rationalize the integrand → ∫[0, ∞] (du /(u^4+4)) By evaluating the last term, we get I=π/8.
Michael, I have to say that this video was disappointing in that it struck me as an exercise in vanity mathematics. You present what is a difficult integral and you do a lot of transformations in order to get to the end result but you never make a comment about convergence. There is no hint of any insight. How many of your viewers noticed that you transformed something that blew up at 1 to something that goes to zero at infinity (10:07)? Mathematica does this integral in 0.710641 seconds and gets Pi/8. That in itself interesting because of the way Mathematica does the processing for this type of integral basically using essentially the stuff the blind Jim Slagle did in his PhD thesis at MIT in 1961 under Marvin Minsky. Mathematica also uses the Risch and Ritt algorithms which build on the 18th-19th century works of Laplace, Louiville and Hardy to name but a few. The symbolic processors can do transformations of indefinite integrals and then perform the limits - Slagle’s system allowed for that and contains some specific examples. I think your audience would benefit from a follow up video explaining the convergence issue.
Gotta say, really don’t like the newer video titles and thumbnails. Makes each video feel more like clickbait, so I don’t watch them as much. I prefer to see thumbnails relating much more directly to the problem addressed in the video
where did that factorization of t^4 + 4 come from? is there some more general principle involved that enables one to factor stuff of the form a^4 + b^2 (a^2)^2 + b^2 = (a^2 + bi)(a^2 - bi).. no good a^n + b^n, for odd n, factors to (b)(a^(n-1) - a^(n-2)b +..-..+.. - ab^(n-2) + b^(n-1)) which is why if this were something like t^6 + 8, i could see rewriting it as (t^2)^3 + 2^3 and factoring it that way no way to write t^4 + 4 as a sum of odd powers (t^2)^2 + 2^2 = (t^2 + 2i)(t^2 - 2i).... is no good. completing the square somehow and getting (t^2 + sqrt(4t^2))(t^2 - sqrt(4t^2)) doesn't foil out properly.. without +2 to each binomial so i guess a^4 + b^2 = (a^2 + b + sqrt(4a^2))(a^2 + b - sqrt(4a^2)) which gives the (t^2 + 2t + 2)(t^2 + 2t - 2) that was used in the video but that would mean you can always factor a sum of squares this way so a^2 + b^2 is actually (a + b + sqrt(2ab))(a + b - sqrt(2ab)) = (a + bi)(a - bi) so only useful if 2ab is a square so we can do this for any a^(2m) + b^(2n) where 2(a^m)(b^n) is a perfect square which seems to work best when a or b is 2. but will probably work in a lot of other cases. is there a name for this?
That's a spooky place to stop.
nice.
Well done
Underrated comment. XD
I thought it was a very satisfying result.
This is my favorite kind of monster integral. Everything you need to solve it would be things taught in a high school AP Calc class, or taught in a college freshman semester of Calc 2. But this integral is far more intricate than anything you would see in a textbook and needs to be processed through multiple layers of manipulation, including non-obvious substitutions you would not likely see in an average elementary Calc class.
The kind of integral you will spend a week on trying different things until you get lucky.
@@yanceyward3689 And that’s week of wasted time.
This kind of integral will keep you busy for a month, and even Carl Friedrich Gauß will probably need 1 hour and 1/2.
There is no way one would think of such grave manipulation of the given integrand in an exam situation.
Just saw this video, and I was totally blown away by how easy this integral is once I let x^2 = tan(t), and this whole thing turned into a nice beta function (√2/16)*int_(0 to pi/2) 2*(sin(2t))^(1/2) (cos(2t))^(-1/2) d(2t) = (√2/16)*B(3/4, 1/4) = pi/8. It only took like 2 minutes.
He really woke up one day and decided to write the funniest descriptions.
No no, that's me. I decided that the videos will have wild descriptions cuz it's fun!
-Stephanie
MP Editor
@@MichaelPennMath You're doing great, Stephanie! I love seeing Dr. Penn get a team to further the channel, it's hard to believe it used to be just him and his chalkboard. You guys have been rocking it lately!
@@MichaelPennMath Thanks for catching those corrections! Keeps the cognitive flow undistracted.
You can also do this as a contour integral using a quarter circle. Taking care to account for a sign change due to the branch cuts, we get the contour integral, as the radius of the circle -> oo and the paths around the branch cuts approach the branch cuts, has value 2*I + 2i*I, where I is the value of our original integral. But by the Cauchy residue theorem and noting the residue at 1/sqrt(2)+i/sqrt(2) is 1/8-i/8, this is 2*pi*i*(1/8-i/8) = pi/4 + i*pi/4. It follows I = pi/8.
That’s a very nice way to do it with complex analysis. There are other ways but yours is the simplest I think
I did the exact same thing lol
At 16:30 it should be 1/16 times that ln thing cause you mult and div by 2
Doesn't change the answer tho
Ah yes, the classic "seamlessly computing a monster integral yet failing to do basic arithmetic operations".
It turns out you can do this with a single substitution: u=sqrt(1-x^4)/x. (Hard to think of or motivate, but then so were the substitutions that Michael used!) It is then routine to check that the original integral is equal to the integral of 1/(u^4+4) from u=0 to infinity, and it is a challenging but classic partial fractions problems to show this equals pi/8.
19:15
disintegration of parts
Integration is a simple game, you do huge calculations and in the end π appears.
u can solve it also with the substitution: u=sqrt(tanx) , this will lead to the integral of sqrt(tanx) from 0 to pi/2 which we already know how to solve this integral.
@ 14:05 The last term of the second factor of t^4+4 should be +2 and not +t.
@17:35 where it says 1/4, I think it should be 1/16
I love these gnarly integrals !
Wow
Thank you, professor
Well done, it's one of the most beautiful integrations I've seen in my life.
Let f(x)=x^2/((1+x^4)*sqrt(1-x^4)) and Int(f(x),x=0..1)=I. It is then not difficult to check that the complex valued integral Int(f(x),x=-infinity+i0..infinity+i0)=2I. In fact you can close the complex integral in the upper half plane since f(R)*R goes to zero in absolute terms as R->infinity. The resulting closed contour integral encloses a pole at x=e^{i*Pi/4} and at x=e^{i*3Pi/4}. Furthermore it encloses a cut from x=i to x=i*infinity. Picking up the residues gives a value of Pi/2 and the closed contour can be written as encircling solely the line (i,i*infinity) counterclockwise plus the Pi/2. Rotating this contour clockwise by 90° with the substitution x=i*u (picking up the phases for the square-root und substituting u=1/t), it is found 2I=Pi/2-2I which gives I=Pi/8.
The subtitution made, is equivalent to substituting x = sort(cos theta), and then using a t = tan(1/2 theta) substitution after.
Well done! Like it.
the crazy thing is this solution means that this indefinite integral is actually doable, it becomes 1/16*log(4x^2/(1-x^4)-4sqrt(x^2/(1-x^4))+2)-1/16*log(4x^2/(1-x^4)+4sqrt(x^2/(1-x^4))+2)-1/8*arctan(1-2sqrt(x^2/(1-x^4)))+1/8*arctan(1+2sqrt(x^2/(1-x^4)))
EDIT: it can also be expressed more compactly as 1/8*(artanh(2x*sqrt(1-x^4)/(x^4-2x^2-1))-arctan(2x*sqrt(1-x^4)/(x^4+2x^2-1))), though getting the right value for the definite integral from this involves choosing the correct branches of the multi-valued inverse trig/hyper functions
The reward of calculating this monster integral is 1/8 of a pie.
Yes. But if you watch it again, then you get another 1/8th of a pie and so on. So if you watch it 8 times, you'll have a full pie.
-Stephanie
MP Editor
If I have this integral problem in a test，I will skip it without even 1 second hesitation🤣
Michael uses the Sophie Germain's identity, but he makes a little mistake when he writes it on the blackboard. Anyway, thank you very much for such an awesome exercise.
Imagine having such a question in an exam ! Nice problem 👍
Again, I love when I learn something new about maths.
Those substitutions feel like hacking.
An amazing elementary method of evaluation! I would have rewritten it as a hypergeometric function and simplified using its special relations, but that feels more like cheating.
I don't think that you are cheating by using hypergeometric functions; in fact it could be awesome do this "elementary" integral using more sophisticated methods.
Impressive ! Very impressive but ... how the hell can you find these two substitution from scratch ?
He explains the first one: take a part and see what the derivative is, then construct a 1 out of it by multiplying it with it's inverse. Half of what you have multiplied in automatically substitutes to du. Nice method - I wonder if that strategy has an unofficial name. XD The second one is more straight-forward, you get a feeling for these after solving a couple similar ones. :)
I guess he start with Integral of t^2/(1+t^4) and use substitution to make it monster Integral
if it was truly a gnarly integral, the video would be 35 minutes long and you'd already have the boards filled with pre-requisite identities that you have to prove first...
plus there would be a smattering of phi's, tree (3) exponents, and congruences equal to 5 (mod pi)
i missed some good old crazy integration video! :D
I wonder who has the idea for crazy substitutions like these ones… :-)
Once we got to the rational function inside the integral contour integration seems like it would be a lifesaver.
a dance with the devil
Call me pedantic, but maybe you could have said a few words about the singularity in the integrand at x=1? I played with this briefly in my Computer Algebra System (Scientific Workplace, now sadly defunct), but not enough. It appears that if you expand the integral as an infinite series you may be getting one of those weird counterexamples to the necessity of Leibnitz' Alternating Series Test. (Don't bet the farm, though, I was not able to work out the pattern in the time that I invested. I suspect things would become clearer by changing the variable to (1 - x). ) Really appreciate your work!
I got so scared when you wrote 1/4 instead of 1/16 in the end after all that work but it ended up canceling to 0 anyways.
isnt there a mistake? At 13:35 when he writes t^2, shouldn't it be sqrt of t ? Because x=t^2 in his substitution.
The substitution is sqrt(x) = t
dx = 2t dt
Thus sqrt(x) dx = t * 2t dt = 2(t^2) dt
This was the perfect closure to my work week - calculus just has an unmatched natural beauty to me.
isnt there a mistake?
he workt out that 1-x²= 2/(u+1) but actually he needed x²-1 wich ist -2/(u+1)
for the hole process it doesnt matter because it gets squared so the negative cancel out
14:00
I think in the main part (right side of the board) it was supposed to be (1-x^2)^2, which would be more consistent with the rest of it. But either way it clearly doesn't affect the result.
The bounds of integration automatically changing at 12:03. 😅
I could have sworn I had put an info box there to point it it'd be fixed. My bad. To be fair, the change is to correct an error.
-Stephanie
MP Editor
It's at 10:57 . Also, if you want to know more minor mistakes, 1/4 at 16:30 should be 1/16, but it doesn't matter at the end.
Something my Calc II professor would have put on an exam
ugh, sounds like fun...
Hold on, the coefficient of the logs at the end should be 1/16 not 1/4, but ig it doesn’t change the answer
It is possible to calculate indefinite integral with substitution
sqrt(1-x^4) = ux
At some point the lower bound of the integral got switched from 1 to 0. Am I missing something, or that’s a typo?
At 12:03, he points out that 1-1/1 is 0, so that's the correct thing to do when switching from u to (the second iteration of) x.
@@samwalko right, my bad, should not watch those at 1AM 😂
Hi Dr. Penn!
I would suggest using x^2=tan@
that integration bounds fix was smooth as fuck.
Monster among intemen
Will we ever get a stream were Michael plays cup head?!
Too long solution, can be destroyed by just substitution x=sqrt(tan(x/2))
The monster is also hirsute! 🙂
While watching video, I've found another but similar way to solve the integral.
I = ∫[0, 1] (x^2 dx /{(1+x^4)sqrt(1-x^4)})
First, divide numerator & denominator by x^2 → ∫[0, 1] (1 /{(1/x^2 + x^2)sqrt(1/x^2 - x^2)})dx/x
Next, let "x^2=y, dx/x=dy/2y" to reduce order of x → ∫[0, 1] (1 /{(1/y + y)sqrt(1/y - y)})dy/2y = ∫[0, 1] ((1/y^2 +1)dy /{2(1/y + y)^2 sqrt(1/y - y)})
Let "1/y - y=z, -(1/y^2 +1)dy=dz" so that "(1/y + y)^2=z^2+4" → ∫[∞, 0] (-dz /{2(z^2+4) sqrt(z)}) = ∫[0, ∞] (dz /{2(z^2+4) sqrt(z)})
Then let "z=u^2, dz/(2sqrt(z))=du" to rationalize the integrand → ∫[0, ∞] (du /(u^4+4))
By evaluating the last term, we get I=π/8.
This was my first idea but I didnt see that x^2=y substitution
Michael, I have to say that this video was disappointing in that it struck me as an exercise in vanity mathematics. You present what is a difficult integral and you do a lot of transformations in order to get to the end result but you never make a comment about convergence. There is no hint of any insight. How many of your viewers noticed that you transformed something that blew up at 1 to something that goes to zero at infinity (10:07)? Mathematica does this integral in 0.710641 seconds and gets Pi/8. That in itself interesting because of the way Mathematica does the processing for this type of integral basically using essentially the stuff the blind Jim Slagle did in his PhD thesis at MIT in 1961 under Marvin Minsky. Mathematica also uses the Risch and Ritt algorithms which build on the 18th-19th century works of Laplace, Louiville and Hardy to name but a few. The symbolic processors can do transformations of indefinite integrals and then perform the limits - Slagle’s system allowed for that and contains some specific examples. I think your audience would benefit from a follow up video explaining the convergence issue.
Thumbnail reminds me of {\displaystyle \cup head}.
？？？？？？
Gotta say, really don’t like the newer video titles and thumbnails. Makes each video feel more like clickbait, so I don’t watch them as much. I prefer to see thumbnails relating much more directly to the problem addressed in the video
Crap that was narly
im the 69th comment. nice !
where did that factorization of t^4 + 4 come from?
is there some more general principle involved that enables one to factor stuff of the form a^4 + b^2
(a^2)^2 + b^2 = (a^2 + bi)(a^2 - bi).. no good
a^n + b^n, for odd n, factors to (b)(a^(n-1) - a^(n-2)b +..-..+.. - ab^(n-2) + b^(n-1))
which is why if this were something like t^6 + 8, i could see rewriting it as (t^2)^3 + 2^3 and factoring it that way
no way to write t^4 + 4 as a sum of odd powers
(t^2)^2 + 2^2 = (t^2 + 2i)(t^2 - 2i).... is no good.
completing the square somehow and getting (t^2 + sqrt(4t^2))(t^2 - sqrt(4t^2)) doesn't foil out properly.. without +2 to each binomial
so i guess a^4 + b^2 = (a^2 + b + sqrt(4a^2))(a^2 + b - sqrt(4a^2)) which gives the (t^2 + 2t + 2)(t^2 + 2t - 2) that was used in the video
but that would mean you can always factor a sum of squares this way
so a^2 + b^2 is actually (a + b + sqrt(2ab))(a + b - sqrt(2ab)) = (a + bi)(a - bi)
so only useful if 2ab is a square
so we can do this for any a^(2m) + b^(2n) where 2(a^m)(b^n) is a perfect square which seems to work best when a or b is 2. but will probably work in a lot of other cases.
is there a name for this?
I solved this using the property of definite integrals x = ((b-a)-x) and then making u = x^2, rest is trivial