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.
@@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.
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.
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.
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.
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)
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
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. :)
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
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!
It is possible to calculate indefinite integral with substitution sqrt(1-x^4) = ux
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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?
