It's nice to see a connection between that integral and the sum of the alternating inverse squares. In particular, π²/12 = η(2), where η is the Dirichlet eta function.
We can evaluate this integral using complex analysis: consider we have f(z)=ln(z)/(1+z^2+z^4). Since 1+z^2+z^4 is even, we can choose upper semicircle as our integration contour. The branchcut is (0;+inf). I_c is a counter integral, I_+ is integral of f(x) from 0 to inf, I_- is integral of f(-x) from -inf to 0 and I_R is integral of semicircle. So, I_c=(I_+)+(I_-)+I_R ln(-x)=ln(x)+i*pi. Hence, (I_+)+(I_-)=2*(I_+)+i*pi*integral of 1/(1+x^2+x^4) from 0 to inf Obviously I_R->0 as R->inf I_c=2*pi*i*sum of residues in the upper part of complex plane After finding the residues, we can just take real part of the equation and then find I Calculation of residues is easy for f(z) because they are just 1st order poles
Tabular integration! I derived that when I was a second-year college student to expand the integral of f(x)g(x) where f(x) is a polynomial. I felt so smart, even though the result is painfully obvious to anyone who's completed an integral calculus course.
You need to evaluate four residues of (log(x))^2/(1+x^2+x^4) that are the Four complex Roots of 1+x^2+x^4=0, which are +-(1/2)+-sqrt(3)*i/2. The keyhole contour integral Will give (log(x))^2-(log(x)+2*pi*i)^2 = -4*pi*i*log(x)+4*pi^2. If you done correctly the sum of residues, it Will give pi^2/6-i*pi^2/sqrt(3), thus the contour are 2*pi*i(pi^2/6-i*pi^2/sqrt(3))= pi^3*i/3+2*pi^3/sqrt(3). Equal the real and imaginary parts, and should give integral(log(x)/(1+x^2+x^4),x,0,inf)=-pi^2/12, and an extra integral(1/(1+x^2+x^4),x,0,inf)=pi/(2*sqrt(3)).
residue theorem and some manipulation of the keyhole contour that has a branch point on the negative x-axis, rounds nicely into 4*pi*i*(Integral) = 2*pi*i*(-pi^2/6) A little quick algebra gives the solution pretty quickly.
Funnily if you remove the x^4 in the denominator, applying the first tool shows that the resulting integral is equal to it's negative and therefore it's zero.
referring to 17:07 , the summation from n=0 to n= infinity of 3/(6n+3)^2 is basically 3 times the summation on the left-hand side (both are the summation of the inverses of squared odd numbers).So if you don't do the factorization it gives a different result ( specifically pi^2 / 4). Why does that happen?
@@vascomanteigas9433 You can do it easier with a half-circle around the upper half plane, with the branch cut down the negative imaginary axis. I did it on my board.
i have also done by taking 8 hours😂 , solution came same to michael penn, also process is quite simiar. He used this summation technique in previous questions so i assumed that technique may be useful to this question also and tried and found out true. i used main technique that is, reducing denominator to numerator. I came to final summation value but i forgot that sum of reciprocal of squares of odd numbers=>pi^2/8.
9:41: do those 2 integrals at the right side of the equation each produce "nice" values (in terms of pi, e, etc) or just some random irrational values that when summed happen to produce the final answer to the problem?
No, The first part (from 0 to 1) is equal to Sum n=0 to n=inf 1/(6n+3)^2-1/(6n+1)^2 The second part (from 1 to inf) Sum n=0 to n=inf 1/(6n+3)^2-1/(6n+5)^2 Even though the first object 1/(6n+3)^2 is easy to solve (as showed on the film) the second objects in both case are not solvable without digamma function (according to Wolframalpha), but when we have integral from 0 to inf, they combined creates this sum of all odd reciprocals (as showed on the film)
@@TheHarryMateuszYT Thanks! The result for the second object in both cases is actually dependant not on the Digamma function, but on the Trigamma function!
I dug deeper and found the result for summation (n=0 to inf) 1/(6n+1)^2 is 2*pi^2 + 10*sqrt(3)*Cl2(pi/3), where Cl2 is the Clausen function of order 2, which is closely related to the Trigamma function.
Quick question. At around 16:25 if all three of those summations represent the reciprocal of the odd integers squared, why do they all collapse into the same sum? Why wouldn't it be 3 times that sum?
Each reciprocal of an odd number can be written in exactly one of the three available forms (1/(6n+5), 1/(6n+3) or 1/(6n+1)) so together the sums exhaust all the odd reciprocals.
I make a video (ua-cam.com/video/mCew821Nem4/v-deo.html) to solve this problem by contour integral, take the "key-hole contour". Integrals vanish on the inner circle and outer circle, and then take the four residues at z1=1/2+sqrt(3)/2i , z2=-1/2+sqrt(3)/2i, z3=-1/2-sqrt(3)/2i, z4=1/2-sqrt(3)/2i. It is done!
I don't get it. (1+x^2+x^4)=x^2(1/x^2+1+x^2) Isn't it? If I pull out x^4 immediately, then I get what is written on the board when the 2nd integral is rewritten.
A lot of it comes from familiarity - the more problems you do, the more you’ll see certain things coming up and you can formalise them in your toolkit. And sometimes you’ll be playing around with a problem and you’ll realise that you can generalise something to use it elsewhere.
Hi, 0:15 , 0:33 , 0:43 : "dx" 0:38 : -"over"- , "of" instead 0:40 : -"times"- , "over" instead 4:24 : -x- , y instead For fun: 1:39 , 2:12 , 12:08 , 16:05 , 16:45 : "ok, nice" (Côte d'azur), 17:50 : would have been a better place to stop, to avoid being hidden by the splash screens.
I don't really like this format, with tools that are proven before the solution. It focuses too much on the apriori knowledge, properties that one should possess and leaves out the intuition behind the problem.
You do need to know some stuff beforehand, at least to know what you are aiming for. Otherwise you can just end up going in circles or worse making the problem more and more complicated. Perhaps the examples here are too specific to just "know", but I don't see the problem proving them ahead of time so they can just be used at appropriate spot, it's a lot cleaner this way.
Michael's tendency to provide proofs for his tools prior to tackling the problem is very much like lemmas being proven at the beginning of a larger, more complex proof. That way once Michael gets to the main problem he can keep the flow moving without having to take a detour.
@@bsmith6276 Maybe it is just a personal preference, but i would sooner have him make occasional detours, rather than have him proove seemingly unrelated lemmas before tackling the problem itself.
i suppose it would be interesting to start on the main problem immediately, then when a tool is needed, mention and prove it in its more general form (so we see it in context) before returning to the main problem
We hope that one day you will demonstrate to us the "dominated convergence" theorem in order to swap integral and summation XD
It's nice to see a connection between that integral and the sum of the alternating inverse squares. In particular, π²/12 = η(2), where η is the Dirichlet eta function.
Thank you, professor. That was indeed a great problem.
5:18 BPRP = he who shall not be named!
using complex variables and Cauchy method leads to a nice and very concise solution. only two poles in the upper half space.
And a very short calculation.
We can evaluate this integral using complex analysis:
consider we have f(z)=ln(z)/(1+z^2+z^4). Since 1+z^2+z^4 is even, we can choose upper semicircle as our integration contour. The branchcut is (0;+inf).
I_c is a counter integral, I_+ is integral of f(x) from 0 to inf, I_- is integral of f(-x) from -inf to 0 and I_R is integral of semicircle. So, I_c=(I_+)+(I_-)+I_R
ln(-x)=ln(x)+i*pi. Hence, (I_+)+(I_-)=2*(I_+)+i*pi*integral of 1/(1+x^2+x^4) from 0 to inf
Obviously I_R->0 as R->inf
I_c=2*pi*i*sum of residues in the upper part of complex plane
After finding the residues, we can just take real part of the equation and then find I
Calculation of residues is easy for f(z) because they are just 1st order poles
At 14:18, the three sums can easily be replaced with two sums, one with the exponent of x being 2n, the other being 6n+2, simplifying the next step.
Tabular integration! I derived that when I was a second-year college student to expand the integral of f(x)g(x) where f(x) is a polynomial. I felt so smart, even though the result is painfully obvious to anyone who's completed an integral calculus course.
That was awesome! When the geometric series got brought up I was like wait how is this gonna help lol
Great! Good place to look forward to tomorrow's video!
Great one, keep it up Michael!!!
I solved this integral last week.. I solved by using another method.. this is also nice way.
17:47 OK, let’s get this done. But first, tell yourself loudly “No negative thoughts are allowed in my brain”! Have a great day.
There can't be no negative thoughts when you doing maths, the most depressing shit ever.
I was too lazy or stupid to get this actually done but wouldn't a more straightforward way of solving the integral be using the residue theorem?
I have been staying away from complex analytic methods to this point. This will change next semester after I add some videos about such methods.
I solved and published this integral last week on my channel by using digamma function.. Micheal’ s solution is also nice.
You need to evaluate four residues of (log(x))^2/(1+x^2+x^4) that are the Four complex Roots of 1+x^2+x^4=0, which are +-(1/2)+-sqrt(3)*i/2.
The keyhole contour integral Will give (log(x))^2-(log(x)+2*pi*i)^2 = -4*pi*i*log(x)+4*pi^2.
If you done correctly the sum of residues, it Will give pi^2/6-i*pi^2/sqrt(3), thus the contour are 2*pi*i(pi^2/6-i*pi^2/sqrt(3))= pi^3*i/3+2*pi^3/sqrt(3).
Equal the real and imaginary parts, and should give integral(log(x)/(1+x^2+x^4),x,0,inf)=-pi^2/12, and an extra integral(1/(1+x^2+x^4),x,0,inf)=pi/(2*sqrt(3)).
residue theorem and some manipulation of the keyhole contour that has a branch point on the negative x-axis, rounds nicely into
4*pi*i*(Integral) = 2*pi*i*(-pi^2/6)
A little quick algebra gives the solution pretty quickly.
I make a video on the contour integral method to solve this problem. ua-cam.com/video/mCew821Nem4/v-deo.html
zeta(-1)*pi^2
Funnily if you remove the x^4 in the denominator, applying the first tool shows that the resulting integral is equal to it's negative and therefore it's zero.
One of your best, really. Thank you!
referring to 17:07 , the summation from n=0 to n= infinity of 3/(6n+3)^2 is basically 3 times the summation on the left-hand side (both are the summation of the inverses of squared odd numbers).So if you don't do the factorization it gives a different result ( specifically pi^2 / 4). Why does that happen?
Interesting that wolfram stuggles with this.
Maxima solves quickly. This integral can bem solve by Residue Theorem with a keyhole contour integral, and this why Maxima made it.
@@vascomanteigas9433 Oooh, I'd like to see that.
Wolfram Mathematica can solve it, just confirmed!
@@vascomanteigas9433 You can do it easier with a half-circle around the upper half plane, with the branch cut down the negative imaginary axis. I did it on my board.
============>Typo Report
Yeah went with Residue Theorem as soon as I saw the thumbnail, cool that Micheal went in a different direction.
i have also done by taking 8 hours😂 , solution came same to michael penn, also process is quite simiar. He used this summation technique in previous questions so i assumed that technique may be useful to this question also and tried and found out true. i used main technique that is, reducing denominator to numerator. I came to final summation value but i forgot that sum of reciprocal of squares of odd numbers=>pi^2/8.
Great explanation
At 5:25 I don't understand why you don't say who uses D I method...
9:41: do those 2 integrals at the right side of the equation each produce "nice" values (in terms of pi, e, etc) or just some random irrational values that when summed happen to produce the final answer to the problem?
yess
No,
The first part (from 0 to 1) is equal to
Sum n=0 to n=inf 1/(6n+3)^2-1/(6n+1)^2
The second part (from 1 to inf)
Sum n=0 to n=inf 1/(6n+3)^2-1/(6n+5)^2
Even though the first object 1/(6n+3)^2 is easy to solve (as showed on the film) the second objects in both case are not solvable without digamma function (according to Wolframalpha), but when we have integral from 0 to inf, they combined creates this sum of all odd reciprocals (as showed on the film)
@@TheHarryMateuszYT Thanks! The result for the second object in both cases is actually dependant not on the Digamma function, but on the Trigamma function!
I dug deeper and found the result for summation (n=0 to inf) 1/(6n+1)^2 is 2*pi^2 + 10*sqrt(3)*Cl2(pi/3), where Cl2 is the Clausen function of order 2, which is closely related to the Trigamma function.
Where did the first term of the sum(1/(2n+1), 0 to infinity) for n=0 go ? I didn’t catch that part
I can`t see it either. I need an explaination
Michael didn't explicitly mention it, but he did combine the two sums into one in the last line (-1 + 1/3 = -2/3)
@@musik350 Sorry, I still can`t see it. I see 1/3 but not -1
@@zeravam The expression is of the form -S + 1/3S
Quick question. At around 16:25 if all three of those summations represent the reciprocal of the odd integers squared, why do they all collapse into the same sum? Why wouldn't it be 3 times that sum?
Each reciprocal of an odd number can be written in exactly one of the three available forms (1/(6n+5), 1/(6n+3) or 1/(6n+1)) so together the sums exhaust all the odd reciprocals.
@@pjmmccann Ohhhh of course. Thank you!
Mathematica can calcurate this integral, so there might be some routine method. Anyway thanks for your exciting method.
neat, this integral is equal to the sum of all natural numbers multiplied by pi squared! /s
I could feel the energy in that last . //
Can you explain how going from the original problem, how could one construct this complicated sequence of permutations to get to the answer?
14:04 Could anyone explain to me why is it okay to switch like that integral and sum?
It's usually okay to do as long as the expressions are convergent/finite
The magic of Mathematics.
I make a video (ua-cam.com/video/mCew821Nem4/v-deo.html) to solve this problem by contour integral, take the "key-hole contour". Integrals vanish on the inner circle and outer circle, and then take the four residues at z1=1/2+sqrt(3)/2i , z2=-1/2+sqrt(3)/2i, z3=-1/2-sqrt(3)/2i, z4=1/2-sqrt(3)/2i. It is done!
I don't get it.
(1+x^2+x^4)=x^2(1/x^2+1+x^2)
Isn't it? If I pull out x^4 immediately, then I get what is written on the board when the 2nd integral is rewritten.
At 4:25 I think you mean x=1, y=0.
How does one find the motivation for the tools on the LHS (or even the knowledge that these tools even exist in the first place)?
A lot of it comes from familiarity - the more problems you do, the more you’ll see certain things coming up and you can formalise them in your toolkit. And sometimes you’ll be playing around with a problem and you’ll realise that you can generalise something to use it elsewhere.
Hello sir, sir can help me please ! Find int sin(x^2)cos(x^3)/(4ln^2(cosx)+1) from x=-inf to inf
Fantastic !!
Integration by parts?
Nice one
666 views as I am currently watching this.
Fantastic
Very nice 👏👏👏
Why not mention the name of the "other youtuber"?
Why did I get this recommended i am too dumb for this video
Great!
jee 2022 aspirant here ig this q is from arihant integral calculus
good stuff
Hi,
0:15 , 0:33 , 0:43 : "dx"
0:38 : -"over"- , "of" instead
0:40 : -"times"- , "over" instead
4:24 : -x- , y instead
For fun:
1:39 , 2:12 , 12:08 , 16:05 , 16:45 : "ok, nice" (Côte d'azur),
17:50 : would have been a better place to stop, to avoid being hidden by the splash screens.
Meraviglioso
14:10 but that’s ok in this case.
This is a rather interesting method with no factoring, partial fraction decomposition, or use of roots of the polynomial.
fraction decomposition is the first thing I thought of doing
Nice！
Are you ok ?
Yes, thanks for asking :)
Wow wow wow
i holic to this video XD
As per usual, the Ways of the UA-cam Algorithm are inscrutable. Alright i try to follow :D Edit: Welp, i failed. Holy F*** thats complicated.
👍🏼🎉🤨🤔😲🤯
I don't really like this format, with tools that are proven before the solution. It focuses too much on the apriori knowledge, properties that one should possess and leaves out the intuition behind the problem.
i like both the elegant solutions and the exploratory ones
You do need to know some stuff beforehand, at least to know what you are aiming for. Otherwise you can just end up going in circles or worse making the problem more and more complicated. Perhaps the examples here are too specific to just "know", but I don't see the problem proving them ahead of time so they can just be used at appropriate spot, it's a lot cleaner this way.
Michael's tendency to provide proofs for his tools prior to tackling the problem is very much like lemmas being proven at the beginning of a larger, more complex proof. That way once Michael gets to the main problem he can keep the flow moving without having to take a detour.
@@bsmith6276 Maybe it is just a personal preference, but i would sooner have him make occasional detours, rather than have him proove seemingly unrelated lemmas before tackling the problem itself.
i suppose it would be interesting to start on the main problem immediately, then when a tool is needed, mention and prove it in its more general form (so we see it in context) before returning to the main problem