I am a frequent viewer of your presentations and I really enjoy them! Two observations a) after splitting in two integrals if you call the first I(a,b) the second is I(a,1) so it is not necessary to repeat the trick and evaluate it , its value isthe value of the first with b=1, and b) if you set a=b to the initial integral it is directly evaluated to (π/2)*ln(a^2)=π*ln(a) which could serve as a special case verification of the final result.
when you cancel the (y-a) terms in the first integral, shouldn't you then have to put a restriction on 'a' such that 'a' cannot be less than 'b' and greater than 1? because if it is, then you would be integrating over a point of discontinuity and you wouldn't be able to cancel both of those factors (since that would be a 0/0 indeterminate form.
I decided to try it myself because I took a break from really doing any math for the past few months, and it took me an hour, but I got it done and I'm pretty proud of myself. I did a completely different method though.
The way I did it is to use differentiation under the integral sign. So we could get (a/2)partial(I,a)+(b/2)partial(I,b) = integral (0,pi/2) 1 dx = pi/2.So we find a function I such that (a/2)partial(I,a)+(b/2)partial(I,b)=pi/2. We could then set I=pi*ln((a+b)/C) where C is some constant. We not that the original integral vanished for a=b=1 => C = 2 so I=pi*ln((a+b)/2)
This (a/2)partial(I,a)+(b/2)partial(I,b)=pi/2 is an inhomogeneous partial differential equation and certainly not trivial to solve. You are guessing the solution!
Second question : At 11:30 how do we know arctan(xy/a) --> pi/2 and not -pi/2 as x --> infinity ? That will depend on the sign of (y/a) - where y goes from 1 to b. We don't know the sign of a or b
I think that since we only care about the integral involving a^2 and b^2 you can set a=+/-sqrt(a^2) and b=+/-sqrt(b^2), but since the calculations are way easier if a, b>0 we choose the positive values
@@cosimodamianotavoletti3513 Yes I reasoned similarly, (although semi-convincingly) since the integral involves squares of a and b, the final answer should be same whether a or b is positive or negative. Hence if choosing a and b as positive, we get a nice answer then that should hold for all values of a or b. But that doesn't answer the question - just that a solution can be found without answering the question. If either or both "a" and "b" are negative, we would run into problems or ambiguity in some of the steps. But I was thinking there may be analytical reasons which can explain properly why we cannot have negative value of a or b in the integration process (may be something related to the anti-derivative's continuity or change of order of integration but cannot explain)
@@angelmendez-rivera351 I thought more about this and u are correct. If the integral was for (p*cos²θ + q*sin²θ) where p & q are positive numbers, I wouldn't have thought about taking a negative square root of p or q in the subsequent calculations. So why think differently for a² and b². Thank u
There is an easier way to substitute for tangent, IMO. Before the substitution, add 0 with -log(a^2*cos^2 theta) + log(a^2*cos^2 theta). Incorporate the first into the big logarithm and simplify to get tangent explicitly. You still eventually end up evaluating the same integral regardless of how you substitute and regardless of going with a double integral or Feynman's technique.
It's great to see different solutions to the same problem My solution was similar until 4:07. Then used the fact that the itegrand is an even function. This allowed me to expand the range of integration to (-infinity, +infinity). Afterwards it was only a small journey through the complex plain and one use of the residuum therom to arrive at the solution But in my opinion your solution is more beautiful because it requires only "simple" methods. So keep up the good work
or you can just linearize cos^2 and sin^2 into cos 2theta, then do a change of variable to get it to an integral of the forn ln cos 2 y, which you did in a previous video
Hey guys, The Pre-Pravega event ‘Gaussian Gamble’ invites you to a journey filled with intriguing and challenging math problems. Participants can form teams of two and compete against each other in a contest of Mathematical rigour and ingenuity, fighting for the top spot on the leaderboard! Please visit our website for more details and register for the event. www.pravega.org/events/science_eve/Enumeration.html The event starts on 21st August, so make sure you don’t miss out! We look forward to your active participation, and hope you will enjoy the event. Irrespective of your area of study, Math is a common language that unites us all ! The exciting problems await your keen interest.
Hi Michael First of all, I would like to emphasize that I am really impressed by your videos and always have the pleasure to follow them. On this one, which the result is impressive honestly, I have a question mark concerning the two simplifications you had done (red box : simplification by y-a, blue box : simplification by y-1). First on the blue box : since the integral is bound by 1 and a, are we Really allowed to do the simplification since we divide by 0 some sense? More fuzzy to me is the simplification of y-a. I may have missed in the video a statement telling that a>b. If 1
About the blue box: The simplification (y-1)/(y²-1) = 1/(y+1) is only problematic for y = 1. However, since {1} is a null set, we can remove it from the domain without changing the integral. In other words, it does not matter whether the domain is [1, a] or (1, a] (or [1, a) or (1, a)). About the red box: If a > b or 0 ≤ a < 1, then (y-a)/(y²-a²) = 1/(y+a) is unproblematic as you have already noticed. If 1 ≤ a ≤ b, then the same reasoning as above applies, since {a} is a null set. One can just integrate over [1, a) ∪ (a, b] instead of [1, b]. Then, when computing the integral with the fundamental theorem of calculus, the domain changed be set back to [1, b] without changing the integral.
@@angelmendez-rivera351 The only thing you'd need to do answer my question is simply prove that you can use fubini's theorem. One example is if you can show me how you can split the function into a product of an integral of f(x) and g(y). Fubini's theorem requires conditions; what's so hard to understand about demonstrating how the conditions are meant?
Bro we are integrating the function in the region x=0 to inf and y=1 to b. As long as b>=0 we can claim that the given function is positive at all points in the region of integration. That means we are essentially integrating the mod of the function. This means that change of order is not a problem.
Considering it as a function of a and b. b times (Partially differentiation wrt a)+ a times (partially differentiation wrt b) is π/2. Can we solve this type of differential equation,and similarly this question too
I’ll have to do it myself but I’m confused for the case where say a and b approach 0+. The integral goes to π/2 but the right side approaches π. I will edit once I figure it out.
@@sahilbaori9052 Oh you were talking about the parametric variables? You can take one of them to be the parameter and let the other be a constant because they're both independent of each other. I'm sure that'll do the job
The original expression is invariant under changes to the sign of a and b, so I'm pretty sure we can take a and b as being positive without loss of generality, except the final a+b should really be |a|+|b|
Im going to do these one at a time. Oh good. While grabbing 2 pieces of chalk and doing both integrals at the same time would be impressive, I wouldn't be able to follow it well.
@@angelmendez-rivera351 Was that stipulated somewhere I missed? Obviously the solution strategy can remain unchanged either way, since the original expression is in terms of a^2 and b^2 only.
It looks like implicitly a and b are assumed to be both positive, because if they were negative there would be some steps that would need to be worked more. Like for ex., when he takes the antiderivatives from 1 to a and 1 to b that give out natural logs, those integrals would need to be separated and an absolute value used because of the asymptote they had (If a and b are both positive you don't need to worry about the asymptotes of the natural log, but for negative values you have to). Which makes in turn like you said, the final general answer not correct if a and b aren't both positive.
Aha the question I was waiting for. Just look at 12:18. Here, while he substituted arctan(xy/a) at the bottom, with π/2. But, clearly this is only true for b>0. For b0 so, go through the video carefully.
I found that an interesting way to evaluate the integral but there're some minor issues with your solution: 1- at 4:26 you say x square but write just x where it should be x square. But you adjust it and write x square in the later steps so it doesn't matter. 2- at 9:12 you wrote the x integral from 1 to infinity where it should be from 0 to infinity. Buuut! You adjust that too in the later steps. So it also doesn't matter.
I recognize this integral from calc. Also if I were to get the problem directly from the thumbnail, I would solved something else entirely, probably depending on the IBP and residue law.
I have an idea for a nice vid, might be a little too easy but you could probably take the idea and go further with it. prove that the antiderivative of 1/z is multivalued without using d/dx(ln(z))=1/z
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It's actually very interesting that in the end the value is closely related to the natural log of the average for a and b!
I am a frequent viewer of your presentations and I really enjoy them! Two observations a) after splitting in two integrals if you call the first I(a,b) the second is I(a,1) so it is not necessary to repeat the trick and evaluate it , its value isthe value of the first with b=1, and b) if you set a=b to the initial integral it is directly evaluated to (π/2)*ln(a^2)=π*ln(a) which could serve as a special case verification of the final result.
The integral is so powerful it broke the thumbnail
What happened in thumbnail?
For more maths problems
ruclips.net/video/igdy05LZj90/видео.html
Wow! Another Michael Penn Tour-de-Force! Again, so many techniques and great tricks. Lots of things to practice in this problem.
when you cancel the (y-a) terms in the first integral, shouldn't you then have to put a restriction on 'a' such that 'a' cannot be less than 'b' and greater than 1? because if it is, then you would be integrating over a point of discontinuity and you wouldn't be able to cancel both of those factors (since that would be a 0/0 indeterminate form.
I decided to try it myself because I took a break from really doing any math for the past few months, and it took me an hour, but I got it done and I'm pretty proud of myself. I did a completely different method though.
You forgot to add one more adjective to the title:
Fantastic.
When you need a complicated way to find the average of a and b (= e^((Integral above)/pi))
The way I did it is to use differentiation under the integral sign. So we could get (a/2)partial(I,a)+(b/2)partial(I,b) = integral (0,pi/2) 1 dx = pi/2.So we find a function I such that (a/2)partial(I,a)+(b/2)partial(I,b)=pi/2. We could then set I=pi*ln((a+b)/C) where C is some constant. We not that the original integral vanished for a=b=1 => C = 2 so I=pi*ln((a+b)/2)
Leibinz's method or rather, Feynmann method
How did you get I=π*ln((a+b)/C)?
This (a/2)partial(I,a)+(b/2)partial(I,b)=pi/2 is an inhomogeneous partial differential equation and certainly not trivial to solve.
You are guessing the solution!
19:45
How? So quick
Hshs Sbvv Videos are released at 8am and 8pm EST so when it’s time I just go straight to the end of the video. That’s about it.
Always so helpful
And thats a good place to stop
Those 2s in the thumbnail just won't cooperate, will they.
just awesome loved it
Second question : At 11:30 how do we know arctan(xy/a) --> pi/2 and not -pi/2 as x --> infinity ?
That will depend on the sign of (y/a) - where y goes from 1 to b.
We don't know the sign of a or b
I think that since we only care about the integral involving a^2 and b^2 you can set a=+/-sqrt(a^2) and b=+/-sqrt(b^2), but since the calculations are way easier if a, b>0 we choose the positive values
@@cosimodamianotavoletti3513 Yes I reasoned similarly, (although semi-convincingly) since the integral involves squares of a and b, the final answer should be same whether a or b is positive or negative. Hence if choosing a and b as positive, we get a nice answer then that should hold for all values of a or b. But that doesn't answer the question - just that a solution can be found without answering the question.
If either or both "a" and "b" are negative, we would run into problems or ambiguity in some of the steps. But I was thinking there may be analytical reasons which can explain properly why we cannot have negative value of a or b in the integration process
(may be something related to the anti-derivative's continuity or change of order of integration but cannot explain)
@@angelmendez-rivera351 I thought more about this and u are correct. If the integral was for (p*cos²θ + q*sin²θ) where p & q are positive numbers, I wouldn't have thought about taking a negative square root of p or q in the subsequent calculations. So why think differently for a² and b². Thank u
There is an easier way to substitute for tangent, IMO. Before the substitution, add 0 with -log(a^2*cos^2 theta) + log(a^2*cos^2 theta). Incorporate the first into the big logarithm and simplify to get tangent explicitly. You still eventually end up evaluating the same integral regardless of how you substitute and regardless of going with a double integral or Feynman's technique.
This was a good mental exercise.
it is rather |a|+|b|
The integral can be reduced to ln(1+lambda*(cosx)^2) which can be evaluated by Feynman trick with parameter lambda
The spice must flow!
It's great to see different solutions to the same problem
My solution was similar until 4:07.
Then used the fact that the itegrand is an even function.
This allowed me to expand the range of integration to
(-infinity, +infinity).
Afterwards it was only a small journey through the complex plain and one use of the residuum therom to arrive at the solution
But in my opinion your solution is more beautiful because it requires only "simple" methods.
So keep up the good work
or you can just linearize cos^2 and sin^2 into cos 2theta, then do a change of variable to get it to an integral of the forn ln cos 2 y, which you did in a previous video
Hey guys,
The Pre-Pravega event ‘Gaussian Gamble’ invites you to a journey filled with intriguing and challenging math problems. Participants can form teams of two and compete against each other in a contest of Mathematical rigour and ingenuity, fighting for the top spot on the leaderboard!
Please visit our website for more details and register for the event.
www.pravega.org/events/science_eve/Enumeration.html
The event starts on 21st August, so make sure you don’t miss out! We look forward to your active participation, and hope you will enjoy the event. Irrespective of your area of study, Math is a common language that unites us all ! The exciting problems await your keen interest.
Sir can you make full lecture series on Multivarible calculus.
The one that you had already I think lacks practicing. Well great content sir🔥🔥
Why’s everybody speaking about your thumbnail? Don’t get it. I’m concentrated on the resolution. Amazing video as always!!! Thanks a lot
Hi Michael
First of all, I would like to emphasize that I am really impressed by your videos and always have the pleasure to follow them.
On this one, which the result is impressive honestly, I have a question mark concerning the two simplifications you had done (red box : simplification by y-a, blue box : simplification by y-1).
First on the blue box : since the integral is bound by 1 and a, are we Really allowed to do the simplification since we divide by 0 some sense?
More fuzzy to me is the simplification of y-a. I may have missed in the video a statement telling that a>b. If 1
About the blue box: The simplification (y-1)/(y²-1) = 1/(y+1) is only problematic for y = 1. However, since {1} is a null set, we can remove it from the domain without changing the integral. In other words, it does not matter whether the domain is [1, a] or (1, a] (or [1, a) or (1, a)).
About the red box: If a > b or 0 ≤ a < 1, then (y-a)/(y²-a²) = 1/(y+a) is unproblematic as you have already noticed. If 1 ≤ a ≤ b, then the same reasoning as above applies, since {a} is a null set. One can just integrate over [1, a) ∪ (a, b] instead of [1, b]. Then, when computing the integral with the fundamental theorem of calculus, the domain changed be set back to [1, b] without changing the integral.
@@usernamenotfound80 thanks a lot for your answer.
Cheers
Nice integral, love the editing
How do you know fubini's theorem is satisifed at 8:42?
@@angelmendez-rivera351 Saying it's trivial doesn't help
@@angelmendez-rivera351 The only thing you'd need to do answer my question is simply prove that you can use fubini's theorem. One example is if you can show me how you can split the function into a product of an integral of f(x) and g(y).
Fubini's theorem requires conditions; what's so hard to understand about demonstrating how the conditions are meant?
Bro we are integrating the function in the region x=0 to inf and y=1 to b.
As long as b>=0 we can claim that the given function is positive at all points in the region of integration.
That means we are essentially integrating the mod of the function.
This means that change of order is not a problem.
super cool method and result!
I Did this Same Integral Using Feynman's Technique. It was A little Lengthy but It didn't took too much Time.
@12:00 you’ve written a one where there should be a zero
1:50 - hypothesis is one?!
Considering it as a function of a and b. b times (Partially differentiation wrt a)+ a times (partially differentiation wrt b) is π/2. Can we solve this type of differential equation,and similarly this question too
residue theorem?
Could you make more videos about the math olympics?
But what if it had any variable in numerator after partial fraction decomposition
I like that you write arctan and not this silly tan^-1
\tan^{-1} \theta = \cot \theta
@@noway2831 yeah, but Wolfram Alpha for example uses tan^(-1)(θ)=arctan, which is silly.
Do you people not write f inverse as f^-1(x) or composition as f^n(x)?
Different in different places. I always learnt it as tan-¹x.
@@cerwe8861 Wolfram Alpha sucks
I’ll have to do it myself but I’m confused for the case where say a and b approach 0+. The integral goes to π/2 but the right side approaches π. I will edit once I figure it out.
Something happened with your thumbnail!! But good video though! :)
Do you mean some sort of flag 😂😂😂
What had happened?
PANDA S the exponents got messed up lol
Tbh I would have just Feynmaned it probably.
There are two variables fam
@@sahilbaori9052 I think they lead to a similar continuation.
@@sahilbaori9052 No, there's just theta
@@pbj4184 Feyman integration does not use the variable which is being integrated.
@@sahilbaori9052 Oh you were talking about the parametric variables? You can take one of them to be the parameter and let the other be a constant because they're both independent of each other. I'm sure that'll do the job
Brilliant.
The final "a+b" should be under absolute value or am I wrong ?
The original expression is identical under changes of sign to either a or b, so I believe you must be correct.
I suppose you assume that at least a is greater then zero, because you use it in oder to compute arcran. Am I right?
The original expression is invariant under changes to the sign of a and b, so I'm pretty sure we can take a and b as being positive without loss of generality, except the final a+b should really be |a|+|b|
Im going to do these one at a time.
Oh good. While grabbing 2 pieces of chalk and doing both integrals at the same time would be impressive, I wouldn't be able to follow it well.
Find integral from 0 to pie/2 1/(a²sin²@+b²cos²@)² d@.
Sir please increase your system volume 🙏
Shouldn't the final answer contain the absolute value of the log argument - that is ln (abs(a+b)) instead of ln(a+b) ?
@@angelmendez-rivera351 Was that stipulated somewhere I missed? Obviously the solution strategy can remain unchanged either way, since the original expression is in terms of a^2 and b^2 only.
I think you mean abs(a)+abs(b). Consider the case where a=-b, for example.
Sir if we replace b with -b the integral doesn't change but the answer will. Can you clarify why this confusion?
It looks like implicitly a and b are assumed to be both positive, because if they were negative there would be some steps that would need to be worked more. Like for ex., when he takes the antiderivatives from 1 to a and 1 to b that give out natural logs, those integrals would need to be separated and an absolute value used because of the asymptote they had (If a and b are both positive you don't need to worry about the asymptotes of the natural log, but for negative values you have to). Which makes in turn like you said, the final general answer not correct if a and b aren't both positive.
Aha the question I was waiting for.
Just look at 12:18. Here, while he substituted arctan(xy/a) at the bottom, with π/2. But, clearly this is only true for b>0.
For b0 so, go through the video carefully.
The classic substitution here is t=tan(theta/2).
I found that an interesting way to evaluate the integral but there're some minor issues with your solution:
1- at 4:26 you say x square but write just x where it should be x square. But you adjust it and write x square in the later steps so it doesn't matter.
2- at 9:12 you wrote the x integral from 1 to infinity where it should be from 0 to infinity. Buuut! You adjust that too in the later steps. So it also doesn't matter.
Antiderivative of 1/(a^2+x^2y^2) is (1/(ay))*arctan (xy/a)
And antiderivative of a^2/(a^2+x^2y^2) is (a/y)*arctan (xy/a)
Note:use variable change
leireeriel1 ya you're right i just checked the arctan rule. I rushed it! I edited it out so it doesn't mislead anyone
I recognize this integral from calc. Also if I were to get the problem directly from the thumbnail, I would solved something else entirely, probably depending on the IBP and residue law.
I have an idea for a nice vid, might be a little too easy but you could probably take the idea and go further with it.
prove that the antiderivative of 1/z is multivalued without using d/dx(ln(z))=1/z
And tomorrow Dr. Penn explaines us how to calculate the Indefinite integral of exp(-x²) ... ;-)
Very clever
12:37 is a bad place to stop
Let's say a²=b² therefore
I=int_0^(π/2) ln(a²) dx or I=int_0^(π/2) ln(b²) dx
I=π/2 ln(a²) or I=π/2 ln(b²)
I=π ln(a+a / 2) or I=π ln(b+b / 2)
by mixing both beacuse yolo I=π ln(a+b / 2) ..problem? :}
Much easier solution
int_0^(π/2) ln((a^2)*(cos^2(x)) + (b^2)*(sin^2(x))) dx;
int_0^(π/2) ln((a^2)*((cos^2(x)) + ((b/a)^2)*(sin^2(x)))) dx;
int_0^(π/2) (ln(a^2) + ln((cos^2(x)) + ((b/a)^2)*(sin^2(x)))) dx;
int_0^(π/2) (2*ln(a) + ln((cos^2(x)) + ((b/a)^2)*(sin^2(x)))) dx;
int_0^(π/2) (2*ln(a)) dx + int_0^(π/2) ln((cos^2(x)) + ((b/a)^2)*(sin^2(x))) dx;
π*ln(a) + int_0^(π/2) ln((cos^2(x)) + ((b/a)^2)*(sin^2(x))) dx;
Let r = b/a, therefore J(r) = int_0^(π/2) ln((cos^2(x)) + (r^2)*(sin^2(x))) dx
;
Let r = 1, therefore, J(1) = int_0^(π/2) ln((cos^2(x)) + (1^2)*(sin^2(x))) dx
;
J(1) = int_0^(π/2) ln((cos^2(x)) + (sin^2(x))) dx
-> J(1) = int_0^(π/2) ln(1) dx ->
J(1) = int_0^(π/2) 0 dx -> J(1) = 0, This is an important information.
dJ/dr = int_0^(π/2) (2r)*sin^2(x)/((r^2)*(sin^2(x)) + (cos^2(x))) dx
;
x = arctan(z); dx = 1/((z^2) + 1) dz; z_bottom = 0; z_top = ∞;
dJ/dr = int_0^(∞) ((2r)*sin^2(arctan(z))/((r^2)*(sin^2(arctan(z))) + (cos^2(arctan(z)))))*(1/((z^2) + 1)) dz;
dJ/dr = int_0^(∞) (2r)(z^2)/(((z^2) + 1)*((r^2)*(z^2) + 1)) dz;
dJ/dr = int_0^(∞) (((2r/((r^2) - 1))/((z^2) + 1)) - ((2r/((r^2) - 1))/((r^2)(z^2) + 1))) dz;
dJ/dr = (2r/((r^2) - 1))*int_0^(∞) ((1/((z^2) + 1)) - (1/((r^2)(z^2) + 1))) dz;
dJ/dr = (2r/((r^2) - 1))*((π/2) - (int_0^(∞) 1/((r^2)(z^2) + 1) dz))
dJ/dr = (2r/((r^2) - 1))*((π/2) - ((π/2)/r))
;
dJ/dr = (r/((r^2) - 1))*(π - (π/r))
;
dJ/dr = (1/((r^2) - 1))*(π*r - π)
;
dJ/dr = (π/((r^2) - 1))*(r - 1)
;
dJ/dr = (π)(r - 1)/((r - 1)*(r + 1));
dJ/dr = π/(r + 1);
J(r) = π*ln(r + 1) + C; J(1) = 0
0 = π*ln(1 + 1) + C -> 0 = π*ln(2) + C -> C = -π*ln(2), therefore;
J(r) = π*ln(r + 1) - π*ln(2); therefore;
int_0^(π/2) ln((a^2)*(cos^2(x)) + (b^2)*(sin^2(x))) dx
= π*ln(a) + π*ln((b/a) + 1) - π*ln(2);
int_0^(π/2) ln((a^2)*(cos^2(x)) + (b^2)*(sin^2(x))) dx
=π*(ln(a) + ln((b/a) + 1) - ln(2)
);
int_0^(π/2) ln((a^2)*(cos^2(x)) + (b^2)*(sin^2(x))) dx
=π*(ln(b + a) - ln(2)
);
int_0^(π/2) ln((a^2)*(cos^2(x)) + (b^2)*(sin^2(x))) dx
=π*ln((b + a)/2)
;
int_0^(π/2) ln((a^2)*(cos^2(x)) + (b^2)*(sin^2(x))) dx
=π*ln((|a| + |b|)/2)
integral six/x
ok, great. 19:48
I was a little sad when you didnt snap your finger for the blue transition part, but otherwise great video xD
Where's my +C ?? =( =( =(
Barney Stinson ? What are you doing here ?
Merch!
U should buy a smart board and make a book out of these questions
👍
Sorry but the tee is cooler that the problem ....
Nice nice nice
This is an IIT practice problem.
It could be more easy
You are doing this too complicated. There are other much simpler ways!
Your hairs again don't looks good! ...