I like the integral and the method presented. However, you can rewrite the integrand as (1-x)/(1-x^6), then use partial fractions. It is messy, but you'll end up with the same exact (no surprise) arctangent evaluation you got.
@@insainsinyou enclosed the first quadrant by making a path along x for zero to R, and then a quarter circle of radius R to meet the y axis. And then down the y axis. You let R go to infinity and hope the contribution on the boundary goes to zero. You can then use the Cauchy residue theorem using the poles in the first quadrant.
Unbelievable... Also the partial fraction could work here if it's possible to convert the integrand into: (1 - 2 x)/(6 (x^2 - x + 1)) + 1/(2 (x^2 + x + 1)) + 1/(3 (x + 1))
I don't understand integration and what you have done on the board but I think that you are a amazing person , thanks you encourage me to learn calculus again ❤
Because of the geometric integral like geometric series title - could we simplify the denominator into an expression using the finite geometric series formula??? - oh wait you would not get an expression that is derivative of x^5 at most on the top or bottom once you simplify the complicated fraction. This be a sneaky trick that ended up giving a nice answer as you said.
How would you do this in general? What if the denominator was 1+x+...+x^n for example? What I might be tempted to do is write the denominator as a geometric series with common ratio x, and then use the sum for a geometric series. The resulting integral should suggest an obvious way forward.
I got it completely differently. For me, the first thing was to use geometric series to get \int_{0}^{\infty}\frac{x-1}{x^{6}-1}dx. sorry about using latex rather than typing it out but i feel like its a little more readable that way due to standardization. I then broke it in 2 parts from 0 to 1 and 1 to infinity: \int_{0}^{1}\frac{x-1}{x^{6}-1}dx+\int_{1}^{\infty}\frac{x-1}{x^{6}-1}dx. then on the second one from 1 to infinity I wanted to expand as a series so i used the u sub for 1/x and got \int_{0}^{1}\frac{x-1}{x^{6}-1}dx+\int_{0}^{1}\frac{x^{-1}-1}{x^{-6}-1}x^{-2}dx. I multiplied top and bottom by x^6 and simplified which gave me \int_{0}^{1}\frac{x-1}{x^{6}-1}dx+\int_{0}^{1}\frac{1-x}{1-x^{6}}x^{3}dx. I realized there were common factors between the 2 so I decided to recombine them and got \int_{0}^{1}\frac{x-1}{x^{6}-1}\left(1+x^{3} ight)dx. I expanded the denominator as a difference of squares and canceled terms which gave me: \int_{0}^{1}\frac{x-1}{x^{3}-1}dx. then I felt i could expand the bottom which i could and lucky for me the 1-x canceled out. \int_{0}^{1}\frac{1}{x^{2}+x+1}dx. from there i factored and did some linear substitutions to arrive at a usable arctan type integral: factoring: \int_{0}^{1}\frac{1}{\left(x+\frac{1}{2} ight)^{2}+\frac{3}{4}}dx. linear sub setup: \frac{4}{3}\int_{0}^{1}\frac{1}{\left(\frac{2}{\sqrt{3}}x+\frac{1}{\sqrt{3}} ight)^{2}+1}dx. then the final arctan type integral: \frac{2}{\sqrt{3}}\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{1}{x^{2}+1}dx. from there it was pluging things in and reduction. \frac{2}{\sqrt{3}}\left(\arctan\left(\sqrt{3} ight)-\arctan\left(\frac{1}{\sqrt{3}} ight) ight). then i used the fact that arctan of 1/x is arccot of x and arccot of x is pi/2 - arctan of x which gave me: \frac{2}{\sqrt{3}}\left(2\arctan\left(\sqrt{3} ight)-\frac{\pi}{2} ight). So then i solved the arctan(sqrt(3)) and simplified: \frac{\pi}{3\sqrt{3}}.
Hello dr payam ! I have a question. What's is difference between multiplication of two function ,convolution of two function and correlation of two question?
Great video as always, Dr! Will you come back to making videos such as this but in other languages? That used to be one of my favorite things about this channel.
I am glad you come again , my question at the minute 2:25 you add I to I =2I and assume that X^3=U^3, but X=1/U? when I solve it the result was very different from your solution ? Please if describe that I will be gratitude , thanks.
That's called dummy variable Like if it is integration of 0 to 1 xdx and it doesn't matter what x= f(y) is We can say that it is equal to integration of0 to 1 ydy or any other variable like tdt . Because we will integrating the same variable with d(variable)
Great comeback Dr Peyam Can we generalize this integral for a n-th degree polynomial function? Say, some kind of Recursion formula? That'll be wonderful Dr.Peyam.
Hey Dr. Peyam! Great video... Could you also plz do a video solving this same integral using complex analysis with concepts like Cauchy's integral formula (using poles, etc)? Thank You! 🍻😃
@arieltabbach4946 that was my first thought, and if that's all it is then I might just need to have some more comfort with that level of notation flexibility. But genuine question if you have a sec to humor this: isn't u = 1/x and, so the denominators {u⁵ + u⁴ + u³ + u² + u} is actually equal to {x⁻⁵ + x⁻⁴ + x⁻³ + x⁻² + x⁻¹} and I am struggling to see how it can just be combined with the other denominator of {x⁵ + x⁴ + x³ + x² + x¹}? Do you see why I am confused? I absolutely assume I am misunderstanding somthing and would love to have this step click ☺️
I tried solving an integral by part the "wrong" way, which develop into an infinity of terms. It turned out that I could factor out a series as a trigonometric function, thus solving it.
That was gorgeous in the end, what a beautiful result and definitely one that I am commiting to memory. The only bit that I found confusing was why we could add the 2I together given that one was in terms of u and the other in x, and change the u³ to a x³? Thanks in advance for any clarification on this.
u is just a dummy variable which could be anything. as long as we properly do the substitutions and change the integral's bounds it works just fine. we just use a different variable to remind ourselves we've done a change of variable.
@@cyberduck027that was my first thought, and if that's all it is then I might just need to have some more comfort with that level of notation flexibility. But genuine question if you have a sec to humor this: isn't u = 1/x and, so the denominators {u⁵ + u⁴ + u³ + u² + u} is actually equal to {x⁻⁵ + x⁻⁴ + x⁻³ + x⁻² + x⁻¹} and I am struggling to see how it can just be combined with the other denominator of {x⁵ + x⁴ + x³ + x² + x¹}? Do you see why I am confused? I absolutely assume I am misunderstanding somthing and would love to have this step click ☺️
@@graf_paperThe integrand is just a function, right? If you think changing back to x is weird, we could just do an x=z and u=z substitution to get: 2I = ∫ (1+z^3)/(z^5+z^4+z^3+z^2+z+1) dz. It feels weird since Dr. Peyam did a u=(1/x) substitution to get f(u). If you can wrap your head around the top part, then he just simply did an x=x and u=x substitution since we're just dealing with 2 functions
@@enerjae7174 ok.. I think I get this. I had to write this out and stare at it for a bit. I don't have an issue with u substitution, or at least I didn't think I did. Somehow it wasn't matching up with the patterns of algebraic manipulation I have seen before. Thanks for taking the time to support 🌻
How are we able to cancel 1+x^3 from both numerator and denominator without any second thought? As math students, we're always cautioned against such cancellations, aren't we? I'm curious.
Amazing, is it not, that we have the integral of the inverse of a simple polynomial function, and out pops π/(3*√3) = π√3/9. What's π got to do with it? 🐈
@@baconboyxy but then arctan also goes to - infinity as well and what happens if i say add n*pi to the angle? we will bounce back and forth from infinity to minus infinity. if we add 2n*pi we will get yet another angle that is also extend to infinity, but it is a different point in time. No, there is no one answer for this, at least not within the method used. We would need a different method that shows a radius of convergence for the function that is -1< r
@@kennethgee2004 arctan(x) is a function so it doesn’t account for all the multiple periods of tan(x). That’s fine tho, because the integral being evaluated to arctan(x), 1/(1+x^2), always uses the standard definition of arctan(x).
@@baconboyxy so arctan(x) is still an angle and there is still radians as that is just a number. You argument is not correct in the slightest. a function must be continuous which arctan is not.
@@kennethgee2004 arctan is continuous, though? And differentiable for that matter. I’d recommend looking at the graph or looking up the integral of 1/(1+x^2) or something
![Queen of Hill Stations Ooty, India 🇮🇳 by Drone [4K] 2021](http://i.ytimg.com/vi/OPIE5j0SN5Q/mqdefault.jpg)
I’m very happy that you’re back. I really did miss you and your videos.
Yay, thank you!
How did this solution possibly cross someone's mind?
Dr Peyam: "It is not too good to be true. It is good and true." I love it😂
He actually looks really cool in this fit,the braclet thingy looks good too
Thank you!! :3
I like the integral and the method presented. However, you can rewrite the integrand as (1-x)/(1-x^6), then use partial fractions. It is messy, but you'll end up with the same exact (no surprise) arctangent evaluation you got.
It’s been a long time, but after writing it as you did, it might be possible t9 use the Cauchy residue theorem to evaluate the integral.
@@chriswinchell1570 How?
@@insainsinyou enclosed the first quadrant by making a path along x for zero to R, and then a quarter circle of radius R to meet the y axis. And then down the y axis. You let R go to infinity and hope the contribution on the boundary goes to zero.
You can then use the Cauchy residue theorem using the poles in the first quadrant.
@@lieslceleste3395 Why would the integral along the y axis disappear? It would stick around.
Nice white board.
As blank as meine Seele
You’re both very nice!
The fact that Dr Peyam is back is GOOD and TRUE!
Given that sqrt(3) = tan(pi/3) the result could also be written pi/3 * cot(pi/3) suggesting potential symmetries.
Great to see you back Dr Payem!! Can't wait to see more amazing maths!!
Your witty humor was sorely missed, good to have you back!
I was overjoyed when I saw a new Dr Peyam video in my recommended feed!
Hope you enjoyed it!
Glad you're back!
Unbelievable...
Also the partial fraction could work here if it's possible to convert the integrand into:
(1 - 2 x)/(6 (x^2 - x + 1)) + 1/(2 (x^2 + x + 1)) + 1/(3 (x + 1))
Dr. Peyam, you are on another level! :)
very happy you are back
You always make me smile Peyam :)
I try 😊
I don't understand integration and what you have done on the board but I think that you are a amazing person , thanks you encourage me to learn calculus again ❤
You are back 😎 your linear algebra videos helped me clutch my exam 🙌
Because of the geometric integral like geometric series title - could we simplify the denominator into an expression using the finite geometric series formula??? - oh wait you would not get an expression that is derivative of x^5 at most on the top or bottom once you simplify the complicated fraction. This be a sneaky trick that ended up giving a nice answer as you said.
Waoh👏 The integral evaluates to the area of a circle with radius of \sqrt(\sqrt(3) /9)
What I particulary apreciate in your videos is the lack of overexplaining! BTW, apparantly the patterns on your shirt are called persian pickles! ☺
I had no idea!! 😍
How would you do this in general? What if the denominator was 1+x+...+x^n for example? What I might be tempted to do is write the denominator as a geometric series with common ratio x, and then use the sum for a geometric series. The resulting integral should suggest an obvious way forward.
Wow what an incredibly simple solution to such a daunting seeming problem. never could i have predicted that would have been the answer.
Amazing!
Cool solution‼️🫶
I got it completely differently. For me, the first thing was to use geometric series to get \int_{0}^{\infty}\frac{x-1}{x^{6}-1}dx. sorry about using latex rather than typing it out but i feel like its a little more readable that way due to standardization. I then broke it in 2 parts from 0 to 1 and 1 to infinity: \int_{0}^{1}\frac{x-1}{x^{6}-1}dx+\int_{1}^{\infty}\frac{x-1}{x^{6}-1}dx. then on the second one from 1 to infinity I wanted to expand as a series so i used the u sub for 1/x and got \int_{0}^{1}\frac{x-1}{x^{6}-1}dx+\int_{0}^{1}\frac{x^{-1}-1}{x^{-6}-1}x^{-2}dx. I multiplied top and bottom by x^6 and simplified which gave me \int_{0}^{1}\frac{x-1}{x^{6}-1}dx+\int_{0}^{1}\frac{1-x}{1-x^{6}}x^{3}dx. I realized there were common factors between the 2 so I decided to recombine them and got \int_{0}^{1}\frac{x-1}{x^{6}-1}\left(1+x^{3}
ight)dx. I expanded the denominator as a difference of squares and canceled terms which gave me: \int_{0}^{1}\frac{x-1}{x^{3}-1}dx. then I felt i could expand the bottom which i could and lucky for me the 1-x canceled out. \int_{0}^{1}\frac{1}{x^{2}+x+1}dx. from there i factored and did some linear substitutions to arrive at a usable arctan type integral: factoring: \int_{0}^{1}\frac{1}{\left(x+\frac{1}{2}
ight)^{2}+\frac{3}{4}}dx. linear sub setup: \frac{4}{3}\int_{0}^{1}\frac{1}{\left(\frac{2}{\sqrt{3}}x+\frac{1}{\sqrt{3}}
ight)^{2}+1}dx. then the final arctan type integral: \frac{2}{\sqrt{3}}\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{1}{x^{2}+1}dx. from there it was pluging things in and reduction. \frac{2}{\sqrt{3}}\left(\arctan\left(\sqrt{3}
ight)-\arctan\left(\frac{1}{\sqrt{3}}
ight)
ight). then i used the fact that arctan of 1/x is arccot of x and arccot of x is pi/2 - arctan of x which gave me: \frac{2}{\sqrt{3}}\left(2\arctan\left(\sqrt{3}
ight)-\frac{\pi}{2}
ight). So then i solved the arctan(sqrt(3)) and simplified: \frac{\pi}{3\sqrt{3}}.
thats a really cool and elegant way to solve it! definitelly will try to remember this way of solving that type of integrals.
is it generalizable?
Awesome video!
oh nice one of my fav math guys is back!
Welcome back!!
Hello dr payam ! I have a question. What's is difference between multiplication of two function ,convolution of two function and correlation of two question?
Very nice, thanks.
Amazing!!!
Great video as always, Dr! Will you come back to making videos such as this but in other languages? That used to be one of my favorite things about this channel.
Awwww I should!!!
I am glad you come again , my question at the minute 2:25 you add I to I =2I and assume that X^3=U^3, but X=1/U? when I solve it the result was very different from your solution ? Please if describe that I will be gratitude , thanks.
That's called dummy variable
Like if it is integration of 0 to 1 xdx and it doesn't matter what x= f(y) is
We can say that it is equal to integration of0 to 1 ydy or any other variable like tdt .
Because we will integrating the same variable with d(variable)
Great comeback Dr Peyam
Can we generalize this integral for a n-th degree polynomial function? Say, some kind of Recursion formula?
That'll be wonderful Dr.Peyam.
fun fact: this value is the same for the integral over 1/1+x+x^2 between 0 and 1.
very good
Never had I thought - judging from a quick look - that these integrals have such a neat connection.
Dr. Peyam, pleeeeeease make video on solving lim(cos(cos(cos(….(cos(n)), where n->inf.
That’t be interesting to see the solution
Nice way to do it 😄
x = -1 is root, (1+x+...) = (1+x)*P4(x).
Hey Dr. Peyam! Great video... Could you also plz do a video solving this same integral using complex analysis with concepts like Cauchy's integral formula (using poles, etc)?
Thank You! 🍻😃
Why not partial fractions??
wow bro was 3 steps ahead
Very nice tnx
Forgive me Dr. Peyam, but can you elaborate more on the geometric part? as in why name the problem as geometric integral...?
very good question
Hey Dr Peyam. How did you add the two integrals together? One is in terms of x while the other is in terms of u.
This is my exact question, and he changed the variable back to x. Id love an explanation here!
@@graf_paper its just a symbol, he can change it to X and it would be the same
@arieltabbach4946 that was my first thought, and if that's all it is then I might just need to have some more comfort with that level of notation flexibility.
But genuine question if you have a sec to humor this:
isn't u = 1/x and, so the denominators
{u⁵ + u⁴ + u³ + u² + u} is actually equal to
{x⁻⁵ + x⁻⁴ + x⁻³ + x⁻² + x⁻¹} and I am struggling to see how it can just be combined with the other denominator of {x⁵ + x⁴ + x³ + x² + x¹}?
Do you see why I am confused?
I absolutely assume I am misunderstanding somthing and would love to have this step click ☺️
@@graf_paper Yeah, I also didn't understand how 2*I equals that.
I suspect wizardry
Great to see you! I followed right along, but I don't have the intuition to say why it is geometrical.
I tried solving an integral by part the "wrong" way, which develop into an infinity of terms. It turned out that I could factor out a series as a trigonometric function, thus solving it.
The integrand = (x-1)/(x^6-1), could we exploit this?
That was gorgeous in the end, what a beautiful result and definitely one that I am commiting to memory.
The only bit that I found confusing was why we could add the 2I together given that one was in terms of u and the other in x, and change the u³ to a x³?
Thanks in advance for any clarification on this.
u is just a dummy variable which could be anything. as long as we properly do the substitutions and change the integral's bounds it works just fine. we just use a different variable to remind ourselves we've done a change of variable.
@@cyberduck027that was my first thought, and if that's all it is then I might just need to have some more comfort with that level of notation flexibility.
But genuine question if you have a sec to humor this:
isn't u = 1/x and, so the denominators
{u⁵ + u⁴ + u³ + u² + u} is actually equal to
{x⁻⁵ + x⁻⁴ + x⁻³ + x⁻² + x⁻¹} and I am struggling to see how it can just be combined with the other denominator of {x⁵ + x⁴ + x³ + x² + x¹}?
Do you see why I am confused?
I absolutely assume I am misunderstanding somthing and would love to have this step click ☺️
@@graf_paperThe integrand is just a function, right? If you think changing back to x is weird, we could just do an x=z and u=z substitution to get:
2I = ∫ (1+z^3)/(z^5+z^4+z^3+z^2+z+1) dz.
It feels weird since Dr. Peyam did a u=(1/x) substitution to get f(u). If you can wrap your head around the top part, then he just simply did an x=x and u=x substitution since we're just dealing with 2 functions
@@enerjae7174 ok.. I think I get this. I had to write this out and stare at it for a bit.
I don't have an issue with u substitution, or at least I didn't think I did. Somehow it wasn't matching up with the patterns of algebraic manipulation I have seen before.
Thanks for taking the time to support 🌻
This makes me curious about the integral from 0 to infinity of the sum of all non-negative powers of x (x^n from n=0 to infinity).
the integrand will go to zero for x > 1 and for x between 0 and 1, just call the denominator 1/(1-x), not very interesting.
This is called power series! The integrand is equal to 1/1-x
@@2piee Yes, I know that. I'm more curious whether there are any interesting techniques one can apply to the integral.
@@mertaliyigit3288 Yes, but that's only within the interval of convergence, [-1,1].
Sorry, I am not convinced: when you added the x and u integrals, there was still the u=1/x substitution to deal with.
Do another u sub after that: x = u with du = dx
@@drpeyam It's not the same x, though. That's my confusion.
There is one property where you can interchange variables. It's the same
How are we able to cancel 1+x^3 from both numerator and denominator without any second thought? As math students, we're always cautioned against such cancellations, aren't we? I'm curious.
As long as it’s nonzero it’s ok!
@@drpeyam Thank you. 🙂
Na cause he looking kinda hot in that shirt(respectfully)
Awwww thank you 😊
Amazing, is it not, that we have the integral of the inverse of a simple polynomial function, and out pops π/(3*√3) = π√3/9. What's π got to do with it? 🐈
Arctan happened!
@@bsmith6276 👍
luces demacrado, cuida tu salud.
Epic
Wow
😮😮
well no that is not correct. the arctan of infinity is approaches a family of answers and not strictly pi/2.
It is though? Arctan(x) has a horizontal asymptote of pi/2 extending to +infinity.
@@baconboyxy but then arctan also goes to - infinity as well and what happens if i say add n*pi to the angle? we will bounce back and forth from infinity to minus infinity. if we add 2n*pi we will get yet another angle that is also extend to infinity, but it is a different point in time. No, there is no one answer for this, at least not within the method used. We would need a different method that shows a radius of convergence for the function that is -1< r
@@kennethgee2004 arctan(x) is a function so it doesn’t account for all the multiple periods of tan(x). That’s fine tho, because the integral being evaluated to arctan(x), 1/(1+x^2), always uses the standard definition of arctan(x).
@@baconboyxy so arctan(x) is still an angle and there is still radians as that is just a number. You argument is not correct in the slightest. a function must be continuous which arctan is not.
@@kennethgee2004 arctan is continuous, though? And differentiable for that matter. I’d recommend looking at the graph or looking up the integral of 1/(1+x^2) or something
Bullshit video ….he replaced one intergral by x^3 and no one challenge him.
?
finally 1st :D
AsymptoticSum[(-Log[1 - t x]/t)/z /. t -> n/z, {n, 1, z}, z -> Infinity]