We calculate the definite integral of ln(cos x) over the interval from 0 to pi/2. Playlist: • Interesting Integrals www.michael-penn.net www.randolphcollege.edu/mathem...
It's amazing, I understand it, but I would never get to these ideas, especially using the trigonometric identities, but that's the reason why I failed my math studies: I was interested but never reached this manditory meta level ... But I like this videos, because even, If I failed, I never broke up with the mathmatics. Greetings from Germany, your vids help me to come to these hard corona times in social distance. Stay healthy :-)
I have the same thing. I love to see such beautiful solutions and I loved to try to find them, but I lacked the (as you called it) meta understanding to find them like that. But I still love math and still love to watch videos and learn more. Recently found this channel and absolutely love it.
Pretty interesting to see such an incredibly easy way to solve this integral! Almost a year ago, at my 2nd uni grade, I've used a parameter differentiate method to calculate it. More later, it was shown how to find a solution using complex calculus:D But the most amazing thing that I've got trying to figure out the answer was a nice series identity for the natural logarithm of 2!
@@ivanmaximenko7227 Similar thing is happening to me. next year im going to learn complex analysis can't wait to find simpler ways than this godforsaken method. Literally just smack your head on the chalkboard and hope to the math gods you dont make a mistake on any step.
Wow...this was really impressive. I've just started Calculus 2 and am learning integration by parts. It's a tricky process that requires so much knowledge in order to do efficiently/effectively. I'm okay on my Pythagorean identities, but I need to get all the double and half angle identities down. Definitely subscribed and will keep watching these cool problems. Thanks Michael!
Have you seen a geometric proof of the double angle formulas (or more generally the sum of angle formulas)? Its simple enough that once youve seen it you can quickly sketch it out again and and rederive the formulas if you forget. Not to mention they're just interesting!
This integral (with sinx instead of cosx) was calculated by Euler more than 250 years ago. You have used a similar idea, but Euler goes backwards, and that makes calculation so much easier. First, he does the substitution x=2t and makes I=ln2*pi/2+ two integrals of 2ln(sint) and 2ln(cost), both from 0 to pi/4. In the second one he does change t=pi/2-u, which makes that an integral of 2ln(sinu) from pi/4 to pi/2. Combining together gives him 2I and all said. It takes literally two lines!
Amazing, im from chile and i was triying to solve that integral for a while and when i give up, UA-cam suggest me this video, is like fallen from heaven
This is probably the first video where I knew exactly where the method was going. I must have seen a trick like this before. Perhaps in solving certain infinite sums that use the same looping substitution / equality trick.
It's awesome, the way you demonstrated, step by step. Thumbs up! I really liked the process which is clearly understandable . Thanks you so much. However i've just one question, I would like to know, is there an other trick more easier and faster than that you've just done to get to the solution ?
From the thumbnail I tried doing it in my head by simply integrating the initial term and my answer was negative infinity (obviously), and when I saw the video my jaw literally dropped (again, obviously) Such an elegant method :D
Thanks for the explanation above. I like this sort of maths but I can never achieved the mental acrobatic that you can do but I enjoy stressing my brain in following it. However, using my calculator, it cannot even come out with lncos(x) but using lnsin(x) I got a complex number answer as -1.0888 - 4.4409x10^16i.
Could you also do this integral by differentiating under the integral sign? Set I(a) = integral of ln(cos ax ) and differentiate with respect to a, then perform a u-sub?
Another way to get rid of ln(z) is to consider the derivative d(z^t)/dt which yield ln(z) at t = 0. Thus it suffices to calculate int_0^{\pi/2} (cos(x))^t dx which results in beta-function after trivial change of variables. After all it will be possible to represent the answer in terms of digamma function. But the simplest way here is to use cos(x) = (e^{ix} + e^{-ix})/2, so ln(cos(x)) = -ln(2) + ix + ln(1 + e^{-2 ix}) = -ln(2) + i x + sum_{n = 1}^\infty (-1)^{n - 1} e^{-2 i n x} / n. As ln(cos(x)) is a real number, we can throw away the imaginary part (because it is zero): ln(cos(x)) = -ln(2) + sum_{n = 1}^\infty (-1)^{n - 1} cos(2 n x)/n As integral from cos(2 n x) within the interval (0, pi/2) yields zero, we get -ln(2) * pi/2.
U-sub was the thing I remember students blowing off in college, because they would cram everything onto one line as well. I was taught to eat the paper vertically, and use a pencil. U-sub helped me a whole lot! It's like getting your own part score transposed for you from the orchestral score, but you're the conductor! Ok, yes, music is important to me, but also it's the quadrivial equivalent to calculus. And I love the seven Liberal Arts!
I am sorry but you need to prove that I exists! Your method IS NOT VALID because lim ln(cosx) = - ∞, when x → (π/2)⁻ (we have also lim ln(sinx) = -∞, when x → 0⁺) You made FORBIDDEN operations on the infinites! You used the same method than when you calculated ∫₀ₐ √sinx dx/(√sinx + √cosx) with a = π/2, which was very very easy to determinate, BUT NOT so easy with a = π/4 : try to do it! Try also to calculate K = ∫₀ₐ ln(cosx) dx with a = π/4 : not so easy, too! To serve you and your followers. Greetings from Paris.
I am sorry but you need to prove that I exists! Your method IS NOT VALID because lim ln(cosx) = - ∞, when x → (π/2)⁻ (we have also lim ln(sinx) = -∞, when x → 0⁺) You made FORBIDDEN operations on the infinites! You used the same method than when you calculated ∫₀ₐ √sinx dx/(√sinx + √cosx) with a = π/2, which was very very easy to determinate, BUT NOT so easy with a = π/4 : try to do it! Try also to calculate K = ∫₀ₐ ln(cosx) dx with a = π/4 : not so easy, too! To serve you and your followers. Greetings from Paris.
You explain how to solve the integral, but you should also explain why you do some of the things you do. For example, you could explain why you made that particular substitution near the beginning. If I knew why you did that it could very well be useful to me for future problems. Just something to consider.
03:09 We can do this because "∫ ln(sin(_) d_" means the same no matter which letter you put into "_" (the same is also for definite integral from any "a" to any "b", for example for definite integral from 0 to π/2). For example "∫ ln(sin(A) dA" means the same as "∫ ln(sin(α) dα" and "∫ ln(sin(u) du" and "∫ ln(sin(x) dx". So we use this property to change letter (to change the variable) from "u" to any letter we want. And in this case we want this "any letter" be "x" :) So instead of writing "∫ ln(sin(u) du" we could write "∫ ln(sin(x) dx". I hope this is helpful and makes it understable :)
Damian Matma but that’s still doesn’t change the fact that the original sub was pi/2 - x, he goes on to use the double angle identity, sin(A+B) = sin(A)cos(B) + sin(B)cos(A) and uses it to get 1/2sin(2x) but this is only true for when A=B which in this case it doesn’t does it? Because he’s saying x = pi/2- x for all x, which isn’t true
Tried using the fact that the definite integral between 0 to pi/2 is same for ln(sin(x)) and ln(cos(x)), then added both leading to the integral being half of the sum of those definite integrals. Then used ln a + ln b = ln ab which yields ln sinx + ln cos x = ln sin 2x - ln 2. This gives the identity I = I/2 - pi/4 ln 2
He's just renaming the variable inside the integral, it doesn't matter if it's called u or x or α or n, it's just a name for "the thing that moves inside the integral". You could also see it as a "substitution with u = x" (which doesn't change the bounds, since you're not doing any change to the variable) if that makes it clearer for you.
I thought of writing cosx as Re(e^ix) and then take the real part all the way to the outside of the integral, so you have Re(integ(ln(e^ix))= Re(integ(ix))= Re(i/2*x^2) from 0 to pi/2 =Re(i*pi^2/8 + 0) =0 Obviously this is wrong but where exactly? Did I use the real part wrong?
Careful, you have written (line 1 - 2): « ln(exp(ix)) = ix » because you thought that ln can be applied to complex exponential like the natural exponential fonction, which is not the case (it more complicated), thus, here lies the error. The ln function you are using is defined in R+* (or ]0; +Infinity[) whereas exp(ix) is a complex.
I was curiosas as to why he used u=x in the second substitution instead of u=pi/2 - x to return to the x world. Wouldn’t using u=x take him into a different variable? I know pi\2 is a constant but it must influence it some way right?
@@brianart8700 I'm missing something too; when he puts the two 'I's together, those x variables represent different things. If they don't, if u = x, then by his first subs x = pi/2 - x, then x = pi/4. The equation of 2 I = ..., this is only true when x = pi/4
Jason Krause Yes, I agree. I do Engineering for a living so I have some math background. On my free time, I like to find these problems and solve them prior to seeing the solution. This is one of those that did not add up.
Brian Art When you are working with an indefinite integral (where the result is a function) you have to change the variable at the end, and you cannot just say u=x for example, and in this context you’re right, but with definite integrals (where the result is a number) you can treat every variable like a dummy variable, because the area under the graph of f(x) and f(u) are the same. Keep in mind that you’re dealing with the change of variable issues when you find dx in terms of u and du, and when you change the bounds from x to u. Hope this helped, i can redo it if you want! If you have other questions feel free to ask!
I am sorry but you need to prove that I exists! Your method IS NOT VALID because lim ln(cosx) = - ∞, when x → (π/2)⁻ (we have also lim ln(sinx) = -∞, when x → 0⁺) You made FORBIDDEN operations on the infinites! You used the same method than when you calculated ∫₀ₐ √sinx dx/(√sinx + √cosx) with a = π/2, which was very very easy to determinate, BUT NOT so easy with a = π/4 : try to do it! Try also to calculate K = ∫₀ₐ ln(cosx) dx with a = π/4 : not so easy, too! To serve you and your followers. Greetings from Paris. P.S : in France: x → (π/2)⁻ means x → (π/2) and x < π/2
I am sorry but you need to prove that I exists! Your method IS NOT VALID because lim ln(cosx) = - ∞, when x → (π/2)⁻ (we have also lim ln(sinx) = -∞, when x → 0⁺) You made FORBIDDEN operations on the infinites! You used the same method than when you calculated ∫₀ₐ √sinx dx/(√sinx + √cosx) with a = π/2, which was very very easy to determinate, BUT NOT so easy with a = π/4 : try to do it! Try also to calculate K = ∫₀ₐ ln(cosx) dx with a = π/4 : not so easy, too! To serve you and your followers. Greetings from Paris.
The u-sub of x=u @ 3:05 seems a little fishy to me. I feel like the x on either side of the equation aren't exactly the same x, which would prevent us from using the the trig identities later. Anyone have any thoughts?
Sorry for answering after so much time, but the variable you use isn't important. In fact, if F is an antiderivative of ln(sin u), that integral is F(u) evaluated from u=0 to u=pi/2, which is F(pi/2) - F(0), but this is also F(x) evaluated from x=0 to x=pi/2. That's why you can use whatever variable and sometimes apply this substitution, just to deal with the same variable. Hope I was clear enough.
@@qing6045 No you're right, we need to bound it from below. So use, cos x > 1 - 2x/pi. So, ln cos x > ln(1 - 2x/pi), And the integral of the term on the right converges.
this is a common trick called dummy variable, the thing is that an integral is independent of its variable, below is an example: ∫ 2x dx = ∫ 2y dy they both mean finding an anti-derivative of f(z)=2z similarly, ∫ ln(sin u) du and ∫ ln(sin x) dx are both the anti-derivative of f(z)=ln(sin z), thus they are technically the same.
I am sorry but you are COMPLETELY WRONG; because I = J = -∞ !!! Your method IS NOT VALID because lim ln(cosx) = - ∞, when x → (π/2)⁻ (we have also lim ln(sinx) = -∞, when x → 0⁺) You made FORBIDDEN operations on the infinites !!! You used the same method than when you calculated ∫₀ₐ √sinx dx/(√sinx + √cosx) with a = π/2, which was very very easy to determinate, BUT NOT so easy with a = π/4 : try to do it ! Try also to calculate K = ∫₀ₐ ln(cosx) dx with a = π/4 : not so easy, too ! You made quite the same error than Ramanujan with his wrong identities, like this one: 1 + 2 + 3 + ..... = - 1/12 To serve you and your followers. Greetings from Paris.
I think probably with some expansion of ln(cos x) as a Fourier series. I am doing a follow up tomorrow where we look at a similar integral and do this expansion.
@@MichaelPennMath I look forward to it. More generally, how do you decide which content to create? It's all very fascinating. I am also generally curious about your background in mathematics. I am pursuing a Masters in Math at the moment and I am now looking at ideals and varieties in the hope of properly understanding commutative algebra and schemes.
Content: I was "flipping" two of my current courses (Calculus 3 and Abstract Algebra) using these videos. Now they have been moved online -- which luckily for me means I am set up to transition more smoothly than most. The other videos are a mixture of things I like (combinatorial identities, my Riemann Zeta function videos, my dilogarithm videos) and things that seem to get views and bring new subscribers ("fancy" integrals and Putnam problems). Background: I am a math professor at a small liberal arts college (Randolph College) in central Virginia. My research is related to the representation theory of infinite dimensional Lie algebras, vertex operator algebras, invariant theory, and mathematical physics. Every year or so I try and learn some modern algebraic geometry -- schemes, sheafs, stacks, etc -- but the abstraction kills me. Maybe I will commit to a video series to "force" myself to learn!
Too much time and effort went into proving that Integral of ln(cos x) equals that of ln(sin x). I think that is trivial! First notice that for each x in the interval [0,pi/2] there corresponds a y = (pi/2 -x) such that ln(cos x) = ln(sin y) ,so that the integrands match for every x, y, respectively . If we now consider the integral as a Riemann sum, the result is obvious! Great video, BTW
I am sorry but you aneed to prove that I exists! Your method IS NOT VALID because lim ln(cosx) = - ∞, when x → (π/2)⁻ (we have also lim ln(sinx) = -∞, when x → 0⁺) You made FORBIDDEN operations on the infinites! You used the same method than when you calculated ∫₀ₐ √sinx dx/(√sinx + √cosx) with a = π/2, which was very very easy to determinate, BUT NOT so easy with a = π/4 : try to do it! Try also to calculate K = ∫₀ₐ ln(cosx) dx with a = π/4 : not so easy, too! To serve you and your followers. Greetings from Paris.
substitution is an equivalent process, so the name of the variable doesn’t matter. it won’t change the value of the integration. he could have changed it to t, or w, or whatever, he just changed it back to x for the sake of consistency.
Strictly speaking it was a "mistake" or bad notation but really this is how you should think of definite integrals. The variable inside is nothing but a dummy variable. Once you get used to this you'll yawn and not keep track of what letters you have already used.
I was wondering something similar and looking down here for answers. How can he say u = 2x and then in the next step say x = u to get the variable back to x? He did this twice. I guess it doesn't matter.
It was cool to see the solution, but it makes me wonder about how Michael chose his strategy for computing it. What I mean is after the substitutions and change of limits, there's still a natural log of a trig function in the integrand. If I were doing this problem, that would make me think I was on the wrong track *unless* I could see several steps "into the future" and realize that this strategy is really all about figuring out a relation for the integral-- and not explicitly computing it. How does one know that this strategy will work? How do you know that by making these substitutions, you will end up with a useable relation for the integral?
At some point you will find in your studies that experience plays a key part in how you solve things after looking at them. He probably pounded away at substitutions for a hot minute before realizing it was going in circles. At this point it reminded me of the special integral of ((e^x)*sin(x)) that one has a very similar methodology
@@auzzffozzie4309 Thanks, I was wondering if there was some clue about this problem that suggested this approach as solution. Is there a name for this approach of making substitutions to arrive at some expression that gives you an equation for the integral (without actually computing the integral)? My approach would have been to attempt to integrate this as series expansion with the aspiration of cancelling all but one term!
@@VIRUS200086 Well, that's not a proof at all. Because he didn't use a limit on an antiderivative, getting an answer doesn't prove convergence. That algebra is true, if and only if the integral converges, but if not, it would just be a logical fallacy. All that algebra would be like dividing by zero--getting an answer means nothing because what you're working with is undefined.
@@christiankotait2954 That doesn't work. Your inequality is true, but both functions are NEGATIVE, so the more negative one has the greater absolute area. Therefore, cosx - 1 converging on that interval doesn't prove convergence of ln(cosx).
I do not understand the substitution that you do at 8:50: why can you transform the integral of ln(sin u) in the integral of ln(sin x), in u is not equal to x but to 2x?
I tried to do this integral using Integration by parts, differentiating ln(cos(x)) and integrating 1 with respect to x. I ended up with having to integrate xtan(x), and that's where I got stuck. I tried to convert tan(x) into its complex form, but that didn't help me a lot. Do you think it's doable this way? Other than that, I loved the video. Your way of explaining the solution is very clear.
I ended up trying to do byparts ending with an Int(xtanx.dx) term also, how did you figure this one out? PS I know that Int(xtanx.dx)= Int(x [ln(secx)]'.dx) but that just brings you back to start again in the original byparts! Finally noticed (as did *@Vijay Bhaskar* above): since cos(x).sin(x) = (1/2)sin(2x), ln(cosx.sinx) = -ln(2)+ln(sin(2x)) => ln(cosx)+ln(sinx) = -ln(2)+ln(sin(2x)) taking def. int both sides: 2I = -(pi/2)ln2 + def.int{ln(sin2x).dx} 2I = -(pi/2)ln2 + I ... since last def. int is same as I or I = -(pi/2)ln2.
It is curious that you may not use Euler's formula to solve this...maybe there is a way...but it must be more complicated. If you take the natural logarithm of Euler's formula it is directly (cos x + i sen x). But your way is the best, well done ¡¡
I would say your method I’d way too long. Is it possible to use integration by parts twice to get the answer much faster? Let me give that a try to see how it turns out. Thanks.
I am a huge fan, but this one seems odd. I feel like michael penn did nearly the same U substitution over and over and over. Is there really not a way to solve this with fewer steps? Michael? Anyone?? .?
No, each substitution may (or may not) be different and lead you to a new expression which is equal. That's why he goes back to x as a variable every time, as a «new starting point» even if the variable is free.
@@Le_Tchouck What keeps us from changing the equation then. Ex: evaluate the indefinite integral: int(xdx). x=u+2, so dx=du. int((u+2)du) so we get 1/2*u^2 + 2u + C. Don't we have to plug in x-2 for x for it to be correct, otherwise we could say u=x. so int(xdx) = 1/2*x^2 + 2x + C. I don't think I quite understand.
@@addisonballif2988 When you perform a change of variables, you have to change the bounds of the integral accordingly If you integrate x dx from 0 to t, and then substitute x = u+2, your integral becomes int (u+2) du from -2 to t-2, because when x goes from 0 to t, u = x-2 goes from -2 to t-2. Both will give the same result
@@Le_Tchouck : I find a lot of people questioning "y = 5, so how can y now equal something else?!". One the variable is free, it can be reused with impunity. Sadly, many programmers--as well as high school algebra students--lack this fundamental concept, riddling their code with superfluous variables, or stepping on their own scope. Nice comment!
@@addisonballif2988 As said, be careful with the boundaries. Integration returns a number, primitive (anti-derivative) returns a function. No "+ C" for integration. Knowing the «aera under the curve» definition, you can feel like the x=u+c variable change type is kind of a translation. x=k*u is homethetie and after that, variable change gets too complicated to be easily seen on graphic. For the variable change, it's kinda like saying f(x)=3x is equivalent to f(u)=3u, except that the integral being a number (when defined) and not a function, what you might think as a variable is not. Because it doesn't expect a single value, it walk on a continuum of values. I hope it helps, Im working on my «maths English» and struggle a bit.
It's amazing, I understand it, but I would never get to these ideas, especially using the trigonometric identities, but that's the reason why I failed my math studies: I was interested but never reached this manditory meta level ...
But I like this videos, because even, If I failed, I never broke up with the mathmatics.
Greetings from Germany, your vids help me to come to these hard corona times in social distance.
Stay healthy :-)
I am happy to hear you like the videos! Luckily my chalkboard is in my basement, so I can make videos without venturing out.
I have the same thing. I love to see such beautiful solutions and I loved to try to find them, but I lacked the (as you called it) meta understanding to find them like that. But I still love math and still love to watch videos and learn more. Recently found this channel and absolutely love it.
Pretty interesting to see such an incredibly easy way to solve this integral! Almost a year ago, at my 2nd uni grade, I've used a parameter differentiate method to calculate it. More later, it was shown how to find a solution using complex calculus:D But the most amazing thing that I've got trying to figure out the answer was a nice series identity for the natural logarithm of 2!
@@ivanmaximenko7227 Similar thing is happening to me. next year im going to learn complex analysis can't wait to find simpler ways than this godforsaken method. Literally just smack your head on the chalkboard and hope to the math gods you dont make a mistake on any step.
Don't you think you could reach that meta level if you keep practicing?
What a sly method. Absolutely amazing
Just watch this impressive Math channel ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
Wow...this was really impressive. I've just started Calculus 2 and am learning integration by parts. It's a tricky process that requires so much knowledge in order to do efficiently/effectively. I'm okay on my Pythagorean identities, but I need to get all the double and half angle identities down. Definitely subscribed and will keep watching these cool problems. Thanks Michael!
Have you seen a geometric proof of the double angle formulas (or more generally the sum of angle formulas)? Its simple enough that once youve seen it you can quickly sketch it out again and and rederive the formulas if you forget. Not to mention they're just interesting!
ua-cam.com/video/y_XwQkchwrE/v-deo.html
Just watch this impressive Math channel ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
Whenever I see a new video like this I say "Ok, great"
Makes you wonder why. Is it faster to compute than an antiderivative for a computer or something?
“U equals zero” damn he’s right
bruh
This integral (with sinx instead of cosx) was calculated by Euler more than 250 years ago. You have used a similar idea, but Euler goes backwards, and that makes calculation so much easier. First, he does the substitution x=2t and makes I=ln2*pi/2+ two integrals of 2ln(sint) and 2ln(cost), both from 0 to pi/4. In the second one he does change t=pi/2-u, which makes that an integral of 2ln(sinu) from pi/4 to pi/2. Combining together gives him 2I and all said. It takes literally two lines!
Amazing, im from chile and i was triying to solve that integral for a while and when i give up, UA-cam suggest me this video, is like fallen from heaven
This is probably the first video where I knew exactly where the method was going. I must have seen a trick like this before. Perhaps in solving certain infinite sums that use the same looping substitution / equality trick.
It's awesome, the way you demonstrated, step by step. Thumbs up! I really liked the process which is clearly understandable . Thanks you so much.
However i've just one question, I would like to know, is there an other trick more easier and faster than that you've just done to get to the solution ?
Great video Michael. If you integrate x/tanx from pi/2 to 0 (using integration by parts) you will use all the properties you've specified here
From the thumbnail I tried doing it in my head by simply integrating the initial term and my answer was negative infinity (obviously), and when I saw the video my jaw literally dropped (again, obviously)
Such an elegant method :D
How did you "simply integrate" that?
What do you mean?
this is what Neuroscientist call it. the " HA" moment. That's when you find the hidden pattern within a pixellized photo for example.
I did it wrong in my head guys
Integral from 0 to pi of ln(sin x) equals 2 times “I” since sin x is an even function with respect to x=pi/2, and so is ln(sin x)
Yes, I thought exactly the same thing but I think his substitution argument is a lot more rigorous.
Thanks for the explanation above. I like this sort of maths but I can never achieved the mental acrobatic that you can do but I enjoy stressing my brain in following it. However, using my calculator, it cannot even come out with lncos(x) but using lnsin(x) I got a complex number answer as -1.0888 - 4.4409x10^16i.
Could you also do this integral by differentiating under the integral sign? Set I(a) = integral of ln(cos ax ) and differentiate with respect to a, then perform a u-sub?
amazing, i believe such beautiful solution comes from countless practices.
Like any language => Art, e.g. math, writing, music--performance, fine, or utilitarian.
And more...
Another way to get rid of ln(z) is to consider the derivative d(z^t)/dt which yield ln(z) at t = 0.
Thus it suffices to calculate int_0^{\pi/2} (cos(x))^t dx which results in beta-function after trivial change of variables. After all it will be possible to represent the answer in terms of digamma function.
But the simplest way here is to use cos(x) = (e^{ix} + e^{-ix})/2,
so ln(cos(x)) = -ln(2) + ix + ln(1 + e^{-2 ix}) = -ln(2) + i x + sum_{n = 1}^\infty (-1)^{n - 1} e^{-2 i n x} / n.
As ln(cos(x)) is a real number, we can throw away the imaginary part (because it is zero):
ln(cos(x)) = -ln(2) + sum_{n = 1}^\infty (-1)^{n - 1} cos(2 n x)/n
As integral from cos(2 n x) within the interval (0, pi/2) yields zero,
we get -ln(2) * pi/2.
Just watch this impressive Math channel ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
Taking the integral to the u world
And back. Twice.
ua-cam.com/video/y_XwQkchwrE/v-deo.html
Just watch this impressive Math channel ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
Will you do the convergence method for the integral?
Could you still do the dummy substitution thing if it is an indefinite integral?
I love Dr Penn's videos, but am I the only one who has to pause the video so my ears can catch their breath?
I had first time understand it's proof. 🤔
In the beginning I just memorized the whole thing.🔥
Thanks for these beautiful and elegant explanation 🥰🔥
Just watch this impressive Math channel ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
That's really great Michael!
ua-cam.com/video/y_XwQkchwrE/v-deo.html
The dummy variables changes back to x . That helps a lot. And you set the integral equal to I , it helps a lot too.
U-sub was the thing I remember students blowing off in college, because they would cram everything onto one line as well. I was taught to eat the paper vertically, and use a pencil.
U-sub helped me a whole lot! It's like getting your own part score transposed for you from the orchestral score, but you're the conductor! Ok, yes, music is important to me, but also it's the quadrivial equivalent to calculus. And I love the seven Liberal Arts!
@@argonwheatbelly637 I don't know why, but I find cool the fact that you related maths with your Passion with music in that way
I am sorry but you need to prove that I exists! Your method IS NOT VALID because lim ln(cosx) = - ∞, when x → (π/2)⁻ (we have also lim ln(sinx) = -∞, when x → 0⁺)
You made FORBIDDEN operations on the infinites!
You used the same method than when you calculated ∫₀ₐ √sinx dx/(√sinx + √cosx) with a = π/2, which was very very easy to determinate, BUT NOT so easy with a = π/4 :
try to do it!
Try also to calculate K = ∫₀ₐ ln(cosx) dx with a = π/4 : not so easy, too!
To serve you and your followers.
Greetings from Paris.
Or simply use 2 properties of definite integration
But it is good that u showed us each step
Very beautiful way to solve this integral
Michael Penn, what makes you know how to solve in this way ? just curious
Blessed UA-cam Recommendation 🙏
ua-cam.com/video/y_XwQkchwrE/v-deo.html
I loved that technique, great!
I am sorry but you need to prove that I exists! Your method IS NOT VALID because lim ln(cosx) = - ∞, when x → (π/2)⁻ (we have also lim ln(sinx) = -∞, when x → 0⁺)
You made FORBIDDEN operations on the infinites!
You used the same method than when you calculated ∫₀ₐ √sinx dx/(√sinx + √cosx) with a = π/2, which was very very easy to determinate, BUT NOT so easy with a = π/4 :
try to do it!
Try also to calculate K = ∫₀ₐ ln(cosx) dx with a = π/4 : not so easy, too!
To serve you and your followers.
Greetings from Paris.
You explain how to solve the integral, but you should also explain why you do some of the things you do. For example, you could explain why you made that particular substitution near the beginning. If I knew why you did that it could very well be useful to me for future problems. Just something to consider.
Just amazing! Thankyou!
ua-cam.com/video/y_XwQkchwrE/v-deo.html
I was wondering where you were headed with this - it seemed you were going around in circles!
厉害啊，这个证明过程真的精彩
Can't we use by parts to solve in wothout any substitution
I've found a new way to do this integral but it involves gamma functions and exponential integrals. = D
ua-cam.com/video/y_XwQkchwrE/v-deo.html
Yup that one is pretty easy
Mathematics ,"feeling so good today" thank you
ua-cam.com/video/y_XwQkchwrE/v-deo.html
Why can we do change of variables back to x x = u, when we have ever used the substitution x = pi/2 - u, thanks for your answer
It isn't super necessary, the main reason is to make everything "look" the same so that the integrals can be easily compared and combined.
03:09 We can do this because "∫ ln(sin(_) d_" means the same no matter which letter you put into "_" (the same is also for definite integral from any "a" to any "b", for example for definite integral from 0 to π/2).
For example "∫ ln(sin(A) dA" means the same as "∫ ln(sin(α) dα" and "∫ ln(sin(u) du" and "∫ ln(sin(x) dx".
So we use this property to change letter (to change the variable) from "u" to any letter we want.
And in this case we want this "any letter" be "x" :)
So instead of writing "∫ ln(sin(u) du" we could write "∫ ln(sin(x) dx".
I hope this is helpful and makes it understable :)
@@damianmatma708 thank you very much
@@MichaelPennMath This only works with definite integrals, right?
Damian Matma but that’s still doesn’t change the fact that the original sub was pi/2 - x, he goes on to use the double angle identity, sin(A+B) = sin(A)cos(B) + sin(B)cos(A) and uses it to get 1/2sin(2x) but this is only true for when A=B which in this case it doesn’t does it? Because he’s saying x = pi/2- x for all x, which isn’t true
Tried using the fact that the definite integral between 0 to pi/2 is same for ln(sin(x)) and ln(cos(x)), then added both leading to the integral being half of the sum of those definite integrals. Then used ln a + ln b = ln ab which yields ln sinx + ln cos x = ln sin 2x - ln 2.
This gives the identity I = I/2 - pi/4 ln 2
But will not I/2 = integral(ln(cos^2x)) which I think will not be equal to ln(cosx)
How do we show this function is integrable?
すげぇ
Amazing, I thought this was an impossible one!!
Wouldn't it be easier if you do u=cos(x) and then do an integration by parts?
how do you calculate "integrale ln(2+cosx) dx"?
is it okay to locate zero in logarithm?
Why does integration by parts udv=uv-vdu not work
Make sense that the identical +ve areas under two similar trig curves 0->pi/2 could end up being the same
If you were to plot the cos(x) and sin(x) (without the log) from 0 to pi/2 one would realize immediately the symmetries.
It's amazing!
At 12:23 when turning u back into x why didn’t the bounds on the integral change too?
He's just renaming the variable inside the integral, it doesn't matter if it's called u or x or α or n, it's just a name for "the thing that moves inside the integral". You could also see it as a "substitution with u = x" (which doesn't change the bounds, since you're not doing any change to the variable) if that makes it clearer for you.
I thought of writing cosx as Re(e^ix) and then take the real part all the way to the outside of the integral, so you have
Re(integ(ln(e^ix))=
Re(integ(ix))=
Re(i/2*x^2) from 0 to pi/2
=Re(i*pi^2/8 + 0)
=0
Obviously this is wrong but where exactly? Did I use the real part wrong?
The Re does not commute with the ln
Careful, you have written (line 1 - 2):
« ln(exp(ix)) = ix »
because you thought that ln can be applied to complex exponential like the natural exponential fonction, which is not the case (it more complicated), thus, here lies the error.
The ln function you are using is defined in R+* (or ]0; +Infinity[) whereas exp(ix) is a complex.
One minor variation at the end. I used u = pi - x instead for the last substitution.
I was curiosas as to why he used u=x in the second substitution instead of u=pi/2 - x to return to the x world. Wouldn’t using u=x take him into a different variable? I know pi\2 is a constant but it must influence it some way right?
@@brianart8700 I'm missing something too; when he puts the two 'I's together, those x variables represent different things. If they don't, if u = x, then by his first subs x = pi/2 - x, then x = pi/4. The equation of 2 I = ..., this is only true when x = pi/4
Jason Krause Yes, I agree. I do Engineering for a living so I have some math background. On my free time, I like to find these problems and solve them prior to seeing the solution. This is one of those that did not add up.
Brian Art When you are working with an indefinite integral (where the result is a function) you have to change the variable at the end, and you cannot just say u=x for example, and in this context you’re right, but with definite integrals (where the result is a number) you can treat every variable like a dummy variable, because the area under the graph of f(x) and f(u) are the same. Keep in mind that you’re dealing with the change of variable issues when you find dx in terms of u and du, and when you change the bounds from x to u. Hope this helped, i can redo it if you want! If you have other questions feel free to ask!
I am sorry but you need to prove that I exists! Your method IS NOT VALID because lim ln(cosx) = - ∞, when x → (π/2)⁻ (we have also lim ln(sinx) = -∞, when x → 0⁺)
You made FORBIDDEN operations on the infinites!
You used the same method than when you calculated ∫₀ₐ √sinx dx/(√sinx + √cosx) with a = π/2, which was very very easy to determinate, BUT NOT so easy with a = π/4 :
try to do it!
Try also to calculate K = ∫₀ₐ ln(cosx) dx with a = π/4 : not so easy, too!
To serve you and your followers.
Greetings from Paris.
P.S : in France: x → (π/2)⁻ means x → (π/2) and x < π/2
This could be done pretty quickly using the complex definitions of cos? The e terms would just cancel?
No, not really, since it would just be ln([e^ix+e^-ix]/2), which =/= ln(e^ix)+ln(e^-ix)
can you solve that for any bounds?
I did it with a bunch of stupidly complicated stuff and it took like 2 hours and here he does it in 13 minutes
I am sorry but you need to prove that I exists! Your method IS NOT VALID because lim ln(cosx) = - ∞, when x → (π/2)⁻ (we have also lim ln(sinx) = -∞, when x → 0⁺)
You made FORBIDDEN operations on the infinites!
You used the same method than when you calculated ∫₀ₐ √sinx dx/(√sinx + √cosx) with a = π/2, which was very very easy to determinate, BUT NOT so easy with a = π/4 :
try to do it!
Try also to calculate K = ∫₀ₐ ln(cosx) dx with a = π/4 : not so easy, too!
To serve you and your followers.
Greetings from Paris.
This problem can be solved using Maclaurin's expansion
Very elegant solution.
In order to prove that ln(cos x) is integrable on the interval we can use the comparison test
Gracias ahora ya pude entender mejor las integrales
And that is a good place to finish this video!
How about integral from 0 to pi/4
Can someone tell me why if I use Re{e^ix} and sub into the equation...which I believe will result in Re{ix} which equals 0 which then integrates to 0.
Re(ln(exp(ix))) is not the same as ln(Re(exp(ix)))
The u-sub of x=u @ 3:05 seems a little fishy to me. I feel like the x on either side of the equation aren't exactly the same x, which would prevent us from using the the trig identities later. Anyone have any thoughts?
Sorry for answering after so much time, but the variable you use isn't important. In fact, if F is an antiderivative of ln(sin u), that integral is F(u) evaluated from u=0 to u=pi/2, which is F(pi/2) - F(0), but this is also F(x) evaluated from x=0 to x=pi/2. That's why you can use whatever variable and sometimes apply this substitution, just to deal with the same variable. Hope I was clear enough.
The real fun is to prove the integral does indeed converge..
Bound it from above with 0, and bound ln(sin x) from below with ln x, which gives a finite integral.
@@Ricocossa1 how to bound to x for x near pi/2
@@qing6045 x > sin x for x between 0 and pi/2
@@qing6045 No you're right, we need to bound it from below. So use,
cos x > 1 - 2x/pi.
So,
ln cos x > ln(1 - 2x/pi),
And the integral of the term on the right converges.
No it’s boring
Very good!
Thanks
Can someone explain how Michael can do a change of variable at 2:56 by defining x=u? I thought he defined x=pi/2-u earlier, why can he change it?
this is a common trick called dummy variable, the thing is that an integral is independent of its variable, below is an example:
∫ 2x dx = ∫ 2y dy
they both mean finding an anti-derivative of f(z)=2z
similarly, ∫ ln(sin u) du and ∫ ln(sin x) dx are both the anti-derivative of f(z)=ln(sin z), thus they are technically the same.
y is definitely an imaginary number in the range of x:p1/2i~3/2pi. What is the value of the definite integral of the range?
I am sorry but you are COMPLETELY WRONG; because I = J = -∞ !!! Your method IS NOT VALID because lim ln(cosx) = - ∞, when x → (π/2)⁻ (we have also lim ln(sinx) = -∞, when x → 0⁺)
You made FORBIDDEN operations on the infinites !!!
You used the same method than when you calculated ∫₀ₐ √sinx dx/(√sinx + √cosx) with a = π/2, which was very very easy to determinate, BUT NOT so easy with a = π/4 :
try to do it !
Try also to calculate K = ∫₀ₐ ln(cosx) dx with a = π/4 : not so easy, too !
You made quite the same error than Ramanujan with his wrong identities, like this one: 1 + 2 + 3 + ..... = - 1/12
To serve you and your followers.
Greetings from Paris.
Could you do this if it were an indeterminated integral?
you mean indefinite?
What do you think is the easiest way to argue why this converges?
I think probably with some expansion of ln(cos x) as a Fourier series. I am doing a follow up tomorrow where we look at a similar integral and do this expansion.
@@MichaelPennMath I look forward to it. More generally, how do you decide which content to create? It's all very fascinating.
I am also generally curious about your background in mathematics. I am pursuing a Masters in Math at the moment and I am now looking at ideals and varieties in the hope of properly understanding commutative algebra and schemes.
Note that cosx>=1-(2/pi)*x for 0=
Content: I was "flipping" two of my current courses (Calculus 3 and Abstract Algebra) using these videos. Now they have been moved online -- which luckily for me means I am set up to transition more smoothly than most. The other videos are a mixture of things I like (combinatorial identities, my Riemann Zeta function videos, my dilogarithm videos) and things that seem to get views and bring new subscribers ("fancy" integrals and Putnam problems).
Background: I am a math professor at a small liberal arts college (Randolph College) in central Virginia. My research is related to the representation theory of infinite dimensional Lie algebras, vertex operator algebras, invariant theory, and mathematical physics.
Every year or so I try and learn some modern algebraic geometry -- schemes, sheafs, stacks, etc -- but the abstraction kills me. Maybe I will commit to a video series to "force" myself to learn!
なるほど。そうやるのか！！Amazingすぎる
I might be missing something,but the log has a singularity at x=0, why does it still workout?
Too much time and effort went into proving that Integral of ln(cos x) equals that of ln(sin x). I think that is trivial! First notice that for each x in the interval [0,pi/2] there corresponds a y = (pi/2 -x) such that ln(cos x) = ln(sin y) ,so that the integrands match for every x, y, respectively . If we now consider the integral as a Riemann sum, the result is obvious!
Great video, BTW
I am sorry but you aneed to prove that I exists! Your method IS NOT VALID because lim ln(cosx) = - ∞, when x → (π/2)⁻ (we have also lim ln(sinx) = -∞, when x → 0⁺)
You made FORBIDDEN operations on the infinites!
You used the same method than when you calculated ∫₀ₐ √sinx dx/(√sinx + √cosx) with a = π/2, which was very very easy to determinate, BUT NOT so easy with a = π/4 :
try to do it!
Try also to calculate K = ∫₀ₐ ln(cosx) dx with a = π/4 : not so easy, too!
To serve you and your followers.
Greetings from Paris.
Ahh now I realize why my teacher wanted me to just remember it
I’m afraid you and your teacher completely missed the point then.
Wowwwwwww
Can someone tell how he repose u = x at min 3
Initially he posed u = pi/2 - x
Wtf
The 'u's are diferent variables in each part.
substitution is an equivalent process, so the name of the variable doesn’t matter. it won’t change the value of the integration. he could have changed it to t, or w, or whatever, he just changed it back to x for the sake of consistency.
Strictly speaking it was a "mistake" or bad notation but really this is how you should think of definite integrals. The variable inside is nothing but a dummy variable. Once you get used to this you'll yawn and not keep track of what letters you have already used.
That is the differential dx=du
I was wondering something similar and looking down here for answers. How can he say u = 2x and then in the next step say x = u to get the variable back to x? He did this twice. I guess it doesn't matter.
So, basically the area under this curve evaluates to a negative value? Does it simply mean that the plot of ln(cosx) lies below the x-axis?
Just watch this impressive Math channel ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
I was able to do this when I was in school.
Love your videos ❤️
Very nice
It was cool to see the solution, but it makes me wonder about how Michael chose his strategy for computing it.
What I mean is after the substitutions and change of limits, there's still a natural log of a trig function in the integrand. If I were doing this problem, that would make me think I was on the wrong track *unless* I could see several steps "into the future" and realize that this strategy is really all about figuring out a relation for the integral-- and not explicitly computing it.
How does one know that this strategy will work? How do you know that by making these substitutions, you will end up with a useable relation for the integral?
At some point you will find in your studies that experience plays a key part in how you solve things after looking at them. He probably pounded away at substitutions for a hot minute before realizing it was going in circles. At this point it reminded me of the special integral of ((e^x)*sin(x)) that one has a very similar methodology
@@auzzffozzie4309 Thanks, I was wondering if there was some clue about this problem that suggested this approach as solution. Is there a name for this approach of making substitutions to arrive at some expression that gives you an equation for the integral (without actually computing the integral)? My approach would have been to attempt to integrate this as series expansion with the aspiration of cancelling all but one term!
Does it apply to the wakanda maths tho ?
this resolution is worthy of the gold diggers who end up finding a nugget !
@@mr.knight8967 ua-cam.com/video/f8yXkqqqeFs/v-deo.html
How about switching the order of the functions? Could you integrate cos( ln x ) ?
You can try by parts
good démonstration!.
Just watch this impressive Math channel ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
I wonder how you could prove that the integral is convergent even though you have no idea what the antiderivative is ?
You have to look at the behavior of the function at the bounds on integration. Looking at a graph of the function with the bounds in mind also helps!
For all u >0, ln(u)
I know it's not what you asked but do you agree that the video itself is a proof that the integral converges?
@@VIRUS200086 Well, that's not a proof at all. Because he didn't use a limit on an antiderivative, getting an answer doesn't prove convergence. That algebra is true, if and only if the integral converges, but if not, it would just be a logical fallacy. All that algebra would be like dividing by zero--getting an answer means nothing because what you're working with is undefined.
@@christiankotait2954 That doesn't work. Your inequality is true, but both functions are NEGATIVE, so the more negative one has the greater absolute area. Therefore, cosx - 1 converging on that interval doesn't prove convergence of ln(cosx).
Verry good.👍
Refreshing !
I do not understand the substitution that you do at 8:50: why can you transform the integral of ln(sin u) in the integral of ln(sin x), in u is not equal to x but to 2x?
Salvatore Di Lorenzo The u's are different variables. If it makes it easier, think of the second u as a w instead, or as u1 and u2.
@@Stop.Arguing Why are they different variables?
Thanks a lot
I tried to do this integral using Integration by parts, differentiating ln(cos(x)) and integrating 1 with respect to x.
I ended up with having to integrate xtan(x), and that's where I got stuck.
I tried to convert tan(x) into its complex form, but that didn't help me a lot. Do you think it's doable this way?
Other than that, I loved the video. Your way of explaining the solution is very clear.
Elyâdés Can’t you just use integration by parts on int x*tanx dx? Doable provided you know how to integrate tanx.
Yes, it is doable that way.
blackpenredpen did it:
ua-cam.com/video/PthIehmcNKA/v-deo.htmlm26s
Thanks, to both of you. I figured it out
@@damianmatma708 your video is the integral of cos(ln x), this one is ln (cosx)
I ended up trying to do byparts ending with an Int(xtanx.dx) term also, how did you figure this one out?
PS I know that Int(xtanx.dx)= Int(x [ln(secx)]'.dx) but that just brings you back to start again in the original byparts!
Finally noticed (as did *@Vijay Bhaskar* above):
since cos(x).sin(x) = (1/2)sin(2x),
ln(cosx.sinx) = -ln(2)+ln(sin(2x))
=> ln(cosx)+ln(sinx) = -ln(2)+ln(sin(2x))
taking def. int both sides:
2I = -(pi/2)ln2 + def.int{ln(sin2x).dx}
2I = -(pi/2)ln2 + I ... since last def. int is same as I
or I = -(pi/2)ln2.
It is curious that you may not use Euler's formula to solve this...maybe there is a way...but it must be more complicated. If you take the natural logarithm of Euler's formula it is directly (cos x + i sen x). But your way is the best, well done ¡¡
How do you go from ln(cos x) to cos x + i sin x ? They are clearly not equal
9:57 You could have simply wrote that 1/2 integral of ln(sin x) dx, from 0 to pi as I
Please do videos about Elliptic integrals.
ua-cam.com/video/y_XwQkchwrE/v-deo.html
Real life most integrals are impossible to solve! Long live Tylor approximations!
Muito bom
I would say your method I’d way too long. Is it possible to use integration by parts twice to get the answer much faster? Let me give that a try to see how it turns out. Thanks.
It won't work with by parts
You will eventually come to the same integral after two integrations
친절하게 영어로 설명해주시지만 영어를 몰라 알아들을 수 없는 것이 아쉽다. 하지만 미적분식은 아주 쉽게 설명해주시는 게 보인다.
This is relaxing on so many levels...because to me, math is one of the Language Arts, not merely a Science; this being a fun story to read.
Would have solved it totaly diffrent, but interesting seeing a norher perspective of the same thing
13:52
I am a huge fan, but this one seems odd. I feel like michael penn did nearly the same U substitution over and over and over. Is there really not a way to solve this with fewer steps? Michael? Anyone??
.?
Wow very nice.....
At 3:01, How can you say that x = u if you already said that x = pi/2 - u. Don't you have to keep the same substitution through the whole problem?
No, each substitution may (or may not) be different and lead you to a new expression which is equal. That's why he goes back to x as a variable every time, as a «new starting point» even if the variable is free.
@@Le_Tchouck What keeps us from changing the equation then. Ex: evaluate the indefinite integral: int(xdx). x=u+2, so dx=du. int((u+2)du) so we get 1/2*u^2 + 2u + C. Don't we have to plug in x-2 for x for it to be correct, otherwise we could say u=x. so int(xdx) = 1/2*x^2 + 2x + C. I don't think I quite understand.
@@addisonballif2988 When you perform a change of variables, you have to change the bounds of the integral accordingly
If you integrate x dx from 0 to t, and then substitute x = u+2, your integral becomes int (u+2) du from -2 to t-2, because when x goes from 0 to t, u = x-2 goes from -2 to t-2.
Both will give the same result
@@Le_Tchouck : I find a lot of people questioning "y = 5, so how can y now equal something else?!". One the variable is free, it can be reused with impunity. Sadly, many programmers--as well as high school algebra students--lack this fundamental concept, riddling their code with superfluous variables, or stepping on their own scope. Nice comment!
@@addisonballif2988 As said, be careful with the boundaries. Integration returns a number, primitive (anti-derivative) returns a function. No "+ C" for integration.
Knowing the «aera under the curve» definition, you can feel like the x=u+c variable change type is kind of a translation. x=k*u is homethetie and after that, variable change gets too complicated to be easily seen on graphic.
For the variable change, it's kinda like saying f(x)=3x is equivalent to f(u)=3u, except that the integral being a number (when defined) and not a function, what you might think as a variable is not. Because it doesn't expect a single value, it walk on a continuum of values.
I hope it helps, Im working on my «maths English» and struggle a bit.