Integral of (ln(cos x))^3

okay in this video we're going to complete the trilogy of calculating the integral of the natural log of cosine to some power so we started off with the natural log of cosine to the first power then we moved on to the natural log of cosine squared and now we're gonna finish it off with the natural log of cosine cubed obviously you could keep going but I think this is a good place to stop because we see all of the tricks that are involved in such a problem once we get to this point so previously in this series we had these following four facts that we've used so the first one is that if we take the integral from zero to PI over two of the natural log of cosine we get minus PI over two natural log of two if we do the same thing with the natural log of cosine squared we get PI over two times natural log of 2 squared plus PI cubed over twenty-four then next we used this fourier series for the natural log of cosine in order to calculate the second integral and let's just recall that that the natural log of cosine was equal to minus natural log of two minus the sum from N equals one to infinity of minus 1 to the N over N cosine of two and X so this looks slightly different than the version that we used in this video in this video we had distributed this minus sign through and we had minus 1 to the n plus 1 but it's the same thing and then finally we had this fact that we used in the proof of this identity up here which is the orthogonality of these cosine type functions in other words if you take the integral from zero to PI over 2 of cosine M X times cosine NX you get PI over 4 Delta M in and let's just recall that Delta MN is equal to 1 if M is equal to n and it's if u equal to 0 otherwise okay so now let's see how we can do this so I'm going to open this portal up and we're going to set this equal to the following thing so we'll have the integral from 0 to PI over 2 and then we're going to replace the natural log of cosine with this guy right here so that's going to give me minus natural log of 2 minus this sum N equals 1 to infinity of minus 1 to the N over N cosine 2 and X and then all of this is going to be cubed and then DX so I've just replaced natural log of cosine with this Fourier series expansion which we calculated that expansion in the previous video ok so now what I want to use is the fact that a binomial cubed has a nice formula so let's add that in right here just so that we can see it so notice in this case we're gonna have minus a minus B cubed so my a term is like natural log of 2 my B term is this big sum but notice that since this is an odd power both of these are negative that's the same thing as minus a plus B cubed but then it's really easy to multiply that out and get minus a cubed plus 3a squared B plus 3a B squared plus B cubed great so that's just a standard binomial expansion ok so that's going to break this thing up into four pieces so let's see we're gonna have a minus out in front of the whole thing and then we'll have the integral from zero to PI over two of the natural log of two cubed and then I'm going to go ahead and separate these out into four integral so that'll be the first integral and then the second integral will be plus three times the integral from zero to PI over two and then this term squared times this term so I'm going to have natural log of two squared times this sum N equals 1 to infinity of minus 1 to the N over N cosine 2 and at DX okay so I've got that and then I have plus three times this sum in sorry this integral from zero to PI over two of the natural log of two times the product of this sum squared so that's going to be N equals 1 to infinity of minus 1 to the N over N cosine of 2 and X DX like that and then that's going to be squared good and then finally Plus this sum and then finally Plus this integral from 0 to PI over 2 of this cube of this sum so N equals 0 to infinity of minus 1 to the N over N cosine 2 and X the whole thing cubed DX okay so now we can close this big bracket off and this is the quantity that we need to calculate so I'm gonna clean up the board put these four integrals in a small as possible space up at the top and then we'll calculate one at a time okay so I've cleaned everything up as and okay so I've made everything as compressed as possible I've put everything at the top and this is what we have so we have our goal integral is going to be equal to negative the sum of these four integrals so we have this first one which is the integral from 0 to PI over 2 of natural log of 2 cubed DX so actually we can go ahead and take care of this one immediately given that the fact that the natural log of 2 cubed is a constant and we're integrating from 0 to PI over 2 so that's just going to be natural log of 2 cubed times pi over 2 so let's switch that ok so there we replace our first integral with its value and then we have these remaining three integrals where the X of the sum is ever-increasing okay so now let's go ahead and take care of one of them at a time so the first one we want to take care of is this one so the first thing that I want to notice is that this sum inside the integral makes up part of this formula over here and in fact notice that we can solve for this sum in terms of natural log of cosine of X and the natural log of two we know what the integral of the natural log of cosine X is so that'll help us out quite a bit so I'm going to go ahead and notice that this sum can be solved for notice we can just add this to both sides and then subtract the natural log of cosine X to both sides and this is the same thing as minus two minus natural log of two minus natural log of cosine of X great and so what that does is that's going to go ahead and make this integral star equal to the following so this is going to be three natural log of two squared so I'm gonna leave all of that out front and now I have the integral from zero to PI over two of minus natural log of two minus natural log of cosine of X DX okay now we can split this into two integrals so notice this is going to be three natural log of two quantity squared and then times the quantity zero to PI over two minus natural log of two DX minus integral from zero to PI over two of natural log of cosine of X DX great but now we know this is just an integral of a constant over the interval zero to PI over two so this is going to give me minus PI over two natural log of two okay good and now we're going to subtract from that the integral from zero to PI over two of natural log of cosine of X but we calculated that previously so that's going to be minus PI over 2 natural log of 2 so notice all of that is 0 so in other words this whole term right here is just going to give us 0 ok so given that I'm going to go ahead and clean up the board erase that term because we no longer need it and then push everything else together ok so we just finished arguing that that second integral was 0 so that leaves us with this first integral which we already calculated down to a number this third and this fourth integral so now what I want to notice is I can use this same trick that I did in the last case in order to calculate this second integral so notice that this guy right here this sum in here can be solved for in terms of natural log of cosine of X and my M natural log of 2 and this is in fact minus natural log of 2 minus natural log of cosine of X and then that whole thing squared so that's pretty easy to see but now notice that we can factor a minus 1 out of each of these terms and then square them and that's going to turn this into a plus so what that means is that that third integral which is now the second integral because the second integral was 0 is going to be equal to 3 times the natural log of 2 I'll keep that constant times the integral from 0 to PI over 2 of Ln 2 plus Ln cosine of X quantity squared but now we can just use the binomial squaring formula so we have a plus B squared is a squared plus 2 a B plus B squared so good and so that's going to turn this thing into 3 natural log of 2 times the quantity and I'll split split this thing up into three integrals so I'm going to have the natural log of 2 where'd DX then next I'm going to have twice the natural log of 2 times the integral from 0 to PI over 2 of the natural log of cosine of X DX so that's the Kroft terms from the foiling and then finally I'm going to have plus the integral from 0 to PI halves of natural log of cosine of X quantity squared DX ok so that's what we get from all of that but each of these have been calculated in these in previous videos so this one was calculated in episode 1 of this saga this one was calculated in episode 2 of this saga so and this one is just a constant so we're good to go there so this is turns out to be 3 natural log of 2 and then notice here we have PI over 2 times natural log of 2 quantity squared so that's easy to see because that's a constant plus 2 Ln 2 and then let's see the integral of natural log of cosine X from 0 to PI over 2 is this guy right here so it's negative PI halves natural log of 2 great and then finally this integral right here which is natural log of cosine of x squared so that's going to be PI halves natural log of 2 quantity squared plus PI cubed over 24 so we get something like that going on so let's see what we've got going on here so we've got the this stuff that can cancel so notice here this two can cancel with this two now we have pi over 2 natural log of 2 squared PI over 2 natural log of 2 squared those are going to add 2 pi times natural log of 2 squared then we've got PI times natural log of 2 squared here so those are going to all cancel each other then we've got 3 natural log of 2 times that well 3 over 24 is obviously 8 so that's going to give us natural log of to will actually let's put it in this order that's going to give us a PI cubed over eight times natural log of two so that means we can replace that second integral with this term okay so I'm going to go ahead and do that then compress everything up at the top and then we're ready to look at this third integral so so far we've simplified the first three integrals down to the sum of these two terms so we've got PI over two times the natural log of two cubed we've got PI cubed over eight times the natural log of two now we've got this thing right here this is the last integral that we need to calculate so you might think well we're just going to do the same thing again we're going to solve this sum for sorry we're gonna solve this equation for this sum and then we're gonna plug it in and use some facts what we know but that's going to quickly arrive at this goal integral and everything else is going to cancel and we're just gonna actually end up with this goal integral equals itself and so that's not going to help at all so what instead we need to do is tackle this guy on its own so let's go ahead and say that this is star and now we're going to have star equals to the following so I'm going to write this as the integral from zero to PI over two and then we're cubing an infinite sum and the way you do that is with the Koshi product formula so we went over the coast product formula a little bit last time we went over it even more carefully in my video about the Catalan numbers so I'll let you guys look at those resources in order to see what we get but what this is going to turn into is the following so we're going to have the sum N equals one to infinity and then the sum M equals 1 to n minus 1 and then the sum L equals 1 to M minus 1 and now we've got this term evaluated at some different spots so it's to be evaluated at L and then it's can be evaluated M minus L and then n minus M and so that's going to give us minus 1 to the L minus 1 to the M minus L and then minus 1 to the n minus M over L m minus L and then n minus M and then we're gonna have the same thing for the cosines so we're gonna have a cosine of 2 L X cosine of 2 m minus L X and then finally cosine of 2 n minus M X DX so that is what we need to calculate okay so now I'm going to go ahead and move this integral inside the sum and that's going to allow us to do some simplification so obviously that only works if you've got absolute converges but what we do in this case but that's again something that we're not going to prove so we have M equals 1 to n minus 1 the sum l equals 1 to M minus 1 and then notice all of this stuff on the top simplifies to minus 1 to the N power so notice we have this minus m and this plus M cancel the L and the - they'll cancel so we get that then this is going to be all over L M minus L and then n minus M and now we have the integral from 0 to PI halves of 2 of cosine of 2 L X cosine of 2 times the quantity M minus L X and then cosine of 2 times the quantity n minus M X DX so the next thing that we're going to use is this so well first of all I'm going to go ahead and call this integral which is inside the some blue star so that I can do some simplification and then furthermore we're going to use this sum angle formula so let's go ahead and use the fact that the cosine of alpha times the cosine of beta is equal to 1/2 cosine of alpha plus beta plus cosine of alpha minus beta okay good and what we're gonna do is we're gonna use that on maybe these two terms right here so let's see what that gives us so that means integral star is going to turn into the integral from 0 to PI halves of I'm gonna bring this 1/2 out front because I can the half that comes from this sum to product identity I should say product to sum identity and now I'm going to have cosine of alpha plus beta so notice alpha plus beta is just going to be 2 m so that's going to give me 2m X and now I'm going to have times this guy right here so because I'm simplifying these terms using this product to sum identity they're still going to be distributed onto this cosine of 2 n minus M so that's going to give me a cosine of 2 n minus M X great and then I have cosine of alpha minus beta so let's see alpha minus beta in this case is going to give me plus cosine of 2 and then I have M minus 2l X where I've used the fact that cosine is an even function to switch the order like that and then finally I have this is a cosine of 2 x n my SM X DX okay so that's how all of this got simplified so now what we can do is we can simplify each of these parts using this guy right here so notice that we know that this part is going to simplify down to PI over 4 times Delta of n minus M comma M right so this entry needs to be equal to this entry but we can actually simplify that that means n minus M needs to be equal to M which is the same thing as in needs to be equal to 2m ok great so we get that and now let's see what we get for this part so this is going to give us PI over 4 times Delta of M minus 2 L comma n minus M ok good so we've got something like that going on but notice that this is exactly the same thing as n minus M equals M minus 2l which is the same as PI over 4 Delta n comma what's the N is going to need to be 2 m minus 2 L so 2m minus 2 L so we get something like that so what that means is that we can go up here and simplify this sum using both of those things that we've just calculated so I'll clean up the board I'll bring this up and then we'll start simplifying that son on the previous board we worked internal star into this quantity so we've got this sum from N equals one to infinity M from 1 to N minus 1 L from 1 to M minus 1 minus 1 to the N over LM minus L in minus m and then we have these two kronecker deltas so Delta n 2 m and Delta n 2 m minus 2 L and we're actually only going to focus on this one this one is going to produce a similar quantity so I'm gonna go ahead and star this one in blue this is the one that we're going to work with and so if we starred this one in blue notice that this is going to set all of the values for n equal to 0 except when N equals to M so that means we can take this sum and we can replace in with 2 m everywhere so that's going to be the sum M equals 1 to infinity now the upper bound of that sum is infinity and then L equals 1 to M minus 1 now minus 1 to the to M that's just going to be 1 because 2m is obviously always going to be even then we have L M minus L and then finally n is 2 m so we have 2 M minus M so that's just going to be M okay great now the next thing that I want to do is notice that this M right here doesn't depend on L so we can just pull it out and that's going to give us PI over 8 the sum M equals 1 to infinity of 1 over m and then the sum L equals 1 to M minus 1 of 1 over L times M minus L ok now the next thing we want to do is a trick from calculus 2 so we're going to do partial fractions on this term we're where we are viewing Elle as the variable so I'm not gonna do the details there but I'll let you guys check that we get the following so this outer Sun is still the same so 1 to infinity of 1 over m and then the inner Sun is l equals 1 to M minus 1 and now we have the following quantity so we have 1 over L times M and then we have 1 over m times M minus L so now we can split this up into two parts so this is going to be PI over 8 times the sum M equals 1 to infinity and notice I can bring this M out I get 1 over M Squared and then the sum L equals 1 to M minus 1 of 1 over L ok so that's what we get for that first part and then the second part we get something super similar we're gonna get the sum M equals 1 to infinity of 1 over M Squared and then times the sum L equals 1 to M minus 1 of 1 over m minus L so we get that ok now the next thing that we can do is we can take this sum right here and we can re index it so will we index it in the following way we will send m to M minus L so notice if M goes to M minus L then M minus L is going to go to M minus M minus L which is l and then the bounds we'll just swap so really what we're doing here is just reversing the order of summation and I should say instead of starting at 1 and going to M minus 1 we're like going from M minus 1 down to m and then re-indexing ok so this is going to give us PI over 8 times the sum well now both of these are the same so it's twice the some M equals one to infinity of 1 over M Squared the sum L equals 1 to M minus 1 of 1 over L but notice that that is going to simplify so that's going to give us PI over 4 and then the sum M equals 1 to infinity of well something over m squared but that something is actually a well known number so whenever you take this partial sum of the harmonic series that's known as a harmonic number so this is the M minus first harmonic number now I'm gonna introduce a little plot hole into our saga and that plot hole is the fact that this harmonic number has a sorry this some of these harmonic numbers over M Squared have a nice formula this is equal to PI over 4 times the Riemann zeta function evaluated at 3 so there's a plot hole in our story maybe that'll be filled in a future video ok so I'll also say that doing the summation with this second Chronicle Delta will give us a very very similar result which will allow us to write this whole thing as 3 over 4 times pi times the Riemann zeta function evaluated at 3 ok so that means that we can replace this integral with that quantity so I'll go ahead and do that so to summarize what we have our goal integral up here which is the integral from 0 to PI over 2 of natural log of cosine quantity cubed is going to be negative of this sum so we have pi over 2 natural log of 2 cubed PI cubed over 8 natural log of 2 and then 3 PI over 4 of the Riemann zeta function evaluated at 3 so let's just recall that this riemann zeta function evaluated at s is defined by the following sum so it's n equals one to infinity of 1 over N to the s so this like immediately converges if the real part of s is bigger than 1 but it can be an analytically continued to almost the entire complex plane so I'll let you guys look into that if you want to and then so the Riemann zeta function evaluated at 3 is the sum of 1 over N cubed so it's that P series so you might think that something like that would have a nice closed form but it doesn't but what I want to do what I want to recall is that while we were calculating this part of the sum which came from this integral down here we actually got this pi cubed as part of a calculation of the sum of 1 n equals 1 to infinity of 1 over N squared which is equal to the Riemann zeta function evaluated at 2 which does have a nice closed form of PI squared over 6 ok so that Ennis that that finishes this trilogy of videos related to integrals of this top type

Channel: Michael Penn

Views: 13,787

Keywords: math, mathematics, number theory, abstract algebra, calculus, differential equations, Randolph College, randolph, Michael Penn, fourier series, interesting integrals

Id: f9pR30LRnzs

Length: 29min 47sec (1787 seconds)

Published: Mon Mar 23 2020