So much to learn/review in one integral!!!

in this video we're going to look at a really interesting integral so it goes like this it's the integral from zero to infinity of the natural log of 1 plus x times the natural log of 1 plus 1 over x squared all over X DX and so along the way we're going to see the following things feature in this solution so the first is going to be transformation into a triple integral then we'll also use integration by parts we'll do an infinite series expansion we'll also see the Riemann zeta darish lay ADA and Dericks like beta functions and finally this thing called the Catalan constant so this integral really provides us an opportunity to review and learn about a lot of different things now along the way we're going to prove these four bullet points in order to make our solution a little more smooth so the first one will be the following integral so 0 to 1 of 1 over 1 plus x times y dy is going to be this function natural log of 1 plus x over X so notice that's that then the integral from 0 to 1 of 2 Z over Z squared plus x squared DZ is the natural log of 1 plus 1 over x squared so that's going to be that function so these two bullet points will really tell us how to transform into a triple integral as we will see then we'll use integration by parts to prove this formula so the integral from 0 to 1 of t to the M natural log of T is 1 over m plus 1 squared and it's negative of that and I should say here that M takes on certain values positive integers but it doesn't really matter in this case as we'll see in our solution and then finally we'll use this fact that the D relay ate a function which is sometimes called the alternating zeta function so notice it looks exactly like the riemann zeta function except for its alternating instead of just being a straight sum and so it is equal to 1 minus 2 to the 1 minus s times the zeta function ok so I'm going to clean up this board and then we're gonna get to proving these bullet points but I want to try to clean up this board in a way that I haven't done in a while let's see if it works oh it seemed to work so let's get to proving this first bullet point which is actually quite simple so we're going to look at the integral from 0 to 1 over 1 plus x times y dy now you can do a little baby use substitution if you want to to do this so maybe we would let u equal x times y and notice here our variable of integration is y so that means when we take D U we will get that this is X dy which tells us that dy is equal to 1 over XD u but now that means this dy turns into 1 over X D U and here notice that this is U and now let's see what happens to our bounds of substitution so recall that these are Y numbers right here so when y equals 0 that tells us that U is also going to be equal 0 and Y is equal to 1 that tells us that U is going to be equal to X so that's going to change this thing into the integral from 0 to X 1 over 1 plus u times D U and then we can put a 1 over x out in front because that is a constant with respect to the integral but notice now we're pretty much home free this is going to be 1 over x times the natural log of 1 plus u evaluated from 0 to X notice evaluating at X gives us natural log of 1 plus x over X and evaluating at 0 gives us the natural log of 1 which is equal to 0 so that means we have proved this first bullet point and now we're ready to move on to the proof of the second one so here we have integral from 0 to 1 of 2 Z over Z squared plus x squared DZ ok so again this is going to be an integration with respect to Z so that makes x squared a constant so let's do a baby u substitution again so let's go ahead and in this case we'll let u equal Z squared plus x squared but notice that is going to make D U equal to two Z DZ so notice this numerator those are my D u earmuffs and then this whole denominator is equal to u great now let's see what happens to the bounds of integration when Z is equal to zero that means U is going to be equal to x squared and then when Z is equal to 1 that's going to make you evil 2 1 plus x squared great so let's see what that does this is going to change it to the integral from x squared to 1 plus x squared of 1 over u D u so that's going to give us the natural log of U evaluated from x squared to 1 plus x squared great but that's going to give us the natural log of 1 plus x squared minus the natural log of x squared but now using logarithm rules that's going to be the natural log of 1 plus x squared over x squared which is the natural log of 1 plus 1 over x squared ok so that means we've established this second bullet point and now we're ready to move on to the third so for the third bullet point we want to look at the integral from 0 to 1 of t to the M natural log of T DT and we can just assume that M is bigger than or equal to 0 here that's actually the case that we're interested in ok so in this problem we'll use integration by parts that so that means we need a U and we need a DV and so we need to choose U so that when we take its derivative becomes simpler so a good choice there will be u equals the natural log of T so let's say U is equal to the natural log of T so notice that's going to make D u equal to 1 over T DT which means V will be equal to t to the M sorry D V will be equal to T to the M DT but that is going to make Z equal to 1 over M plus 1 T to the M plus 1 ok perfect now if we recall our integration by parts formula we have u DV is equal to UV minus VD u where we plug in the bounds of integration as appropriate ok so let's see what we get from that so we're gonna get u times V so that's going to be a natural log of T over m plus 1 times T to the N plus 1 evaluated from 0 to 1 and then we will subtract from that V D u so notice that is going to be 1 over n plus 1 I can bring that out of the entire thing and then the integral from 0 to 1 of T to the M DT so notice we have t to the n plus 1 over T which just makes T to the M ok good now notice if we plug one into here we get zero because natural log of 1 is 0 if we plug 0 into here we also get zero because obviously t 0 is 0 and you might say well if you plug 0 into the natural log of T that's actually tending towards negative infinity but it's an easy l'hopital's rule problem to check that the limit as T approaches 0 is in fact zero in this case so I won't do that ok so the long and short of it is this whole thing goes to zero which leaves us with this integral so this is going to be minus M plus 1 now we can take the antiderivative which puts another n plus 1 in the denominator and then we're evaluating T from 0 to 1 which obviously gives us 1 over n plus 1 squared and that whole thing is negative ok great so now we've established this third bullet point and we're ready to move on to this fourth which involves the Dirac slay Adah function in other words the alternating riemann zeta function and the riemann zeta function itself so let's see how this is going to go so we can say well let's just start with left hand side so ADA of s is equal to the sum N equals 1 to infinity of negative 1 to the N over into the s so notice this is the sum over all even and odd terms sorry this should be into the money in -1 up there so now let's split it up into the sum over all even terms and the sum over all odd terms so I can do that in this way so this is going to be the sum N equals 1 to infinity and now I'll just put odd here to say that I'm summing over all odd terms of negative 1 to the N minus 1 over in to the S and now I have plus the sum N equals 2 to infinity and I'll put even here to say that on something overall even terms and now I have minus 1 to the n minus 1 over N to the S ok now we can do some simplification notice if we're summing over all odd terms n is odd which makes n minus 1 even which means we can turn all of these into a 1 great positive 1 I should say and then here n is even which makes n minus 1 odd which means we have negative 1 to an odd power so we can turn all of these to a negative 1 and I'll do that in the following way so I'll just bring the minus sign out front now from here what I want to do is add and subtract this second term to this right-hand side of the equation so in other words I'm going to add the sum N equals 2 to infinity over the even terms of 1 over N to the S and then I'm also going to subtract N equals 2 to infinity over the even terms of 1 over N to the S ok great but now notice if I add all of the odd terms with all of the even terms and it's no longer alternating then that's just going to give me the sum of everything so I'll put the sum N equals one to infinity and I'll put all here just to reiterate that I've got the even terms and the odd terms one over in to the s power and now I have minus 2 times the sum over all the even terms so that's going to be the sum N equals 2 to infinity over all of the even terms of 1 over N to this but notice I can re-index this thing since I'm going over only the even numbers that means I can replace n with 2 N and then sum over everything so here's how I'll do that I'll erase even I'll put a 1 here we're gonna sum over everything and then I'm gonna change this to 2 in to the S power and notice that's exactly what I had before it's just I've indexed it slightly differently okay great but now what I want to do is notice that this guy right here is just the riemann zeta function evaluated at s and then I'll simplify this a little bit more notice I can bring it to to the s out of the denominator here which gives me 2 to the 1 minus s I've got a 2 to the 1 there and then a 2 to the s in the denominator and now I have the sum N equals 1 to infinity of 1 over N to the S but again this guy is just a Riemann zeta-function so this gives me the Riemann zeta function evaluated s minus 2 times 1 minus s the Riemann zeta function evaluated s but I can easily factor a Riemann zeta function out of that and that gives me 1 minus 2 to the 1 minus s and then the riemann zeta function evaluated at s good and that establishes this fourth bullet point and now we're ready to move on to our main result so the first thing that we'll do is we will change this function natural log of 1 plus X over X into that integral which is on the left and the same thing for the natural log of 1 plus 1 over x squared so that's going to change the single integral into a triple integral and it'll look like this so we'll have the integral from 0 to infinity 0 to 1 and then another integral from 0 to 1 of now it looks like the integral of this times this so that's going to give us 2 Z over 1 plus X Y and then Z squared plus x squared and then let's get the order right so this is going to be dy DZ DX ok fantastic now the next thing that we want to do is change the order of integration to put the infinite integral on the interior and so there's like a theorem that says when you can change like the order of integration in this case and I'll let you guys look that up if you're interested so that's going to give us the integral from 0 to 1 the integral from 0 to 1 the integral from 0 to infinity of twice Z over 1 plus X Y and then Z squared plus x squared and now we have DX dy DZ so we've got something like that going on now the next thing that I want to do is look at this integrand as a function of X and notice that it's a rational function in X and so that really screams that maybe we should do some sort of partial fraction decomposition so that's what we'll do so do a partial fractions so let's take to Z over 1 plus X Y times Z squared plus x squared and notice we're gonna want to write that as a over 1 plus XY plus BX plus C over Z squared plus x squared so recall here we are viewing X as our variable so we have this linear term in X right here which means we need a constant on the top then we have the quadratic term in X right here which means we need this linear term on the top just by standard kind of calculus 2 tricks now what I'll do is I'll clear the fractions and that will lead me towards some sort of equation that I can solve so clearing the fractions and then I'm going to change the left and the right-hand side of this we get the following so we're going to get a times Z squared plus x squared plus BX plus C times one plus XY equals two times Z okay so we get something like that but now notice we can go ahead and multiply out this left-hand side and then look at the coefficient of x squared the coefficient of the linear term and then the coefficient of the constant so let's just look at what the x-squared coefficient is first so notice the x-squared coefficient times first is a because we get that from a times x squared and then it's going to be B times y we get that from be x times XY so B times y there's our coefficient of the x squared term now let's go ahead and look at the coefficient of the X term so all the coefficients from the X term we get from this bit right here so notice the X term we'll get from be x times 1 so that will give us B and then we'll also have C times y for the next bit so we get that okay good but now let's go ahead and look at the constant term so for the constant term will have a Z squared so that's from this multiplication right here we're called that Z is a constant with respect to this decomposition and then we have plus C from the second term okay great so that gives us all of this is equal to 2z but now what we want to do is equate coefficients so let's think about this right hand side as 2z plus 0 times x plus 0 times x squared ok great so I'm going to clean up the board and then we'll kind of look at what system of equations we need to solve based on this large polynomial equation so picking up from the last board we were using this partial fractions decomposition and we did all the multiplication out and that gave us a large polynomial I have extracted the x squared term the X term and the constant term and that gives me these three equations so it looks like the best way to do this may be so since this is our most complicated equation it has a and C in it maybe we'll take these two equations and eliminate the B so it has to do with a and C so maybe we'll do equation 1 minus y times equation 2 so in other words this equation minus y times this equation so let's see what that gives us that's going to give us a minus C times y squared equals 0 because notice our B Y minus B Y will cancel so we get a minus C y squared equals 0 but notice that tells us that a equals C y squared which gives us a way that we can solve for C pretty easily just plugging it into here so let's go ahead and plug this into this equation and that is going to give us C y squared Z squared plus C equals 2 Z but we can factor a C out of this and we get Y squared plus Z squared sorry y squared times Z squared plus 1 equals to Z in other words we have C equals 2 Z over 1 plus y squared Z squared great so we have that value for C so I'm going to go ahead and write that over here and so we can erase the board and continue on with the rest of the calculation so Y squared Z squared so great we've got our value of C and now I'll erase the board and we can use that to find the values of a and B so on the last board we figured out what C is but now notice that's going to very easily allow us to figure out what a is by this equation that we got on the last calculation so notice a is going to be this guy times y squared in other words it'll be 2y squared Z all over one plus y squared Z squared ok good and then notice that by this equation B is equal to negative C Y but that means that we can write B equal to negative two Y Z over one plus y squared Z squared so we've got something like that going on but now what we can do is take all of these values for a B and C and plug them in to this partial fraction decomposition and that will allow us to change this box into something that's maybe a little bit simpler okay so I'll erase the board and I'll do that all in one step okay so we just did an extensive partial fraction decomposition that decomposed this guy into this kind of larger product which seems like it might be a little harder to work with but actually it's quite a bit simpler given that we have some easier functions to integrate ok so let's first notice that our innermost integral is with respect to X so that means X is the variable and also notice that this big term out front only depends on y&z so that's a constant with respect to the innermost integral ok so now I want to tackle these one at a time so I've got the integral from 0 to 1 then it'll go from 0 to 1 and then we need to integrate this thing right here so let's see what that is so notice 2y squared Z is a constant with respect to X so we only need to integrate 1 over 1 plus x times y and we actually did that as we were proving this bullet point so I won't bother with it but what we'll end up getting is a new Y in the denominator which will cancel that Y out and that will give us 2y Z times the natural log of 1 plus x times y ok so I'm going to go ahead and underline this in purple and this in purple just to see where it came from and let's notice if you take the derivative of this with respect to X then the interior the natural log goes downstairs and we've got to multiply by the derivative of that which is Y which gives us a y squared so we are good to go here ok great now let's look at this next one so this is going to be plus twice and then this is actually kind of a well-known integral which is on many integral charts or you can derive this on your own this is an inverse tangent and it's in fact the arc tan of x over Z so again I'll let you guys check the depth that those are that's the correct antiderivative ok so now let's look at this one but notice that's also going to be a natural log because if we take the derivative of this denominator we get something that looks like the numerator so this will in fact be minus y times Z times the natural log of x squared plus Z squared so let's make sure that that is right so notice if we take the derivative of that natural log of x squared plus Z squared a 2 X is going to come out in the numerator then x squared plus Z squared are going to go downstairs ok so that's what we get and now we need to evaluate this from x equals 0 to infinity and then do a dy DZ integral for it so moving on I'm going to combine this purple term in this yellow term together and I'm going to first notice that they have a Y Z as a common coefficient this one has a 2 as an extra coefficient but I can bring that in and square using natural log rules so this is going to give me the integral from 0 to 1 0 to 1 I still have my 1 over 1 plus y squared Z squared out front and then here I'll have Y times Z times the natural log of 1 plus XY quantity squared over x squared plus Z squared ok and then this is going to be plus twice inverse tangent x over Z ok good and now we need to evaluate this from 0 to infinity obviously some limits are going to occur as we do that but we'll talk through those so X approach as X approaches infinity so I'll circle that in purple the interior of this becomes essentially x squared y squared over x squared great but notice that x squared and the numerator and the denominator will cancel which tells us that this whole thing charges off to the natural log of y squared ok good now let's look at the second term so as the argument of the inverse tangent goes to infinity it's well-known that the output is PI over 2 so I'll just go ahead and write that that's PI over 2 okay so now I'll circle in yellow what's happening as we let X go to 0 so as X goes to 0 notice this thing is going to turn into the natural log of 1 over Z squared so that's even easier to see and then here as X goes to 0 this term is going to go to 0 ok good so now I'll put all of those into their respective places and bring that to the top and then we'll continue on I've plugged in all the appropriate limiting values and we have the following integral to work with so it's a double integral 0 to 1 0 to 1 1 over 1 plus y squared Z squared and now we have Y Z times the quantity natural log of Y squared minus to login one over Z squared plus PI recall that this was two times pi over two from that inverse tangent limit but I've multiplied that together okay now the firt the next thing that I want to notice is that this natural log of Y squared can be written as two times the natural log of Y using exponent rules and then also this natural log of one over Z squared well I can turn that to a plus and then this is going to be two times the natural log of Z for a similar reason notice that's really the natural log of Z to the minus two but I can bring that minus two out front and that cancels that from a minus to a plus okay so now let's see what we have so that's going to turn into the integral from 0 to 1 0 to 1 of 1 plus y squared Z squared and now we have 2y Z natural log of y plus 2y Z natural log of Z plus pi dy DZ now the next thing that I want to notice is that this guy right here switches in to that guy right there if you switch Y with the Z so in other words if we take this term and this term and we switch Y with Z one of them turns into the other one so that means there's some symmetry between how Y is acting inside of this term and how Z is acting within this term which means when you do that their integration they will be the same let's just notice that so after integrating these give the same value so in other words we might as well only deal with one of them so let's go ahead and keep this one because Y is before Z in alphabetical order although there's no real reason but that means this 2 is going to turn to a four and we're left with the following integral so 0 1 2 0 to 1 of 1 over y squared plus Z squared plus 1 and now we have 4 y Z natural log of Y plus pi and then dy DZ ok great now the next thing that I want to do is change this term right here using geometric series into an infinite sum so let's recall maybe that a 1 over 1 minus u is equal to the sum N equals 0 to infinity of U to the N and so that's what you use here but in this case u will be equal to minus y Z quantity squared great so let's see what that gives us that's going to give us the integral from 0 to 1 the integral from 0 to 1 of the sum N equals 0 to infinity of minus 1 to the N Y squared to the N in other words Y to the 2n z to the 2 and for the same reason and then we have this as x 4y z natural log of y plus pi and then we have dy DZ ok so the next thing I'll do is I'll bring that to the top do a very very slight simplification along the way and then we'll go off to the next step so I've brought that previous expression up from the bottom of the board and I've factored some constants out I've exchanged the order of summation and integration which we can do because this is absolutely convergent and I've broken it apart into two sums another thing that I've noticed is that the integrand for this first double integral is a function of Y times a function of Z so notice it's y to the 2n plus 1 times natural log of Y that's our function of Y and then it's Z to the 2n plus 1 that's our function of Z and likewise the integrand here is also a function of Y times a function of Z but that's a little bit easier to see even so anytime you have an integrand that's like this inside the double integral you can split that apart into a product of two single integrals so let's see what that gives us so this is going to give us four times the sum N equals zero to infinity of minus 1 to the N now let's look at our y integral that'll be the integral from 0 to 1 of Y to the 2n plus 1 times natural log of Y dy great so I'm going to go ahead and put purple brackets around this this is our Y integral and then we'll also have a Z integral which will be from 0 to 1 of Z to the 2n plus 1 DZ great so now we've got to like calculus to type integrals which are fairly simple now this is going to be plus pi times the sum N equals 0 to infinity minus 1 to the N but notice if we pull those apart into two integrals they're actually the same integral with just a different variable so that means we might as well just write it in the following way this is the integral from 0 to 1 of Y to the 2n dy and then times that same integral with just y's replaced with Z's but that means they're the same so I can just change that to this integral squared ok great but now I've got some fairly simple stuff to do so this is going to be 4 times the sum N equals 0 to infinity minus 1 to the N now notice this purple integral is actually our third bullet point so we can use that directly where M is replaced with 2n plus 1 so like I said M is 2n plus 1 so that's going to give us minus 1 over 2n plus 1 plus 1 in other words 2n plus 2 so that's what we get for this purple guy so here maybe I'll put that in purple and then notice here I just do an antiderivative which is really the power rule that gives me 1 over 2 n plus 2 Z to the 2n plus 2 evaluated from 0 to 1 good so that's actually easy to do so I'll just write that here kind of in the margin that is going to go to 1 over 2n plus 2 notice when you plug in 1 that's what you get when you plug in 0 you get 0 so that's all you have okay now here we're going to have this as plus pi the integral from N equals 0 to infinity sorry the sum minus 1 to the N and then this is very very clearly 1 over 2n plus 1 squared so now we can start to simplify this I'm going to bring this minus sign out front that's going to give me a minus 4 now I'm going to have the sum N equals 0 to infinity of minus 1 to the N over 2n plus 2 to the third power okay then I'm gonna have this as plus pi times that sum right there but that some actually has a name it doesn't really have a closed form unless you use this name for the closed form and that is G where G is Catalans constant so this is this is also the Dara Schley beta function evaluated at 2 if you want to look if you want to look that up ok good so now we can simplify this a little bit so first thing I want to do is print bring a 2 out so notice how if I bring a 2 out of the denominator I have 2 cubed which is 8 that's gonna cancel with this to give me a minus 1/2 so here we have minus 1/2 the sum N equals 0 to infinity of minus 1 to the N over n plus one plus pi times G okay but notice that this guy right here is just a very very simple re-indexing of this darish lei ada function and in fact it's this D relay ADA function evaluated at three great so that means here I get minus one half the dear sleigh ADA function evaluated at three plus pi times G but now using our final formula that put it puts it in terms of the zeta function that's going to be minus one half 1 minus 1/4 because notice we have 1 minus 3 so that's going to be 2 to the minus 2 of the zeta function evaluated at 3 plus pi G in other words this thing is pi times the Catalan constant minus 3 over 8 times the zeta function evaluated at 3 and that is the value for our goal integral which finishes this video

Channel: Michael Penn

Views: 42,420

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

Id: 12oIcTcp1Y4

Length: 36min 39sec (2199 seconds)

Published: Mon Mar 30 2020