An interesting integral with the floor function.

our goal for this video is to evaluate this really interesting integral so we have the definite integral from zero to one of negative 1 ^ the 4 of 1 over X DX so before we get started let's just explore a little bit so notice that if X is between 1/2 and 1 what that tells us is that 1 over X is between 1 & 2 notice we have to flip the inequality when we reciprocate but what that tells us is that the floor of 1 over x equals 1 so on this interval 1/2 to 1 the floor is a constant of 1 so that's interesting to know we could maybe split this up into some different intervals and then furthermore if you take the sub interval one third to one half on this sub interval we have X is between 2 & 3 which tells us that the floor of 1 over x equals 2 so in other words on this sub interval this floor is a constant of 2 so that's the main idea behind this is to break this interval 0 1 up into a bunch of pieces infinitely many pieces in fact we're on those pieces this floor is constant but that's not what we're going to do I heard you guys to play with this solution in order to get a solution in that direction we're actually going to do a substitution which makes this a little bit easier to work with where we will deal with just this sub interval directly instead of having to take the reciprocal okay so I'll clean up the board and then we'll get going with that substitution okay so on the last board we argued that we're gonna have to split this up into infinitely many integrals really a sum of infinitely many integrals and then I noted that it's going to be easier to do that with a substitution that it is directly so let's see what substitution we should use well there really only one choice and that's 1 over X you might think well what about just substituting for that whole floor of one over X but that's not going to work too well because we need to take a derivative of something and there's not really a good way to take a derivative of something like that so let's go ahead and we'll let u equal 1 over X but notice that's going to mean that x equals 1 over u but that makes DX equal to minus 1 over u squared great so that is going to be our main driver for our substitution along with obviously this which built the whole thing now we need to figure out what happens to the bounds of integration so notice these are X values right here so and x equals 0 is really an improper part of the integral because we have 1 over X so really we have X approaching 0 from above so notice if X approaches 0 from above that is going to make you approach positive infinity so that's pretty obvious by this definition right here and then if x equals 1 we don't need to do any sort of limit there because this function is ok then that's going to make you equal to 1 so that allows us to rewrite this integral completely in terms of U like this we'll have the integral from infinity to 1 notice 0 goes to this positive infinity and 1 goes to 1 and now we have minus 1 to the floor of you great and then we've got DX which is negative 1 over u squared so I'm going to go ahead and put this minus sign out here and then I've got au squared in the denominator D U okay good now the next thing that I want to do is maybe let's take this minus sign and use it to flip the bounds of integration so we could take this minus sign and flip the bounds of integration to one to infinity now what we want to do is know that the four of you is constant on all of these subintervals like 1 to 2 2 to 3 3 to 4 if we take those as open subintervals or closed from below so that's exactly what we're going to do notice we can write this as an infinite sum so this is the sum N equals 1 to infinity of the integral from n to n plus 1 so notice our first one is going to be the integral from 1 to 2 the second one is from 2 to 3 the next one is 3 to 4 and so on and so forth and now we're going to have minus 1 to the floor of u over u squared D u ok so we've got something like that going on so now the next thing that I want to do is notice that this 4 of U is going to be a constant on that sub interval so once point that out so if u is between n it can be equal to n and strictly less than n plus 1 we know that the floor of U equals n great so that's going to allow us to change this we can change this floor of you here just to end but now since that doesn't depend on what we're integrating with respect to we can just pull it right out of an integral so here we have this is the sum N equals 1 to infinity of minus 1 to the N and now we have the integral from n to n plus 1 of 1 over u squared D u but this thing it's fairly easy to find an antiderivative that's going to give us the sum N equals 1 to infinity minus 1 to the N the antiderivative of this thing is going to be minus 1 over u and now we need to evaluate this from n to n plus 1 ok but what I like to do here is I'm going to go ahead and take this minus sign turn it into a plus and then switch the bounds of evaluation so I've got n plus 1 as my bottom bound and as my top bound so notice that's going to give me the following some I have the sum N equals 1 to infinity of minus 1 to the N and now I've got a 1 over N minus 1 over n plus 1 but now each of those parts confer converge because they're basically the alternating harmonic series starting at a different point so that means I can rewrite this as the sum N equals 1 to infinity of minus 1 to the N 1 over N minus the sum N equals 1 to infinity of minus 1 to the N over n plus 1 ok but what I like to do here is take this minus sign and turn it into a plus by distributing this through and making that an N plus 1 right there ok so now we're in a good spot I'm going to go ahead and clean up the board I'll bring this up here and then we'll continue so I cleaned up the board and this is where we left off so notice these sums look very very similar they're just indexed a little bit differently so what I'll do is I'll reindex this first one so that it goes in line with this second one so let's see what we get if we do that so here I'm going to take in and replace it with in plus one so that's going to make my bottom bound start at 0 instead of 1 so let's just keep that in mind and as we change this ok so let's go ahead and write this out here we're going to have the sum N equals 0 to infinity of minus 1 to the n plus 1 over n plus 1 plus the sum N equals 1 to infinity of minus 1 to the n plus 1 over n plus 1 now these are almost exactly the same but what I'm missing here is the 0th term so what I'd like to do is add and subtract the 0th term into this but notice the 0th term here is negative 1 over 1 so in other words it's negative one so what I'll go ahead and I'll do I'll subtract one and I'll add one and then the subtracting one I'll combine with that so make it the N equals zero term and then I've got this one added on the outside one so let's see what we get so we're going to get one that's from this guy right here plus this sum N equals 1 to infinity of minus 1 to the n plus 1 over n plus 1 but now we actually have two of those sums because we can take this minus one pull it inside and make that thing start at zero good and that sum should have started at zero as well okay so now we're in a really good spot the next thing that I want to do is actually pull out this plus one and make this a minus sign that's not strictly necessary but it'll make everything a little more obvious in the next step so that's going to be 1 minus 2 times the sum N equals zero to infinity of minus 1 to the N over n plus 1 but now I'm going to say that this is the same thing as adding an X to the n plus 1 here if we evaluate that X from x equals 0 to x equals 1 great so obviously evaluating at x equals 0 gives us 0 and then 1 to the power of n plus 1 is clearly 1 so that doesn't change anything but it does make it look like an antiderivative of something notice we've got X to the n plus 1 over n plus 1 that looks like the antiderivative of X to the N so notice here we get 1 minus twice the sum N equals 0 to infinity we have minus 1 to the N and now we're going to have the derivative the integral from 0 to 1 of X to the N DX so notice this definite integral is going to give us this guy right here which is equal to what we had now we're going to change the order of summation and integration which we can do because this is absolutely convert so we have 1 minus 2 times the integral from 0 to 1 of the sum N equals 0 to infinity of minus 1 to the N times X to the N but that's really minus X to the N DX great but that's a geometric series notice it's a geometric series with common ratio of negative X so that gives us 1 minus twice the integral from 0 to 1 of 1 over X plus 1 again because it's a geometric series with common ratio minus X now taking the antiderivative of that that's going to give us 1 minus twice natural log of X plus 1 now we need to evaluate that thing from 0 to 1 so let's see what we get if we plug in 0 we get natural log of 1 which is 0 and then if we plug in 1 we're getting at natural log of 2 and that gives us our final answer so it's 1 minus 2 natural log of 2 so that is the final answer for this goal integral and that finishes this video

Channel: Michael Penn

Views: 73,117

Keywords: math, mathematics, number theory, abstract algebra, calculus, differential equations, Randolph College, randolph, Michael Penn, integral, series, calc 2, logarithm, natural log, ln(2), ln, alternating series, improper integral, floor function

Id: 50upJvdVCXU

Length: 12min 10sec (730 seconds)

Published: Thu Apr 09 2020