 This video will be about some rather strange integrals known as Borewine's integrals, which have some rather odd properties. So the Borewine integrals are integrals of the sync function. So we recall that that sync of X is just sign of X over X and the sync comes from the signals processing community and it's graph, it sort of looks like just decreasing oscillations. So it's graph looks something like this and its value here is one and it has the basic property that the integral from minus infinity to infinity of sync of X, DX is just pi. This is a common exercise in either complex analysis or Fourier transform theory. And Borewine's integrals are a sort of generalization of this. So this is the first one. The next one you look at the integral from minus infinity to infinity of sync of X times sync of X over three, the X and this is still pi and then you can do the integral of X times sync of X times sync of X over three times sync of X over five and this is still pi. And you can go on like this up to the integral from minus infinity of sync of X times all the way up to sync of X over 13 and DX and this is still pi. But when you get to 15, it suddenly changes. It turns into this rather impressive number here. So here you notice this is very nearly equal to one, the first 10 digits of the top and the bottom are the same but it's very, very slightly less than pi. And John Borewine apparently sent this in as a bug report to the computer algebra system he'd been using to evaluate these integrals. And there are actually two slightly different versions of this story. One says that he honestly thought it was a bug and was just being helpful in reporting this bug. The second version is that he knew perfectly well this wasn't a bug and was sending this in as a sort of practical joke. As he was playing on the vendors of the computer software. I think the second version is more likely because Borewine was a smart guy and probably knew perfectly well what was going on. So I'm going to explain why this sequence of integrals suddenly changes its behavior when you go from 13 to 15. So the explanation is as follows. First of all, we look at what is the Fourier transform of the sync function? Well, the Fourier transform is just a sort of almost a, it's a very simple function. It looks like this. So it just looks like a sort of block. Now you come along zero here and then you go up to something there and then down back to there. Now you probably want to know what this value is here and what these points are. Now I'm not going to tell you. And the reason is there are three or four different ways of normalizing the Fourier transform and they all give different answers for what you get here. And I don't really, I don't want to spend my time worrying about different conventions for Fourier transform. So I'm just going to sort of draw this without being too precise about it. Well, what about sync of X over three? Well, sync of X over three means you're sort of squishing up means you're sort of changing the sync function by the constant factor three. And the effect on the Fourier transform is that it squeezes it by a factor of three and multiplies its height by a factor of three. So you get another step function like this. So this is the Fourier transform for sync of X over three. And if you do the Fourier transform of sync of X over five, it gets even taller. So it's Fourier transform kind of looks like this. It goes way up there and then back down. And now what we want to do is we really want this integral from minus infinity to infinity of sync of X times sync of X over three and so on, the X is just the value of the Fourier transform of this expression here at the point zero. Fourier transform zero is just the integral from minus infinity to infinity. Dr. Constance, depending on how you normalize these things. And this is, the Fourier transform of this is just the convolution of F one, F two, F three and so on. Where are these functions here? So what we do is we want to take the convolution of this infinite, of this sort of finite sequence of functions and take its value at zero. And there's an easy way to work out the value at zero of a convolution of a step function with another function. So I suppose I've got some sort of function that looks like this. So here I've got a purple function and suppose I've got a blue function and the support of the blue function is going to be contained in the support of the purple function. And the purple function has this very simple form. Then the value of the convolution at zero turns out to be just the height of the purple function times the integral of the blue function. That's quite easy because you just sort of write down the integral for the convolution and you notice that as long as the support of the blue function is contained in the support of the purple function, the value at zero is just going to be this value here. Times the integral. So what we want to do is we're looking at, we're taking the support of, we're taking the first function here. So this is going to be the function f1. And then we've got the functions. Here we get f2 convoluted with f3 convoluted with all the way up to fn. And the integral of this is just one. That's because the integral is the value of sink x over three times sink x over five and so on at zero. And these all have value one at zero. So we see that the number we want, which is the value of the convolution of f1 and this at zero is going to be, so the value we want is just pi if the support of f3 convoluted with f5 convoluted with f7 and so on is contained in the support of f1. And if the support of this sticks out, then the value will actually be a little bit less than pi. So all we have to do is to work up what the support of this convolution is. And that's quite easy. So the support of f3 star f5 and so on is some constant times that you just get minus a third, minus a fifth, minus a seventh all the way up to a third, plus a fifth, plus a seventh and so on. And the support of f1 is just this constant times minus one to one. So the integral is pi if a third plus a fifth, plus seventh, plus all the way up to whatever you go up to is less than or equal to one. And we notice that a third plus a fifth plus all the way up to plus a thirteenth happens to be less than or equal to one. It's about 0.96. And if we add up a third plus a fifth plus all the way up to one over 15, this is slightly greater than one. It's about 1.02. So that's why the integral changes behaviors because the sum of the reciprocals is suddenly bigger than one at that point. You notice there's nothing special about these numbers a third, a fifth and so on. If you work out the integral from minus infinity to infinity of sinck of ax times sinck of bx times sinck of cx and so on, this is equal to pi over a if a is greater than or equal to b plus c plus and so on and all these bi's and ci's are greater than zero. More generally, you can see that the value of this integral will sort of change its behavior whenever some subset of all these numbers a, b and c is equal to the sum of the leftover terms. So if a is equal to the sum of all the remaining terms b plus c and so on, then it sort of changes behavior. For example, you can check that the integral from minus infinity to infinity of sinck of ax times sinck of bx is equal to pi over a if a is greater than or equal to b is greater than or equal to nought and it's pi over b if b is greater than or equal to a is greater than or equal to zero. By the way, you might want, you notice in particular that this becomes non-differentiable in a at these points and you may wonder why you can't find the derivative at the point a just by differentiating under the integral sign and taking the derivative with respect to a. Well, the problem is if you do that, the integral becomes divergent. So if I take the derivative with respect to a of sin ax over ax this turns out to be cosine of ax over a minus sin of ax over a squared x and you notice the integral from minus infinity to infinity of this dx diverges. I mean, this doesn't sort of oscillates and doesn't decrease enough for its integral to be convergent. So if you try and find the integral of cosine of ax over a times sin of bx over bx dx at a equals b this integral actually diverges which is what's causing the problems.