CHAPTER 07.07: IMPROPER INTEGRATION: Multiple Segment Gauss Quadrature Rule for Singular Integrands: Part 1 of 2
In this segment, we're going to talk about an example of improper integration. How do we numerically solve improper integrals? And in this example, we are looking at singular integrands, so this example is corresponding to integrands which are singular. So the example which we want to solve the integral for is given as this. So in this case, what you are already finding out is at x equal to 3, the integrand is going to infinity, so, but this integral is finite, so far as the integral is concerned, but the integrand is infinite at x equal to 3, and we want to be able to find the approximate value. In the previous example, we used the two-point Gaussian quadrature rule. So in this case, we still want to use two-point Gauss quadrature rule, but over two equidistant segments. What I mean by that is that, in the previous example, we took two-point Gaussian quadrature rule, applied it from 0 to 3, and found the value of this integral. In this example, what you are asked to do is, because we got a very large percentage error of about, I think 25 percent, we are thinking that maybe what we should do is break this interval into two equidistant segments, and then apply two-point Gaussian quadrature rule over each of those segments. So basically we'll have four quadrature points, but we're going to break it up into two equidistant segments, and apply the two-point Gaussian quadrature rule over each segment, and see that whether we get a better approximation. So in this case, what we are trying to do is we are taking the integral going from 0 to 3, 5 x, divided by square root of 9 minus x squared, dx, we're going to break it up from 0 to 1.5, 5 x, divided by square root of 9 minus x squared, dx, and then the next one we're going to take from 1.5 to 3, 5 x, divided by square root of 9 minus x squared, dx. We're going to apply the two-point Gaussian quadrature rule on this integral, we're going to apply the two-point Gaussian quadrature rule on that integral, and see that what we get from there. So I'm going to skip some steps, because we have been showing these steps for every . . . for many cases in Gaussian quadrature rule, and also in the previous example for this one, that the first thing which we have to do is that we have to convert the integral going from a to b to an integral going from -1 to +1, to be able to apply the two-point Gaussian quadrature rule formula, and which is given by this formula right here. So this is how you are able to convert any integral going from . . . any definite integral with finite limits of integration to an integral with going from to -1 and +1, and also, we also know that the integral from -1 to +1, f of x dx is approximated by the two-point Gaussian quadrature rule by this particular formula here, where c1 is 1, c2 is 1, x1 is -0.5773, and x2 is 0.5773 . . . +0.5773. So we're going to apply these two formulas. So first we're going to convert it into an integral going from -1 to +1, and then we're going to apply this approximation of this two-point Gaussian quadrature rule for each of those integrals. So you might think that I'm skipping steps here as I'm going to show you the next step, it's mainly because we have covered this in the previous example, and also in the examples in the Gaussian quadrature rule, which you are more than welcome to see. So the first integral is basically going to become 0.75 the integral going from -1 to +1, f of 0.75 x, plus 0.75, times dx. And the other one is going to become b minus a, divided by 2, which will be 0.75 also, but the integral . . . integrand will have the arguments of 0.75 x, plus 2.25, dx. So what that means is that this is the change which we are doing to convert these integrals from . . . from 0 to 1.5 is this integral, and from 1.5 to 3 is this integral here, so that we can apply the Gaussian quadrature formula. |