Calculus: Integration,Convergence,Divergence Limit

This lesson has not been reviewed.

Please purchase the lesson to review.

Improper Integrals
Improper Integrals
Infinite Limits of Integration, Convergence, and Divergence Page [1 of 3]
Okay, so let me try to formalize some of these ideas of improper behavior. So I want to basically take a look at more improper behavior, but no spitting please. So just to give some sort of formality to these ideas that we've been thinking about with respect to improper integrals, let me just remind you that if you have some function, and you really are interested in figuring out the area, starting with a and going all the way out to infinity, how do you make that improper thing really rigorous? Suppose that this is actually some function y = f(x). Well, the answer is to inch up to it, using the idea of a limit.
So what you would do, basically, is just say, okay, what I'll do is - to find that area - I'll just stop at a point here. I'll call it b. And I'll find that area. Now, that's actually easy to do. That's just the integral from a to b of f(x)dx. But I want b to get bigger and bigger and bigger and bigger. So what I'll do is I'll take the limit as b goes to infinity. So, in fact, that integral we just evaluate normally. And then once you have the answer to that definite integral, just let the parameter, b, drift off to infinity. If this is infinite, then we say the integral diverges. If, in fact, you get something that's actually a number, then we say the integral converges and then the answer is that number.
So, in fact, this is the formal definition of what we mean - we don't have to put quotes around it anymore - by this. So if you see integral from a to infinity, what it really means is a to some big number, b, evaluate the integral, and then let b race off to infinity and see what you get. That's the meaning of this.
And similarly, by the way, of course now you can see that if I wanted to look at this integral, how would I make some sense out of this? Well, what I would do is I would first just take an integral from a to b, instead of negative infinity way off on the left-hand side. But then at the very end, I would take the limit as a approaches negative infinity. So again, it's the same idea.
In fact, if you really want something exciting, you can even imagine what it would be like to integrate from minus infinity to plus infinity. How would you do that? Well, it's the exact same idea. What you do is you break this up into two pieces. You would first take the limit as c approaches infinity of minus infinity to c. Well, in fact, you can just do this, and then use the method you discovered here to do that.
So in fact, taking integrals from infinity to something or minus infinity to something and so forth - not a big deal. You just use this idea. And really the thinking is exactly the same. It's just sort of a formal way of thinking about it. Really what you do is just the idea of you just figure out the integral and then just let the parameter that you want to go to infinity or negative infinity go to infinity or negative infinity and see what happens.
What about the other type of improper behavior - improper behavior where you have some sort of spike? And if you've got some sort of spike - the thinking here, suppose you have a spike at this point here, c. And you want to go from a to b and you have this function and you've got this spike. It's an asymptote right at c. And you want to find this area right here. The way to proceed is to break this up into two pieces. And the way to break it up into two pieces - so if you want to look at the integral from a to b of f(x)dx - and you've got the spike there, the spike means that, in fact, the function's not defined there. So in fact, I don't want to write - you see, I'd be tempted to say, what I'll do is I'll write the integral. I'll first go up to c. So from a to c, you see, I'm tempted to write that.
But I can't plug in c into this thing, because it's undefined there. So instead, what I do is I inch up to it. I take the limit as approach c. So I'll put in here something like an e, and take the limit as e approaches c. But actually, I'm going to approach it just this side, because I'm going to be coming in from the left, so I'm approaching it from the left, which means just a one-sided limit. And then plus, I do a similar thing here. I want to go from c up to b, but instead, I'd better go from someplace else. I'll call it d. And then I take the limit as d approaches c, but now from the right. I approach it this way.
So it's a very formal way of thinking about it. The real way of thinking about it and doing these things is just to realize you have to go from here up to something that's approaching c, and then from here, something that's approaching c up to b. Then you're home free.
By the way, why do I use e and d? Because, of course, it spells out my name. And you can use your own name if your name if your name happens to have two letters. And if not, then you're in trouble. So I guess if your name, was like, what? What other names have two letters? Al! If your name is Allison, and you go by Al, as a shorthand, then you can one a and one l. The problem with that is if you usually think of the left-hand point a, then it could be a problem, then good luck. So if you're Al, you're great, and if you're like Janice, then you're just really out of luck.
Okay, anyway. Let's take a look at some examples, and you can see these things really in practice. There's too much formality. If there's too much formality, then no one's getting anything. So let's do some examples.
Let's find the integral for minus infinity to infinity. Let's just do a real big one, of dx. Let's look at this integral and see what this looks like. Okay, first I want the picture of this where you can graph this thing. I'll just do it really fast here - really fast sketch. It sort of looks like a belly kind of curve - not like a belly like stomach, but like a bell type curve. Looks sort of like this. Looks kind of like that. And what I want to do is figure out the area, all the way from all the way off the left horizon, forever, and then all the way off to the right horizon. So it's two infinite pieces here. What I technically have to do is go from negative infinity up to some place and then from some place up to positive infinity. And you do all that right, and let's see what happens.
The first thing I do is I integrate this thing. And what's the integral of that? Well, that's one of those new things that we've seen. You're in calculus. This is actually the arc tangent - arc tan - of x. And now I'm going to evaluate this, and on the one side, I'm going to approach minus infinity, and on the other side, I'm going to approach infinity. I'll use these symbols, which I'm literally inventing on the fly, to just tell you that - of course, you can't just plug in that. It's not a number. But we're going to approach it. Now let's see what happens.
If we plug in infinity into arc tan, what do we get? All right, now arc tangent - that means I'm trying to figure out the angle whose tangent is approaching infinity. Well, that's actually . So in fact, this is . And I subtract from that. When I approach negative infinity, what's the angle whose tangent is approaching negative infinity? And remember, where arc tangent is defined. It's defined between and . So this is actually . And so look. This improper integral actually just equals + , which is . So the area under this whole curve - isn't that really cool? It's . All you need is red, and if you have a red , you can always spit it right into this infinitely long region. That's pretty cool. That's a nice example.
Let's try one where the functions are sadly vertically impaired - vertically challenged. So how about this? How about if we look at f(x) = . And I want us to find the area under that curve - let me draw you a little picture, maybe, of it - from zero to 1. So it looks something like this - has that sort of hyperbolic look to it. So from zero to 1 - so it's sort of undefined at zero. It's an asymptote. So that's the improper part, and I'm going up to 1. What would that be? Well, you just set it up, so that would be the limit as a approaches zero of the integral of a to 1 of dx. Why is this whole funny thing here? Because the thing is not defined at zero, so I have to inch up to it. Now technically, if you're going to really be pedantic, you have to put a plus sign here. I'm inching up to it from this side. You see, I'm just coming in from this side. And so if you like to be pedantic, then please pedant away.
All right, so anyway, so we have to take this limit, and so what's this limit? Well, this limit is limit as a approaches zero from the right. And then I have to now evaluate this integral. Now what's that integral? Well, okay, let's do a little side calculation really fast. So , that equals x^-1/2. The one-half power gives me square root. The minus sign gives me the flip of the one over. So that's what I'm looking at and so what's the integral? The integral would be what? I add 1 to that, and so I'd see x^1/2. And I divide by , which would give me a 2. The integral would be 2. Right?
And you can see that if you check. Take the derivative. You bring this down. The would cancel with the 2. And then I'd see - 1, which is -. So this actually is 2 evaluated from a to 1, which equals the limit as a approaches zero of 2 minus, and then 2.
Now what happens as a goes to zero? As a goes to zero, this term goes to zero and so this thing equals 2. So, in fact, what we see is that this area here is, in fact, 2. Great! So, in fact, even though this goes on forever, in fact, it equals 2.
All right, let me just advertise a question that I'd like for you to think about right now. So here's an advertise question. So here's a little ad that you try this one. Find the area under the curve f(x) = from zero to 3. See if you can do it. I'll give you a hint. I think that if you do this, you've got to be a little bit careful, because from zero to 3, there's going to be a problem, because this is undefined at 1. So what I want you to do is to, first of all, consider the region from zero up to 1 and consider that area separately, and then consider the region from 1 up to 3. Okay, so there's a hint. There's a bad point in here, and the point is at 1, so you've got to consider this two separate problems and evaluate that integral and evaluate that integral and see what happens. I'll leave this is a cliffhanger, great challenge for you, as you think about the improper. See you at the next lecture.