Improper Integrals

Start   Yes, well I have no control over that. So it's before the physics exam we're going to do two review sessions all day, one Sunday afternoon and professor Bonifant will do one Tuesday evening. Um, we're sort of constrained by getting rooms large enough to hold more than 100 people. So the review session that we are gonna do on the Tuesday before , I think, what is the
0:33the last slot that ends at eight. So 6:50, right? Yeah. So this is expecting and I still need to get in on Sunday afternoon. Yes, there are other classes, there are no other exams.
1:00You'll have a class at that time me to make arrangements not to go. So this time was announced when you use this time and the time of the other one was scheduled last January So even before I knew I was teaching this class, we had no control over that yeah?
oh, the review I can review the sunday, uh, I can do the sunday one
1:32Im not so sure about the tuesday one.
the video for the Sunday. Um, it probably won't be up until Tuesday or so. Maybe Monday.
Do people watch the videos?
yeah? once in a while like cheese again, OK.
2:03so for those of you that are using laptops or phones for clickers, I know that networking in this in this room is a little problematic. Um, they, they wrote wolfinet secure and then they tried to fix it until you've got messages about it and kind of sort of works if you force it. Um, sometimes it works, sometimes it doesn't. Uh, you could also try wolfie net guest
2:35I don't think will feed out opening system is supposed to do ok, anyway Ok, so lets go on from, from where we were So what we were talking about was this idea of improper integrals
3:02which is if we have either an unbounded domain So, so for example, so, so we can define so I said this last time the integral from anything to infinity of some function just be the limit as the upper bound goes to infinity of that function
3:40so, this is just saying if I wanted to integrate some function that stretches off that way I just cut it off further and further out. Cut Cutoff off to infinity. So if this limited this and then we say it converges, and
4:08if not, then we say it diverges OK. So I did an example of that last time at the end of the class when we integrated e to the minus x from zero to infinity and the answer is one Everybody OK with this idea? is relatively straight forward.
4:36So, um, so let's move on so the integral from 0 to infinity of 1 over 1 plus x square dx is
5:08A) divergent B) 1 C) pi D) pi/2
6:01so, most of you got it right well, half of you so, 31 percent think this is the right answer 17 percent think this is the right answer 50 percent think this is the right answer, and this must be something smaller, I dont know
6:33So, this is supposed to be relatively easy because what we do is, we can do this integral, I hope we all can do this integral this is the arctangent so this is the limit of something going to infinity of the arctangent of x evaluated from 0 to m
7:04which is the arctangent of m take the limit as m goes to infinity minus the arctangent of zero so the inverse tangent of zero is zero and the inverse tangent of, if you think of the graph of the inverse tangent
7:32because the tangent of pi over 2 is infinity so this pi over 2 minus zero, so the answer is pi over 2 So this is, I thought fairly easily, but I guess not so much for you. You took a long time So half of you got the right answer and a lot of you thought it was divergent
8:03OK, let's try another one. Um, do we have time?
So, let's try the integral from one to infinity of one over x dx so, Im gonna ask you again A) Divergent B) 1
8:31C) ln 2 D) e^2 and E) convergent but not one of the above
9:13how many people need more time?
nobody?
you need more time?
let's give it thirty seconds
10:05ok, so this time most of you got it wrong!
so a lot of people, very few people say this one, so less than 5 percent for this, this, and this so now is between this two so, how do we do this?
we do this just the same way the limit as m goes to infinity
10:30what is the integral of one over x?
right? the log so, its log of m minus log of 1 log of 1 still zero this is zero and now we need, what's the limit as m goes to infinity of the log?
what is the graph of the log look like?
11:01goes like that! it doesn't have a limit, it grows forever the log is what exponent you need to raise e to to get that big number so you take that big number and raise e to that exponent, so this is infinity minus zero and this can't, this does not exist 47 percent thought, no, sorry, 41 percent of you thought it diverged and 47 thought it converged
11:50OK, one more. You should get the hand of this by now
12:27so, same choices
12:40ok, Im gonna stop this now ok, oh well, some people came in at the last moment and got it wrong before it was like 85 percent right answer to this problem. So now it is 53
13:02ok, so anyway, lets just do this problem so the integral of one over x squared is minus one over x because this is x to negative two, add one to the power and get a minus one, and then divided by the power so this is, this guy evaluated from one to m take the limit as m goes to infinity so this is the same as
13:30minus one over m minus minus one limit so as m goes to infinity, one over m goes to zero, and minus minus one is one so this is the right answer and 63 of you thought this was right answer
14:00and, second prize goes to this so, its good that you know it converges, but it'd be better if you knew it converges to one ok, so these things I think are relatively straight forward. It's just a matter of doing an integral as before but one of the limits you just let go to infinity it's a little, yeah, so it's a little
14:33so, now if we have something like, let's just do this if we have something like this where we are going on both sides so, this is saying, you graph the function one over one plus x squared it's like the bell curve, it's not the bell curve but it looks like that and you were saying both sides. We can do this in two ways.
15:01One way we can notice that this is even function and so this is the same thing as twice the integral from zero to infinity of one over one plus x squared dx you can do that, because it is even and it's the are over here the same as the area over here but, even, if it weren't an even function we could still do this by splitting it up, by splitting it
15:37by splitting it up as say the integral from minus infinity to pick any place, how about zero plus the integral from zero to infinity.
And now we can do this in the same way we take the limit as let's say as l goes to minus infinity on this side
16:02and m goes to plus infinity on this side so this is the arctan of zero minus the arctan of l, that is the first integral plus the limit as m goes to infinity of the arctan of m minus zero
16:36so anyway, this is pi over two plus pi over two what I'm saying. I can do it in two ways. One way I can notice that the function is even do half of the integral start at zero, go to infinity total, just did it, its right here, I got pi over two
17:04so it is twice what I got before so even if this function is not even squared, I can still always split it in half I split it at zero, I do one side and then I do the other side so, even means symmetric with respect to the y axis, odd means symmetric with respect to zero
17:36or, it even but it is flipped on the other side there's certainly plenty of functions that are neither even nor odd such as e to the x, log, most polynomials, but not all. Things like that.
most functions are neither even nor odd just like most numbers are neither even nor odd, most are fractions if it is odd, you can play same trick, you just change the sign
18:02if the function is odd, then I can still split it, but I only have to do one half of the integral because I know the other one which is negative OK, Um, so again, not a whole lot to these, other than wrapping your head around the idea that something with infinite domain can have a finite area, but as we see, we have
18:37another, another thing that can go on is we might have an integral like no, let's do
19:00something like this this. This has a little bit of an issue at zero. It's OK at one, but it has a problem at zero but it's the same trick. The graph of one over the square root of x. looks like this the area that we're looking at here could be infinite because it blows up. So we can
19:30do the exact same trick.
you just take this to be the limit as A goes to zero from above of a to 1 x to the negative one half dx and then we can do this integral as before. The integral of x to minus one half is x to the plus one half
20:00I have to divide by one half so this is limit A goes to zero from above of x to the plus one half and we divide by one half (i.e. multiply by two) from A to 1 are we ok with this?
ok, so then, this is...
20:33I take the limit of 2, the square root of 1, minus 2 times the square root of A as A goes to zero so this is two.
21:00now, once in a while, these things can be hiding.
let me just keep the same-- no, I can't do the same, sorry.
Suppose I have something like the integral from 0 to 2 of 1/(1-x)
21:44suppose I have something like that now, this looks like it is OK so, here is a wrong answer
22:00you just do the integral, the integral of 1 over x minus 1, do a substitution make u equals x minus one so this is the log Three minus the log of one half is wrong. Why is this wrong? Right? So this is just
22:37a number, and this is wrong because when x is one we have trouble since the graph here since the graph here from one half to three so, one over a half is two, so it blows up at one
23:07oh wait, sorry it is x minus one so it blows up here at 1 Did I do this wrong?
23:38so the graph is like this, it has a problem here at x equals 1 so we can't do it this way
24:08doesn't matter, it is still wrong so anyway it's all wrong because it completely wrong, but the point that I'm trying to get out, because you can't just hold this and in fact notice that the integral from one to three
24:31one over x minus 1 dx these must be equal if they exist at all to the integral from one half to one x over x minus one dx plus the integral from 1 to 3 and here this diverges because
25:18so, here, um, ok, so this guy we can do this is the limit as A goes to 1 from below
25:30of the natural log of x minus one evaluated from one half to A plus the limit as B goes to 1 from above of the natural log of x minus one evaluated from B to three
26:02and so this is the natural log one half minus 1 is one half when you take absolute value minus the natural log of, the limit as A goes to 1 minus of natural log of A minus 1... and this blows up
26:32and so does this same way and so, this diverges
27:06so, you have to be a little bit careful you have to make sure that when you integrating you function is defined on the whole domain as if it blows up up somewhere within the domain, you have to split it there and do an improper now sometimes
27:31so for example, suppose that you are going to do something numerically, probably a good idea to know whether your integral make sense or not before, even if you can't do the integral in this way so one of the questions that you might ask is, does the integral exists? or not
28:05so, If I ask something like the integral from zero to infinity of e to the negative x squared dx Might want to know does this converge or does not converge?
Before we try and do it numerically This we cannot do by any means except numerically we already know this
28:35this is the bell curve but we know that this makes sense. How do we know this makes sense? even though we can't calculate because I told you?
so what's the biggest this could possibly be? one.
29:10somebody said one, one is good, why one?
because it is not seven and I keep writing one in the board?
just because you know oh, there you are, yeah?
29:33oh well this is not quite the bell curve. You have to divide by pi over two so the real one from statistics its over two here and there is two pi pi over two or two pi over two, ... So, how can we. So people know this is true, that's true. But why?
so, it is one over e to the x squared, that is true
30:14ok, but If I add up a lot of things less than one, they can add up too big so, I can add up, for example, one, one is not bigger than one, plus a half, plus a fourth, plus an eight, and so on
30:35or I can add one plus a half, plus a third, plus a fourth, and to infinity, so so, just because everything under here is less than does not mean this is less than one but, you have the right idea even though maybe you don't know someone else has an idea?
31:01but, it's a number. How can I take the derivative?
take the derivative of... that would tell what the critical point is I can tell you right now the critical point is at zero perhaps if this function that is inside here looks like this ok, so this is not completely obvious
31:34ok, so that is true if x is bigger than one right? maybe it is equal to e to the negative x and we already know what the integral of this is the integral from 1 to infinity of e to the minus x is 1 we already did that
32:01so, maybe 1 is not quite big enough, maybe I should So the idea here is we know that this graph e to minus x square eventually slips underneath e to minus x, they cross at 1, and this guy, the shaded area that we want
32:33sits underneath the area of e to the minus x. That means that whatever this area is, if has to be less than this piece so this has to be less than, or maybe equal the integral from 1 to infinity of e to the minus x dx
33:03and we did this yesterday, I mean, not yesterday last week, but this is the limit as m goes to infinity of minus e to the minus x from 1 to m which is 1 minus 1 over m limit is 1 so, because the function here is always smaller than the function here
33:35the integral here has to be smaller than the integral here even if you can calculate it so, the function here is always smaller than the function here so the integral here has to be smaller the integral here If I have a room that is higher, then the smaller room
34:02I can fill up the taller room and more stuff than I can fill the smaller So if I were to build, if I were to fill this room halfway full, then we'd have less stuff if I'd this room all the way full So we know that if I fill this room halfway full of those little balls and you play around, so you fill with those little red and yellow and green balls, the volume of the balls is less than the entire volume of this room. So we can say that a little more formally...
34:43call this the comparison theorem
35:09so If I have one function that's bigger than another function and if the big function (I don't care where it starts) converges,
35:32then so does the little function.
and it converges to something less than the big function
36:02so If we have an upper bound that we know There is, there is another part to this which says that if the little function doesn't converge
36:41so, again, I still have f(x) greater than g(x) and the integral diverges
37:01then so does the big one So if if g(x) is going to the moon and you're sitting on top of g(x), guess where you are going
37:31right? that guy is bigger than that one, this guy, the area goes to infinity, then the are of that has to be infinity