 In this video, I wanna take a look at some more improper integrals. But this time, the bounds of the integral do not approach positive, negative, infinity. We have a different type of improper integral. That is to say another concern that makes a definite integral improper is when the function has a discontinuity somewhere along the interval of integration. Now, if this is a removable discontinuity, it has no effect on the integration whatsoever, so you can basically ignore it. If it's a jump discontinuity, you can just break it up at the jump, and that's not a big deal as well. The issue, really, we're gonna worry about is, what happens if there's a vertical asymptote, either at the end points or in the middle of the interval? If you have a vertical asymptote, that is, we're in the situation where x is approaching plus, sorry, not x, as y. What happens as y approaches plus or minus infinity? What does what do in such a situation like that? Well, you still treat it like a limit of any kind, so let's look at this one. The integral is zero from zero to one of the natural log of x. The issue is that as x approaches zero, the natural log has a vertical asymptote, so we still approach this as a limit. Take the limit as a approaches zero from the right. You're gonna integrate from a to one, the natural log of x, dx. So even though this is a limit, you can do most of this using the fundamental theorem of calculus, right? So the anti-derivative of the natural log, that one's a little bit of a tricky one. You have to use integration by hope to calculate it, but we've done this before. You get x times the natural log of x minus x as we go approach a and one, plug it in one. Notice that you're gonna get one times the natural log of one. That's a zero, so that's just gonna vanish. So you're gonna get a negative one right there for the first piece, one. And then we're gonna subtract from that the limit as x approaches, sorry, as a approaches zero from the right. You're gonna get a times the natural log of a minus a. And so this one right here, this a times the natural log of a, this is gonna have the indeterminate form, zero times negative infinity. This is the indeterminate form, zero times infinity. The negative a is just gonna disappear because it'll just go off towards zero. And so to resolve this indeterminate form, we, I'm gonna rewrite this as a product. We're gonna bump, I'm gonna rewrite it as an x now. We're gonna bump the a to the bottom. So we get the natural log of x sitting over one over x. So by L'Hopital's rule, because this now has the form of infinity over infinity. By L'Hopital's rule, this will look like one minus the limit as x approaches zero from the right. The derivative of the natural log is one over x. The derivative of one over x is negative one over x squared. That simplifies just to be a negative x. And so as x approaches zero, oh, this is, I should mention, this is a negative one. I forgot the negative sign there. Negative one, negative one. And so as x goes to zero, negative x will go to zero as well. So this thing vanishes and you're left with just the limit with the integral of choosing is gonna be negative one. So this is an example of a convergent a convergent improper integral. So as you get closer and closer to the vertical asymptote, turns out the area under this curve is finite. Kind of an interesting thing if you think about it, if you think of your natural log function, the area between the y-axis and the function is actually it's finite and it's gonna equal negative one. Kind of cool. Let's look at another example. Let's take the integral from two to five of one over the square root of five, the square root of x minus two. Notice there's a concern here at the lower limit again because as x approaches two, the denominator here goes to zero. And so there is a vertical asymptote on the graph. Now to do this function, a u substitutions nexus area take x minus two, du would then be dx. So this thing would look like our function. We're gonna get the square root of, well, one over the square root of u. So we're gonna get u to negative one half power du. And as we are doing u substitution, you can switch the bounds as well. So as x goes from five to two, what happens to u? Well, if you plug in five for x, you're gonna get three for u. And if you plug in two for x, you're gonna get zero for u. So we'll just change the bounds as well. Zero and three, like so. It's still an improper integral because as u goes to zero, we still have a vertical asymptote right here. Taking the antiderivative, we're going to get two times u to the one half power as you go from zero to three. And so when you plug these things in here, you're gonna get this two times the square root of three minus two times the square root of zero. And so you'll notice that in this situation, the limit, even though we're taking a limit as x is approaching two, or in this case as u approaches zero, when you get to the antiderivative, there's no concern. I mean, it's just continuity. Plug in zero, your limits are, the integrals can be two times the square root of three. So this is likewise a function which has a finite area near this vertical asymptote. This is another example of a convergent function. Sorry, a convergent integral. As a last example in this vein, consider the integral from zero to three. Now you'll notice that there's no issue at zero or three. If you plug in zero or three, there's no vertical asymptote here. The issue, of course, is happening at x equals one, which is between zero and three. So if there's any vertical asymptote in the middle, you have to break it up according to that. So we have to break it up from zero to one, dx over x minus one, plus the integral from one to three, dx over x minus one. So if you have a discount in the middle, you have to break it up. And in order for this thing to be convergent, both pieces have to be convergent. So what we want is that both are convergent. If any one of the pieces is divergent, the whole thing will be divergent. So let's take a look at this thing here. If we do the first one, zero to one, dx over x minus one, the antiderivative by a very simple u substitution will be the natural log of x minus one as we go from zero to one. When you plug these numbers in here, you're gonna get the natural log of one minus one. You're gonna also get the natural log of, and there are absolute values here, negative one, right? So we get the natural log of zero minus the natural log of one. And the natural log of one of course is zero, but the natural log of zero, assuming we're approaching it, we're approaching it from the right direction. This should be like a zero plus. We're approaching negative infinity. And so we see that this integral right here is divergent. We don't need to bother computing the other one because as one piece is divergent, the whole thing is divergence. Divergence here is a cancer. Once any part of it is divergent, the whole thing would be divergent. But had we started with this side, we likewise would have found out that this integral was divergent as well. So improper integrals are not just those where the limits go off towards positive negative infinity. That is the bounds. Sometimes if there's a vertical asymptote either at the end points or between, we have to also pay attention to the convergence or divergence of the integral in that situation.