how would you evaluate this integral let's say if we wish to calculate the integral from 1 to infinity of 1 over x dx how can we get the answer and this particular integral is it convergent or is it divergent how do we tell well if we get a finite number l let's say if the limit exists that would mean that it converges to a value however let's say if we don't get a finite number let's say we get infinity then we can say the integral is divergent so this is called an improper integral when you see infinity either in the lower or in the upper limit of the integral sign so how do we evaluate this well first we need to replace infinity with some variable which in this case will be t so we're going to express it as a limit so we're going to say the limit as t approaches infinity of the integral one to t one over x dx now what is the anti-derivative of one over x well we know it to be ln x but first we need to rewrite the limit expression so we have the limit as t approaches infinity ln x evaluated from 1 to t so now based on the fundamental theorem of calculus we can rewrite this expression as ln t we're going to replace x with t first and then we're going to replace x with one so it's lnt minus ln one so we have the limit as t approaches infinity ln t and then minus the limit as t approaches infinity ln one now ln one is simply zero so we can get rid of this expression that becomes zero so all we have is just the limit as t approaches infinity for the natural log of t and the natural log of infinity well that's going to be infinity so this is the answer because we don't have a finite number we could say this particular improper integral is divergent it doesn't converge now let's try another example so let's integrate one over x squared from one to infinity feel free to pause the video if you want to work out this example so let's begin by converting this expression into a limit expression so we have the limit as t goes to infinity for one to t and this is one over x squared dx now let's begin by finding the antiderivative of one over x squared so what's the first thing we need to do here we need to rewrite the expression so we need to take the x variable move it to the top so it becomes x raised to the negative two the 2 becomes negative once you move the variable to the top now we need to use the power rule for integration so if we add 1 to negative 2 it becomes negative 1 and then we're going to divide by that result so rewriting the expression taking the x variable and bringing it back to the bottom we're going to have negative one over x plus c so the anti-derivative of one over x squared is negative one over x so we're going to have the limit as t goes to infinity and this is going to be negative one over x evaluated from 1 to t so this is going to be the limit as t goes to infinity and then first we're going to replace x with t and so that's going to be negative one over t and then minus now we're going to replace x with one so it's minus negative one over one so let's go ahead and simplify what we have so this is the limit as t goes to infinity and so it's negative one over t and then negative times negative one that simply becomes positive one so now let's apply this expression the limit as t goes to infinity of negative one over t what is that equal to so if we have negative one over infinity what should we get when you divide by a very large number you're going to get a small number so this will approach zero so this is going to be zero plus one thus the answer is positive one so because we have a finite number we could say that this integral converges and so that's it for this problem the last two examples can be explained by something known as the p series and you're going to hear more about this series as you progress in calc 2. but let me give you the gist of it so let's say we have this particular improper integral now if p is greater than 1 and this integral will be convergent and if p is less than or equal to one this particular improper integral will be divergent and we saw that to be true in the last two examples so in the first example where we had one over x let me write the limits of integration in this example p was equal to one and we saw that it was diverging because our answer was infinity so this case applied in the second example we had this problem but this time p was equal to 2. and in this case we got 1 as our answer so we can see that if p is greater than 1 the series will be convergent but if it's less than or equal to 1 it's a divergent as we saw in the first example now let's work on another example so go ahead and determine if this particular improper integral is convergent or divergent so we have one over three x plus one squared now based on the p series would you say it's convergent or divergent now this expression is very similar to the integral of one over x squared dx where p is two so just by comparison if this particular improper integral is convergent then it's safe to assume that this one is also converging but now let's go ahead and prove it by getting the answer so we shouldn't get infinity as the answer we should get a finite number like five eight or one half or something like that so let's begin by rewriting this as a limit so we have the limit as t approaches infinity for the integral one to t 1 over 3x plus 1 squared dx now i'm going to take a minute and focus on just finding the antiderivative of this expression so we're going to have to use u-substitution so let's set u equal to three x plus one that means d u the derivative of three x plus one is three times dx and now if we divide both sides by 3 dx is going to be du over 3. so we can substitute 3x plus 1 with the u variables so we're going to have 1 over u squared and then we can replace dx with du over 3. so what i'm going to do is move the 3 to the front and move the u variable to the top so this becomes one third integral u to the negative two d u so the antiderivative of u to the negative two it's going to be u to the negative one divided by negative one and if we rewrite it it's going to be negative one over three times u if we bring the u variable back to the bottom so now our last step is to replace u with three x plus one so we're gonna get negative one over three times three x plus one plus c so that's the antiderivative of this expression so now what we have is the limit as t goes to infinity and so we can replace this with what we have here so it's negative one over three times three x plus one and we don't need to worry about the constant c and so we're going to evaluate this from one to t so this becomes the limit as t goes to infinity and we can replace x with t so it's three times three t plus one and then it's going to be minus now we need to plug in 1 into this expression and so don't forget there's another negative sign it's very easy to forget that so replacing x with one is just going to be three and then plus one so what i'm going to do here is distribute the three so this is now negative 1 over 3 times 3t so that's 90 and then we have 3 times 1. now these two negative signs will cancel so it's going to be plus 1 and here we have 3 plus 1 which is 4 times 3 that gives us 12. now what is the limit as t goes to infinity for this expression negative one over nine t plus three now we know that's going to become zero and so it's going to be zero plus one over twelve so the final answer is one over twelve and because it's not infinity this is convergent so that's the answer for those of you who may want to show why this is equal to zero here's what you can do so first you need to keep in mind that the limit as t goes to infinity for one over t is zero so starting with this expression what you need to do is multiply the top and the bottom by one over t and so you're going to get the limit as t goes into infinity negative one over over t divided by so here 9 t times 1 over t the t's will cancel and so you're just going to get 9 and then plus 3 over t so as t goes into infinity we know that one over t becomes zero so this becomes i guess you could say negative zero which is the same as zero and then nine plus this is three times one over t if you uh see it this way and so that becomes three times zero and so you get zero over nine which is zero so that's how you could show that the limit goes to zero for this example but for the problem it was zero plus one over twelve so the final answer is one over twelve