In Exercises 35–68, use integration, the Direct Comparison Tes...

Question

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.

∫ from 0 to ∞ of (dθ / (θ² - 1))

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Determine whether the improper integral converges or diverges. The integral evaluated from zero to infinity of d x divided by x 2 minus 9. OK, so it appears for this particular problem we're given an improper integral, and we're asked to determine whether or not this improper integral will converge or diverge. So, it either will be A converges or B diverges. So now that we know what we're ultimately trying to solve for. Our first major step that we need to take based on our prior knowledge when we look at the integral that is provided to us, we should be able to recognize that the integrand has a vertical singularity. At X is equal to 3. Since x 2 minus 9 is equal to parentheses of x minus 3 multiplied by x + 3 in parentheses, which will lie inside the interval square bracket of 0, infinity. Fantastic. So, with that in mind, and we need to note that the interval will be unbounded, so let's make a note that it's gonna be UB unbounded. At plus infinity, positive infinity. So, with that in mind, we will therefore need to split the integral at X is equal to 3. So with that in mind, we need to take the interval evaluated from 0 to infinity of dx divided by x 2 minus 9 will be equal to the limit as b approaches 3 from the left of the integral evaluate from 0 to b of dx divided by x 2 minus 9. Plus the limit as B or well, rather A, so, so we didn't get, so we don't mix up our limits. So we'll say this is plus the limit as A approaches 3 from the right. Of the integral evaluated from A to positive infinity. Of DX divided by. So DX divided by. X to the power of 2 minus 9. Fantastic. And we need to note, based on our prior knowledge that both one-sided limits must be finite in order for convergence, in order for it to converge. So we need to note once again, it's very important that both one side limits have to be finite in order for it to converge. So with that in mind. We need to now find the antiderivative, and to find the antiderivative, we will need to recall and use the partial fractions method. So you go ahead and write that 1 divided by x 2 minus 9 will be equal to A divided by X minus 3 plus B divided by X + 3. Fantastic. So that will mean that 1 will be equal to, when we do a little bit of rearranging to have 1 be on one side of our equal signs for our expression, you will find that, and once again, you can use algebra to do this. So 1 will be equal to a multiplied by the parentheses of X + 3 plus B multiplied by the parentheses of X minus 3. Fantastic. So now we need to let X be equal to 3, so we could find the value of A. So 1 will be equal to a multiplied by parentheses of 3 + 3 plus B multiplied by parentheses of 3 minus 3. And when we rearrange this expression to isolate and solve for A, we will find that A is simply equal to 1/6. And I've skipped a few steps, but you're just using algebra to isolate and solve for the A variable. And similarly, we need to use a similar methodology to solve for B. And to solve for B, we need to set x equal to -3 in order to find B. So we'll take 1 equal to a multiplied by parentheses of 3 + 3 plus B multiplied by parentheses of 3 minus 3. Which will mean when we isolate and solve for B, that B will be simply equal to -16. Thus, we can go ahead and write that 1 divided by x 2 minus 9 will be equal to 1 divided by 6 multiplied by parentheses of X plus 3. Or rather X minus 3, we have to make sure we have our signs correct. So it's gonna be 1 divided by 6 multiplied by the parentheses of X minus 3. -1 divided by 6 multiplied by X + 3. Which will be equal to 1 divided by 6 or 1/6 multiplied by parentheses, 1 divided by X minus 3 minus 1 divided by X + 3. OK. So We will note that the integral of dx divided by x to the 2 minus 9 will be equal to 1/6 multiplied by the integral of 1 divided by x minus 3 minus 1 divided by x plus 3. D X, which will be equal to. So drum roll, this will be all equal to 1/6 multiplied by square brackets of the natural log of the absolute value of X minus 3 minus the natural log of the absolute value of X + 3 plus C. Which will be equal to 1 divided by 6 multiplied by the natural log of the absolute value of x minus 3 divided by x + 3 plus C, where C denotes some constant variable. So now, at this stage, we need to evaluate the limit as B approaches 3 from the left of the integral, evaluate from 0 to B of DX divided by X 2 minus 9. So with that in mind, we will find that this will be equal to the limit as b approaches 3 from the left of square brackets of 1 divided by 6 multiplied by the natural log of the absolute value of x minus 3 divided by x plus 3 evaluated from drum roll 0 to b. Fantastic, and this will be equal to the limit as b approaches 3 from the left of square brackets of 1 divided by 6 or 1/6 multiplied by parentheses of the natural log of the absolute value of. B minus 3 divided by B + 3. Minus, and then we're gonna subtract the natural log of the absolute value of 0 minus 3 divided by 0 + 3, fantastic. And this will be. Equal to. So essentially what we're doing is we're simplifying as we do our, or as we perform our evaluation. So we're trying to simplify down to our most simple form. I'm going to skip a few steps, but when we do simplify down to its most simple form, we will find that will be equal to square brackets of 1/6 multiplied by the natural log of 0 from the right. Which will be equal to negative infinity when we perform the calculation. So because this one-sided limit is finite, the, or rather it's not finite, it's infinite, and because the one-sided limit is infinite, that means therefore the improper integral cannot converge to a finite value. Because there is no further computation of the right piece is necessary to conclude divergence, because we confirm that already one limit leads to an infinite answer, so we don't have to worry about the other limit. Thus, without a doubt, we could say, without with certainty that this improper interval diverges, and that is our final answer. Hooray, we did it. So going back to look at our multiple choice answers, that will mean that our final answer without a doubt will be B diverges, and that's it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.∫ from 0 to ∞ of (dθ / (θ² - 1))

Watch next

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.
∫ from 0 to ∞ of (dθ / (θ² - 1))