Section 7.9 : Comparison Test for Improper Integrals
5. Use the Comparison Test to determine if the following integral converges or diverges.
\[\int_{6}^{\infty }{{\frac{{{w^2} + 1}}{{{w^3}\left( {{{\cos }^2}\left( w \right) + 1} \right)}}\,dw}}\]Show All Steps Hide All Steps
The first thing that we really need to do here is to take a guess on whether we think the integral converges or diverges.
The numerator of this function is a polynomial and we know that as \(w \to \infty \) the behavior of polynomials will be the same as the behavior of the largest power of \(w\). Also the cosine term in the denominator is bounded and never gets too large or small.
Therefore, it looks like this integral should behave like,
\[\int_{6}^{\infty }{{\frac{{{w^2}}}{{{w^3}}}\,dw}} = \int_{6}^{\infty }{{\frac{1}{w}\,dw}}\]Then, by the fact from the previous section, we know that this integral diverges since \(p = 1 \le 1\) .
Therefore, we can guess that the integral,
\[\int_{6}^{\infty }{{\frac{{{w^2} + 1}}{{{w^3}\left( {{{\cos }^2}\left( w \right) + 1} \right)}}\,dw}}\]will diverge.
Be careful from this point on! One of the biggest mistakes that many students make at this point is to say that because we’ve guessed the integral diverges we now know that it diverges and that’s all that we need to do and they move on to the next problem.
Another mistake that students often make here is to say that because we’ve guessed that the integral diverges they make sure that the remainder of the work in the problem supports that guess even if the work they do isn’t correct.
All we’ve done is make a guess. Now we need to prove that our guess was the correct one. This may seem like a silly thing to go on about, but keep in mind that at this level the problems you are working with tend to be pretty simple (even if they don’t always seem like it). This means that it will often (or at least often once you get comfortable with these kinds of problems) be pretty clear that the integral converges or diverges.
When these kinds of problems arise in other sections/applications it may not always be so clear if our guess is correct or not and it can take some real work to prove the guess. So, we need to be in the habit of actually doing the work to prove the guess so we are capable of doing it when it is required.
The hard part with these problems is often not making the guess but instead proving the guess! So let’s continue on with the problem.
Recall that we used an area analogy in the notes of this section to help us determine if we want a larger or smaller function for the comparison test.
We want to prove that the integral diverges so if we find a smaller function that we know diverges the area analogy tells us that there would be an infinite amount of area under the smaller function.
Our function, which would be larger, would then also have an infinite amount of area under it. There is no way we can have an finite amount of area covering an infinite amount of area!
Note that the opposite situation does us no good. If we find a larger function that we know diverges (and hence will have a infinite amount of area under it) our function (which is now smaller) can have either a finite amount of area or an infinite area under it.
In other words, if we find a larger function that we know diverges this will tell us nothing about our function. However, if we find a smaller function that we know diverges this will force our function to also diverge.
Therefore we need to find a smaller function that we know diverges.
Show Step 3Okay, now that we know we need to find a smaller function that we know diverges.
So, let’s start with the function from the integral. It is a fraction and we know that we can make a fraction smaller by making the numerator smaller or the denominator larger. Also note that for \(w > 6\) (which we can assume from the limits on the integral) we have,
\[{w^2} + 1 > {w^2}\]Therefore, we have,
\[\frac{{{w^2} + 1}}{{{w^3}\left( {{{\cos }^2}\left( w \right) + 1} \right)}} > \frac{{{w^2}}}{{{w^3}\left( {{{\cos }^2}\left( w \right) + 1} \right)}} = \frac{1}{{w\left( {{{\cos }^2}\left( w \right) + 1} \right)}}\]since we replaced the numerator with something that we know is smaller.
Show Step 4It is at this point that students again often make mistakes with this kind of problem. After doing one manipulation of the numerator or denominator they stop the manipulation and declare that the new function must diverge (since that is what we want after all) and move on to the next problem.
Recall however that we must know that the new function diverges and we’ve not gotten to a function yet that we know diverges. To get to a function that we know diverges we need to do one more manipulation of the function.
For this step we know that \(0 \le {\cos ^2}\left( w \right) \le 1\) and so we will have,
\[{\cos ^2}\left( w \right) + 1 < 1 + 1 = 2\]Therefore, we have,
\[\frac{1}{{w\left( {{{\cos }^2}\left( w \right) + 1} \right)}} > \frac{1}{{w\left( 2 \right)}} = \frac{1}{{2w}}\]since we replaced the denominator with something that we know is larger.
Show Step 5Finally, putting the results of Steps 3 & 4 together we have,
\[\frac{{{w^2} + 1}}{{{w^3}\left( {{{\cos }^2}\left( w \right) + 1} \right)}} > \frac{1}{{2w}}\]and we know that,
\[\int_{6}^{\infty }{{\frac{1}{{2w}}\,dw}} = \frac{1}{2}\int_{6}^{\infty }{{\frac{1}{w}\,dw}}\]diverges. Then because the function in this integral is smaller than the function in the original integral the Comparison Test tells us that,
\[\int_{6}^{\infty }{{\frac{{{w^2} + 1}}{{{w^3}\left( {{{\cos }^2}\left( w \right) + 1} \right)}}\,dw}}\]must also diverge.