Improper Integral Comparisons

Section: 
7.8
Date: 
Monday, September 13, 2010 - 14:00
AttachmentSize
fall2010math2300_improper.comparison.pdf164.5 KB

Math 2300 Section 005 – Calculus II – Fall 2010

Comparison of Improper Integrals Examples – September 13, 2010

Comparison of Improper Integrals

  1. \int _{1}^{\infty}\frac{1}{\sqrt{x^{3}+5}}\, dx

    Solution: This is example 1 from Section 7.8 in the text.

  2. Converge or Diverge?

    \int _{{10}}^{\infty}\frac{1}{\ln(t)}\, dt

    Solution: Notice that for t\ge 10 we have

    \frac{1}{\ln(t)}>\frac{1}{t}

    and so

    \int _{{10}}^{\infty}\frac{1}{\ln(t)}\, dt>\int _{{10}}^{\infty}\frac{1}{t}\, dt=\infty

    and so this integral diverges. Notice that starting the integral limit at t=10 makes no difference here. The simple fact is that we know

    \int _{1}^{\infty}\frac{1}{t}\, dt=\int _{1}^{{b}}\frac{1}{t}\, dt+\int _{b}^{\infty}\frac{1}{t}\, dt

    is infinite, while the integral

    \int _{1}^{b}\frac{1}{t}\, dt

    is finite for any b satisfying 1<b<\infty. It is really the integral on the far left, the one with the infinite limit, that causes the problem here.

  3. Converge or Diverge?

    \int _{{10}}^{\infty}\frac{1}{\ln(t)-1}\, dt

    Solution: This is really the same as the previous example. In this case, we have

    \frac{1}{\ln(t)-1}>\frac{1}{\ln(t)}>\frac{1}{t}

    and so the comparison test says that

    \int _{{10}}^{\infty}\frac{1}{\ln(t)-1}\, dt>\int _{{10}}^{\infty}\frac{1}{\ln(t)}\, dt>\int _{{10}}^{\infty}\frac{1}{t}\, dt

    where we already knew that the integral on the far right diverges, and the previous example showed that the integral in the middle diverges. Both with work to show that the integral on the far left diverges.

  4. Converge or Diverge?

    \int _{{1}}^{\infty}\frac{\cos x+2}{\sqrt{x}}\, dx

    Solution: Notice that

    \frac{1}{\sqrt{x}}\le\frac{\cos x+2}{\sqrt{x}}\le\frac{3}{\sqrt{x}}

    simply because the smallest and largest values that \cos x can take on are -1 and 1, respectively. So, we have

    \int _{1}^{\infty}\frac{1}{\sqrt{x}}\, dx\le\int _{1}^{\infty}\frac{\cos x+2}{\sqrt{x}}\, dx\le\int _{1}^{\infty}\frac{3}{\sqrt{x}}\, dx=3\int _{1}^{\infty}\frac{1}{\sqrt{x}}\, dx.

    Of course, the side that we're interested in here is the left, since we know that the integral of 1/\sqrt{x} from 1 to \infty is divergent. Thus, the integral in question diverges.

  5. Converge or Diverge?

    \int _{1}^{\infty}\frac{\sin x+\cos x+2}{x^{2}}\, dx

    Solution: This one is calculated much along the same lines as the first, but the result is very different. First, we wish to have an accurate an approximation to bound the integrand by as possible. While it is true that both \sin x and \cos x have a maximum and minimum value of 1 and -1 respectively, we should try to find the maximum and minimum value of their sum… and since each of the individual functions has a max/min at different parts of the unit circle, the max and min of their sum will not be 2 and -2, repsectively.

    The first derivative of f(x)=\sin x+\cos x is \cos x-\sin x, which has zeros at x=\pi/4 and x=5\pi/4. The second derivative is -\sin x-\cos x, which (by the second derivative test) tells us that \sin x+\cos x has a minimum at 5\pi/4 and a maximum at x=\pi/4, up to periodicity of the function. This allows us to state that

    (2-\sqrt{2})\int _{1}^{\infty}\frac{1}{x^{2}}\, dx=\int _{1}^{\infty}\frac{-\sqrt{2}+2}{x^{2}}\, dx<\int _{1}^{\infty}\frac{\sin x+\cos x+2}{x^{2}}\, dx<\int _{1}^{\infty}\frac{\sqrt{2}+2}{x^{2}}\, dx=(\sqrt{2}+2)\int _{1}^{\infty}\frac{1}{x^{2}}\, dx.

    Which of these bounds are we actually interested in? The one of the left tells us nothing, since it only says that the integral we are considering is greater than one that is known to converge. In order to get information about convergence, the comparison test says that we much bound our integral above by one that we already know to converge… which is what the right side of the above equation does. So, the integral in question converges. (And we have relatively little idea what it may be equal to.)

  6. Converge or Diverge?

    \int _{0}^{1}\frac{x}{13\sqrt{x^{3}}}\, dx

    Solution: Simplifying gives

    \int _{0}^{1}\frac{x}{13\sqrt{x^{3}}}\, dx=\int _{0}^{1}\frac{x}{13x^{{3/2}}}\, dx=\frac{1}{13}\int _{0}^{1}\frac{1}{\sqrt{x}}\, dx

    We know already that the integral on the right converges and we can calculate it using the previous section on improper integrals. I.e., in this case we are lucky and don't need to use the comparison test at all. But, what if we had

    \int _{0}^{1}\frac{x}{13\sqrt{x^{\pi}}}\, dx

    to start with? This looks much more challenging, but we have

    \int _{0}^{1}\frac{x}{13\sqrt{x^{\pi}}}\, dx=\frac{1}{13}\int _{0}^{1}x^{{1-\frac{\pi}{2}}}\, dx

    and since -0.75<1-\pi/2<-0.5 we have

    \frac{1}{13}\int _{0}^{1}\frac{1}{x^{{1/2}}}\, dx<\frac{}{}\int _{{0}}^{1}\frac{x}{13\sqrt{x^{\pi}}}\, dx<\frac{1}{13}\int _{0}^{1}\frac{1}{x^{{3/4}}}\, dx

    and so the right side tells us that this integral converges.

© 2011 Jason B. Hill. All Rights Reserved.