Improper Integrals

Wednesday, September 8, 2010 - 14:00 - Thursday, September 9, 2010 - 15:00
fall2010math2300_improper.integrals.pdf109.2 KB

Math 2300 Section 005 – Calculus II – Fall 2010

Improper Integrals Examples – September 8-10, 2010

  1. Limits of Integration Not Finite

    \int _{1}^{\infty}\frac{1}{x^{p}}\, dx

    Solution: We find that

    \displaystyle\int _{1}^{\infty}\frac{1}{x^{p}}\, dx \displaystyle=\lim _{{b\rightarrow\infty}}\int _{1}^{b}\frac{1}{x^{p}}\, dx
    \displaystyle(p\neq 1) \displaystyle=\lim _{{b\rightarrow\infty}}\left[\frac{1}{-p+1}x^{{-p+1}}\right]_{1}^{b}
    \displaystyle(p\neq 1) \displaystyle=\frac{1}{-p+1}\left(\lim _{{b\rightarrow\infty}}b^{{-p+1}}-1\right).

    This will converge if and only if p>1, meaning that it will diverge when p\le 1. We showed this in class in somewhat painful detail.

  2. Function Not Finite

    \int _{0}^{1}\frac{1}{x^{p}}\, dx

    Solution: In order to use the Fundamental Theorem of Calculus, the function needs to be continuous (and so not infinite). We accomplish this by taking values of the integral limit that cause problems (x=0), and approaching them with a limit. We find the following.

    \displaystyle\int _{0}^{1}\frac{1}{x^{p}}\, dx \displaystyle=\lim _{{a\rightarrow 0^{+}}}\int _{a}^{1}\frac{1}{x^{p}}\, dx
    \displaystyle(p\neq 1) \displaystyle=\lim _{{a\rightarrow 0^{+}}}\left[\frac{1}{-p+1}x^{{-p+1}}\right]_{a}^{1}
    \displaystyle(p\neq 1) \displaystyle=\frac{1}{-p+1}\left(1-\lim _{{a\rightarrow 0^{+}}}a^{{-p+1}}\right)

    In this case, we find that the integral converges if p<1, and diverges otherwise. We also did this example with detail in class, but from this point on the calculations are rather straightforward. The idea is that when p becomes larger, the exponent will be a high power in the denominator, forcing the quantity in the limit to become very large.

  3. Limits of Integration Not Finite

    \int _{0}^{\infty}e^{{-3x}}\, dx


    \displaystyle\int _{0}^{\infty}e^{{-3x}}\, dx \displaystyle=\lim _{{b\rightarrow\infty}}\int _{0}^{b}e^{{-3x}}\, dx
    \displaystyle=\lim _{{b\rightarrow\infty}}\left[-\frac{1}{3}e^{{-3x}}\right]_{0}^{b}
    \displaystyle=\frac{-1}{3}\left(\lim _{{b\rightarrow\infty}}e^{{-3b}}-1\right)
  4. Limits of Integration Not Finite

    \int _{0}^{\infty}\frac{x}{e^{x}}\, dx

    Solution: Using integration by parts with the functions u=x and v^{{\prime}}=e^{{-x}}, and using L'Hopital's Rule:

    \displaystyle\int _{0}^{\infty}\frac{x}{e^{x}}\, dx \displaystyle=\lim _{{b\rightarrow\infty}}\int _{0}^{b}\frac{x}{e^{x}}\, dx
    \displaystyle=\lim _{{b\rightarrow\infty}}\left(\left[\frac{-x}{e^{x}}\right]_{0}^{b}+\int _{0}^{b}\frac{1}{e^{x}}\, dx\right)
    \displaystyle=\lim _{{b\rightarrow\infty}}\left(-\frac{b}{e^{b}}+\left[-\frac{1}{e^{x}}\right]_{0}^{b}\right)
    \displaystyle=\lim _{{b\rightarrow\infty}}\left(-\frac{b}{e^{b}}-\frac{1}{e^{b}}+1\right)
  5. Limits of Integration Not Finite

    \int _{{-\infty}}^{\infty}e^{{-|x|}}\, dx


    \displaystyle\int _{{-\infty}}^{\infty}e^{{-|x|}}\, dx \displaystyle=2\int _{0}^{\infty}e^{{-x}}\, dx
    \displaystyle=2\lim _{{b\rightarrow\infty}}\int _{0}^{b}e^{{-x}}\, dx
    \displaystyle=2\lim _{{b\rightarrow\infty}}\left[-\frac{1}{e^{x}}\right]_{0}^{b}
  6. Function Not Finite

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


    \displaystyle\int _{0}^{1}\sqrt{\frac{1}{x}}\, dx \displaystyle=\lim _{{a\rightarrow 0^{+}}}\int _{a}^{1}\sqrt{\frac{1}{x}}\, dx
    \displaystyle=\lim _{{a\rightarrow 0^{+}}}\int _{a}^{1}x^{{-1/2}}\, dx
    \displaystyle=\lim _{{a\rightarrow 0^{+}}}\left[2\sqrt{x}\right]_{a}^{1}

© 2011 Jason B. Hill. All Rights Reserved.