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Difficult Integral: Evaluating the Limits

  1. Nov 30, 2007 #1
    1. The problem statement, all variables and given/known data

    Ok, I am evaluating the following integral,

    [tex]
    {{\int_{0}^{\infty}}{\frac{{R_{0}}ds}{\left({{s}^{2}+{R_{0}}^{2}\right)^{\frac{3}{2}}}}}
    [/tex]

    Following through with trigonometric substitution I have the following,

    [tex]
    {\left[{{{\frac{1}{R_{0}}}{\cdot}{\frac{s}{(s^2+{R_{0}}^2)^{\frac{1}{2}}}}\right]_{0}^{\infty}}}
    [/tex]

    However, I am not quite sure what the result will be when I evaluate the integral.

    2. Relevant equations

    Trigonometric Substitution Techniques for evaluating Integrals.

    3. The attempt at a solution

    [tex]
    {\left[{{{\frac{1}{R_{0}}}{\cdot}{\frac{(\infty)}{({(\infty)}^2+{R_{0}}^2)^{\frac{1}{2}}}}}\right]-{\left[{{{\frac{1}{R_{0}}}{\cdot}{\frac{(0)}{({(0)}^2+{R_{0}}^2)^{\frac{1}{2}}}}}\right]
    [/tex]

    [tex]
    {\left[{{{\frac{1}{R_{0}}}{\cdot}{\frac{(\infty)}{({(\infty)}^2+{R_{0}}^2)^{\frac{1}{2}}}}}\right]-{\left[0\right]}\right]
    [/tex]

    However, how I am supposed to reduce the expression with the value of infinity plugged in, how would I reduce that expression?

    Any help is appreciated.

    Thanks,

    -PFStudent
     
    Last edited: Nov 30, 2007
  2. jcsd
  3. Nov 30, 2007 #2

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Are you sure you got the right result!? Did you check that if you differentiate you get back the integrand??

    (the limit will be zero since it goes to s/s^3 = 1/s^2 = 0 for large s)
     
  4. Nov 30, 2007 #3
    Hey,

    Thanks, for the quick reply. Actually my power for the denominator was incorrect, instead of [tex]{\frac{3}{2}}[/tex] it should have been, [tex]{\frac{1}{2}}[/tex].

    Yea.., I'm still not sure about what to do about that expression with infinity in it.

    Thanks,

    -PFStudent
     
  5. Dec 1, 2007 #4

    Gib Z

    User Avatar
    Homework Helper

    I Would first start by taking Out the constant and telling us the actual substitution you used!
     
  6. Dec 3, 2007 #5
    Hey,

    Thanks for the reply, Gib Z. When you refer to the "actual substitution," are you referring to the substitution used to integrate or used to evaluated the integral at the listed limits?

    If you are referring to the actual integration, the integral can be evaluated by first normalizing the binomial and then letting,

    [tex]
    {tan\theta} = {\frac{s}{R_{0}}}
    [/tex]

    From, here the integration should be pretty straight forward.

    If you are referring to the substitution used to evaluate the integral, then it as stated before,

    [tex]
    {\left[{{{\frac{1}{R_{0}}}{\cdot}{\frac{s}{(s^2+{R_{0}}^2)^{\frac{1}{2}}}}\right]_{0}^{\infty}}}
    [/tex]

    [tex]
    {\left[{{{\frac{1}{R_{0}}}{\cdot}{\frac{(\infty)}{({(\infty)}^2+{R_{0}}^2)^{\frac{1}{2}}}}}\right]-{\left[{{{\frac{1}{R_{0}}}{\cdot}{\frac{(0)}{({(0)}^2+{R_{0}}^2)^{\frac{1}{2}}}}}\right]
    [/tex]

    Also, after looking over this problem for a while...--am I correct to say that since,

    [tex]
    {\infty}>>{R_{0}}
    [/tex]

    Then, the [tex]{{R_{0}}^{2}}[/tex] term can be neglected, leaving me with,

    [tex]
    {\left[{{{\frac{1}{R_{0}}}{\cdot}{\frac{(\infty)}{({(\infty)}^2)^{\frac{1}{2}}}}}}\right]}
    [/tex]

    Which reduces to,

    [tex]
    {\frac{1}{R_{0}}}{\left(1\right)}
    [/tex]

    [tex]
    {\frac{1}{R_{0}}}
    [/tex]

    Is that right?

    Thanks,

    -PFStudent
     
    Last edited: Dec 3, 2007
  7. Dec 3, 2007 #6
    For a rigorous solution, avoid "plugging" in infinity. Infinity is a concept, not a number. You can't "plug" it into a variable and compute with it without being careful with indeterminate forms. You are trying to use the Fundamental Theorem of Calculus (FTC), but that theorem only applies to finite domains (i.e. non-infinite intervals). To get around that, you would use limits (which also answers your question regarding [tex]\infty >> R_0[/tex]. This may seem picky, but you should practice being precise with your writeup. Since the (FTC) requires finite domains, just integrate from 0 to some finite upper limit b, and then let b go to infinity.

    [tex]
    \int_0^\infty \frac{R_0}{(s^2+R_0^2)^{3/2}}\ ds = \displaystyle \lim_{b\to \infty}
    \int_0^b \frac{R_0}{(s^2+R_0^2)^{3/2}}\ ds
    = \lim_{b\to \infty} \left[ \frac{1}{R_0} \cdot \frac{s}{\sqrt{s^2+R_0^2}} \right]_0^b
    [/tex]
    [tex] = \lim_{b\to \infty} \left[ \frac{1}{R_0} \cdot \frac{b}{\sqrt{b^2+R_0^2}}
    - \frac{1}{R_0} \cdot \frac{0}{\sqrt{0+R^2}} \right]
    = \lim_{b\to \infty} \frac{b}{R_0\sqrt{b^2+R_0^2}} = \frac{1}{R_0}
    [/tex]

    The last equality can be achieved using basic methods for computing limits.
     
  8. Dec 3, 2007 #7
    Improper integral

    The type of integral you are evaluating is called in improper integral (which includes integrals with infinite limits and integrals whose function is discontinuous within the limits of integration). You cannot apply the FTC directly, but you can get around the problems by adjusting your domains, and then taking appropriate limits. This method is thoroughly (or should be) described in any decent calculus text.
     
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