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2 (simple) Improper Integrals I'm stuck with

  1. Oct 16, 2006 #1
    Hi everyone

    Here are two improper integrals and I have to find the values of [itex]p[/itex] for which they are convergent

    1. [tex]
    \int_{0}^{\infty} \frac{1-e^{-x}}{x^{p}}dx

    2. [tex]
    \int_{0}^{\infty} \frac{t^{p-1}}{1+t} dt

    As I do not know the answers and am somewhat stuck with the reasoning, I would be grateful if someone could check these for me and let me know the correct answers

    First Problem

    [tex]\int_{0}^{\infty} \frac{1-e^{-x}}{x^{p}}dx = I_{1} + I_{2}[/tex]


    [tex]I_{1} = \int_{0}^{1} \frac{1-e^{-x}}{x^{p}}dx[/tex]
    [tex]I_{2} = \int_{1}^{\infty} \frac{1-e^{-x}}{x^{p}}dx[/tex]


    [tex]\frac{1-e^{-x}}{x^{p}} \leq \frac{x}{x^{p}} = \frac{1}{x^{p-1}}[/tex]

    So by comparison test (or limit comparison test) [itex]I_{1}[/itex] converges for [itex]2-p > 0[/itex] or [itex]p < 2[/itex].

    Similarly, [itex]I_{2}[/itex] converges for [itex]p > 1[/itex]. Putting these together, the original integral converges for [itex]1 < p < 2[/itex]. Is this correct?

    Second Problem

    \int_{0}^{\infty} \frac{t^{p-1}}{1+t} dt = I_{1} + I_{2}


    [tex]I_{1} = \int_{0}^{1} \frac{t^{p-1}}{1+t} dt[/tex]
    [tex]I_{2} = \int_{1}^{\infty} \frac{t^{p-1}}{1+t} dt[/tex]


    [tex]\frac{t^{p-1}}{1+t} \leq t^{p-1}[/tex]

    by comparison test, we can see that [itex]I_{1}[/itex] will converge for [itex]p > 0[/itex]. But this reasoning applied to [itex]I_{2}[/itex] yields a wrong answer. Whats going wrong here?

    I think the correct answer to second problem is [itex] 0 < p < 1[/itex].
  2. jcsd
  3. Oct 16, 2006 #2
    1. This is correct

    For 2 how did you conclude that it [tex] I_{1} [/tex] will converge for [tex] p > 0 [/tex]? Use the limit comparison test.
    Last edited: Oct 16, 2006
  4. Oct 16, 2006 #3
    Thanks :smile:...please...could you also have a look at the second one.
  5. Oct 16, 2006 #4
    My question then is two-fold.

    First if I compare with [itex]t^{p-1}[/itex] then

    [tex]\lim_{t \rightarrow 0}}\left(\frac{t^{p-1}}{1+t}\right)\left(\frac{1}{t^{p-1}}\right) = 1[/tex]

    So the integral converges if [itex]\int_{0}^{1}t^{p-1}dt[/itex] converges, i.e. for [itex]p > 0[/itex]. (I think something's going wrong here...can you suggest a more general method? I think that by comparing with a power of p, I am forcing only that solution for which the right hand integral is convergent, which depends on p!)
  6. Oct 16, 2006 #5
    This is for [itex]1 < t < \infty[/itex]. What about the first integral?
  7. Oct 16, 2006 #6
    it converges for [tex] p>0 [/tex]. so the whole integral diverges because [tex] p < 0 [/tex] and [tex] p > 0 [/tex] cant both be true.
    Last edited: Oct 16, 2006
  8. Oct 16, 2006 #7
    No, it actually converges:

    Suppose I take p = 0.5, then the integral is simply,

    [tex]\int_{0}^{\infty} \frac{1}{\sqrt{x}(1+x)}dx = \int_{0}^{\infty} \frac{2}{(1+t^{2})}dt [/tex]

    which converges to [itex]\pi[/itex].
    Last edited: Oct 16, 2006
  9. Oct 16, 2006 #8
    indeed, you are correct, it converges for [tex] 0<p<1 [/tex]. I used the same argument as in problem 1.

    you have to set [tex] 1-p < 1 [/tex] for [tex] I_{1} [/tex], and [tex] 1-p > 0 [/tex] for [tex] I_{2} [/tex]
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