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Homework Help: Improper Integral

  1. Jul 31, 2005 #1
    I have to analize the convergence of the following integral:

    [tex]\int_0^1 \frac {x^2+1} {\sqrt x * (1-x)^{5/4}}[/tex]

    I tried to divide it between 0-1/2 and 1/2-1 and on the first one i reached to:
    [tex]\int_0^{1/2} \frac {x^2+1} {\sqrt x * (1-x)^{5/4}}<=\int_0^{1/2} \frac {x^2+1} {x^{14/4}}[/tex]
    can i say that this integral converges and therefore the orgininal converges???, and more important, how would i justify that the last integral converges in an exam????
    please correct any mistakes that i probably had made, and forgive me for me awful english.

    Many Thanks, Paul.
     
  2. jcsd
  3. Aug 1, 2005 #2

    HallsofIvy

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    How did you get that "14/4". If you just add the powers in the denominator you get 13/4. Are you arguing that [tex]\frac{x^2+1}{x^{\frac{13}{4}}}< \frac{x^2+1}{x^{\frac{14}{4}}}?

    In any case, you can't just add the exponents like that. If x is close to 0, then
    1- x is close to 1 and x2+1 is close to 1. For x close to 0, the integrand is close to [tex]\frac{1}{\sqrt{x}}= x^{-\frac{1}{2}}[/tex]. And, of course, that integral exists.

    To look at what happens for x close to 1, it might help to make the substitution
    u= 1- x. Then x2+ 1= 2- 2u+ u2 and the integrand becomes [tex]\frac{2- 2u+ u}{\sqrt{1-u}u^{\frac{5}{4}}[/tex]. What is that like when u is close to 0?
     
  4. Aug 1, 2005 #3
    The first integrand varies as 1/sqrt(x) near zero ( I checked what the expression becomes for small values of x by means of the power series expansion.) .So, the integral will converge or diverge according as the integral of 1/sqrt(x).It converges.
    Similarly,to put in awful technical language, the integrand becomes
    (1^2 +1) /sqrt(1)*( 1-x)^5/4 near 1 & diverges because the integral
    dx/(1-x)^5/4 diverges if taken over a neighbourhood of 1.
    These ideas must be put in terms of inequalities; e.g. in the first case, constants p & q could be found out such that p/sqrt(x) < the integrand < q/sqrt(x)
    holds in a neighbourhood of zero.
    I am, with great respect,
    Einstone.
     
  5. Aug 1, 2005 #4

    HallsofIvy

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    Don't know what happened to my first response- and no "edit" button to fix it!

    What I was saying is that since x2+1 and 1- x are both close to 1 for x close to 0, the integrand is close to x-1/2 which is integrable.

    If you let u= 1- x, then x= 1- u so x2+1= u2- 2u+ 1 and the integrand becomes [tex]\frac{u^2- 2u+ 1}{\sqrt{1-u}u^\frac{5}{4}}[/tex].
    For u close to 1 that is close to [tex]\frac{1}{u^\frac{5}{4}}[/tex] which does not converge.
     
  6. Aug 1, 2005 #5
    thank you

    First of all, thank you for your prompt and clear reply. I think i now understand how to do this exercise, but i dont know how to explain on a test why i can replace the integrands with another integrand that behaves the same way when x is close to 0 in the first part of the integral for example. I really dont understand the part of your explanation where you say that it can be found two constants p and q such that p/sqrt(x)<integrand<q/sqrt(x). I think that you are trying to show that i can replace the integrand by 1/sqrt(x)
    but i dont know if it would be enough justification on a test.
    Many Thanks, Paul.
     
  7. Aug 2, 2005 #6

    saltydog

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    Would this be sufficient:

    [tex]\frac{x^2+1}{\sqrt{x}(1-x)^{5/4}}>\frac{1}{\sqrt{x}(1-x)^{5/4}}>\frac{1}{\sqrt{x}(1-x)^2}[/tex]

    for all [itex]x\in (0,1)[/tex]

    Making the substitution [itex]u=\sqrt{x}[/itex] and using partial fractions, we obtain:

    [tex]\int \frac{dx}{\sqrt{x}(1-x)^2}=\frac{1}{2}\left[\frac{1}{1-\sqrt{x}}-ln(1-\sqrt{x})-\frac{1}{1+\sqrt{x}}+ln(1+\sqrt{x})\right][/tex]

    which diverges as x goes to 1.

    Thus:

    [tex]\int_0^1 \frac{x^2+1}{\sqrt{x}(1-x)^{5/4}}[/tex]

    also diverges.
     
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