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A couple of Improper Intergrals i have problems with

  1. Feb 5, 2013 #1
    INTEGRALS​


    1.[itex]\int^{\frac{\Pi}{2}}_{0} \frac{x}{sin(x^2)} \, dx[/itex]

    2.[itex]\int^{\infty}_{-\infty} \frac{1}{t} \, dt[/itex]

    3.[itex]\int^{2}_{1} \frac{1}{xln(x)^4} \, dx[/itex]

    RESOLVING THE INTEGRALS​


    1. I firts resolved the definte Integral and then applied the
    [tex]\lim_{b\rightarrow +0} -\frac{1}{2}ln(cotg(\frac{x^2}{2})|^{\frac{\pi}{2}}_{b} = \infty[/tex]

    So, i come up to the resault that the integral does not converge. But i dont know if this procedure is ok, and if there is some way to do it by comparison.

    2.So, this one i couldn't finish, you see, i come up to the next limit:

    [tex]\lim_{b\rightarrow +\infty} -ln |-b|+ ln |b| = ?[/tex]

    So, i dont know how to resolve the limit or how to analyse the integral by comparison.

    3.Well, with this one i just dont know what to do...

    Thanks, and sorry for my english.
     
    Last edited: Feb 5, 2013
  2. jcsd
  3. Feb 5, 2013 #2

    Mark44

    Staff: Mentor

    Your antiderivative is wrong here. Show us how you got from the indefinite integral to your antiderivative. We can worry about the definite integral when you get the integration right.
     
  4. Feb 5, 2013 #3
    To be honest i started de Integral by doing the subtitution
    [tex]u = x^2[/tex][tex]du = 2x[/tex]
    So,

    [itex]\int \frac{x}{sen(x^2)} \, dx = \frac{1}{2}\int^{\frac{\pi^2}{4}_{1}\frac{1}{sen(u)} \, dx[/itex]

    Then I looked up in the web "Wolfram|Alpha" how to proceed but they just jump from that integral i wrote, to the one you say it`s wrong. Later with the help of a PF MENTOR i came up to this Integral:

    [itex]-\frac{1}{2}\int^{1}_{-0.78}\frac{1}{1-v^2} \, dv[/itex]

    v = cos(u)
    dv = -sen(u)
     
    Last edited by a moderator: Feb 5, 2013
  5. Feb 5, 2013 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    1/sin(u) can also be written csc(u). Probably best to just look up the integral of that. Or ask Wolfram to get the indefinite integral. Wolfram will also tell you the original integral diverges, but don't use that part. You should figure out why it diverges yourself.
     
  6. Feb 5, 2013 #5
    Yes, of course. Well that was what actually did, i mean i subtituted u=x2, du=2xdx to get [itex]\frac{1}{2}\int\frac{1}{sen(u)} \, du[/itex] which is the integral of csc(u) and looked it up on Wolfram and it gives me the funcion [itex]f(x) = ln(sen(\frac{u}{2}))-ln(cos(\frac{u}{2})) + C[/itex].

    So now i apply
    [tex]\lim_{b\rightarrow 0} {ln(sen(\frac{x^2}{2}))-ln(cos(\frac{x^2}{2})) + C}|^{\frac{\pi}{2}}_{b}[/tex]

    Then
    [tex]\lim_{b\rightarrow 0} {ln(0.94) - ln(sen(\frac{b^2}{2})) - ln(0.34) + ln(cos(\frac{b^2}{2}))}[/tex]

    And so that is equal to [itex]\infty[/itex].

    That mean that does not converge but i wish there was another way to prove it does not converge like by comparison ( i don't know if its called like that in english) and not to do all the integral.

    Sorry for my english and if there is some mistakes its reaaly hard to write math equation in here.

    Thanks again
     
  7. Feb 5, 2013 #6

    Mark44

    Staff: Mentor

    I don't know where you're getting what you show above. This is how the integral usually appears.
    $$ 1/2 \int csc(u) du = -(1/2) ln|csc(u) + cot(u)| + C$$

    I think you might be making errors in your work.
     
  8. Feb 5, 2013 #7
    YES, your absolutly right..Im getting it from Wolfram the thing is that what i'm writing is the equivalent for resticted u values. I'm sorry. So...

    [itex]\int^{\frac{\pi}{2}}_{0}csc(u) \, du = -\frac{1}{2}ln(csc(u) + cot(u)) + C |^{\frac{\pi}{2}}_{0}[/itex]

    Then,
    [tex]\lim_{b\rightarrow 0} {-\frac{1}{2}ln|csc(u) + cot(u)| + C |^{\frac{\pi}{2}}_{b}}[/tex]

    [tex]\lim_{b \to 0} -\frac{1}{2}ln|1.60 - 1.25| + \frac{1}{2}ln|csc(b^2) + cot(b^2)|[/tex]

    So that is equal to

    [itex]-\frac{1}{2}ln|1.60 - 1.25| + \frac{1}{2}ln|\infty| = \infty[/itex]

    Thanks again...
     
    Last edited by a moderator: Feb 5, 2013
  9. Feb 5, 2013 #8

    Mark44

    Staff: Mentor

    No, those aren't the right limits of integration. Those are values of x, but u = x2.

    You can do either of the following:
    1. Change the x limits to the appropriate values of u.
    2. Undo the u substitution.
     
  10. Feb 5, 2013 #9
    I actually wrote the wrong limits of integration but the math i did it right. i just wrote it wrongly,but yes it would be
    [itex]\int^{\frac{\pi}{2}}_{0}csc(x^2) \, du = -\frac{1}{2}ln(csc(x^2) + cot(x^2)) + C |^{\frac{\pi}{2}}_{0}[/itex]

    Then
    [tex]\lim_{b\rightarrow 0} {-\frac{1}{2}ln|csc(x^2) + cot(x^2)| + C |^{\frac{\pi}{2}}_{b}}[/tex]
    [tex]\lim_{b \to 0} -\frac{1}{2}ln|1.60 - 1.25| + \frac{1}{2}ln|csc(b^2) + cot(b^2)|[/tex]
     
  11. Feb 5, 2013 #10

    Mark44

    Staff: Mentor

    And it looks like both terms in the 2nd ln expression approach infinity.
     
  12. Feb 5, 2013 #11
    So the integral does not converge after all...
     
  13. Feb 5, 2013 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Wolfram Alpha does give log(sin(u/2))+log(cos(u/2)) as the antiderivative instead of the more usual -log(csc(u)+cot(u)). There's nothing wrong with SclaP's work. I suggested looking it up rather than deriving it. Since that's what I would do and the main point is to evaluate the limit. Both give the same result.
     
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