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Definite Integral of ln*algebraic function

  1. Feb 7, 2005 #1
    Hello...

    I got a simple looking integral to solve but unfortunately couldn't do it

    [tex]I = \int_{0}^{1}\frac{x^2lnx}{\sqrt{1-x^2}}dx[/tex]

    Any suggestions?

    Thanks and cheers
    Vivek
     
    Last edited: Feb 7, 2005
  2. jcsd
  3. Feb 7, 2005 #2
    Here's what I did:

    Let [itex]J = \int \frac{x^2lnx}{\sqrt{1-x^2}}dx[/itex]. Integrating J by parts,

    [tex]J = lnx\int \frac{x^2}{\sqrt{1-x^2}}dx - \int \int \frac{x^2}{\sqrt{1-x^2}}dx \frac{dx}{x}[/tex]

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

    So [tex]J = lnx(\frac{1}{2}(sin^{-1}x-x\sqrt{1-x^2}+) - \frac{1}{2}\int \frac{\sin^{-1}x-x\sqrt{1-x^2}}{x}dx[/tex]

    One of the integrals to be further computed is [tex]\int \frac{\sin^{-1}x}{x}dx[/tex]. This leads us nowhere :cry:

    I can't think of any property of definite integrals (thanks to the ln function) which I can apply here. I know that this is an improper integral as such so I'll have to take limits after I've evaluated the corresponding indefinite integral somehow.

    Please help.....

    Mathematica gives the answer as [tex]\frac{-\pi}{8}(-1+ln4)[/tex] but I had to do this by hand...

    Thanks and cheers
    Vivek
     
    Last edited: Feb 7, 2005
  4. Feb 7, 2005 #3
    in your second line of equations with the double integral, are you sure you can bring that 1/x inside? I didnt think you could do that.
     
  5. Feb 7, 2005 #4

    HallsofIvy

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    That's NOT integration by parts! and
    [tex]\int \int dx \frac{dx}{x}[/tex]
    makes no sense at all- you can't do a double integral over the SAME variable!
     
  6. Feb 7, 2005 #5

    Curious3141

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    I've thought of a way you can evaluate the indef. integral. I haven't written out the working, but I'll give you the gist. Be warned, it's tedious. But fairly elementary.

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

    Integration by parts, udv + vdu = uv

    Let [tex]\frac{x}{\sqrt{1-x^2}}[/tex] be dv, and [tex](x\ln x)[/tex] be u

    Then v will be [tex]-(1 - x^2)^{\frac{1}{2}}[/tex]

    And du will be [tex](1 + \ln x)[/tex]

    Now we need to integrate vdu wrt x.

    vdu can be expressed as the sum of [tex]-(1 - x^2)^{\frac{1}{2}}dx[/tex] and [tex]-(1 - x^2)^{\frac{1}{2}}(\ln x)dx[/tex]

    The first term can be easily evaluated by making the substitution [itex]x = \sin \theta[/itex]

    Let's integrate the second term. Make the same substitution [itex]x = sin \theta[/itex].

    The second term becomes [tex]-\cos^2(\theta) \ln(|\sin \theta|) d \theta[/tex]

    Integrate that by parts. Use [tex]\cos^2(\theta) = dv[/tex] and [tex] \ln(|\sin \theta|) = u[/tex]

    Then vdu becomes [tex]-\frac{1}{2} \left( \cot \theta \cos(2\theta) + \cot \theta \right)d\theta[/tex] which can be simplified to [tex](\frac{1}{2}\sin 2\theta - \cot \theta)d \theta[/tex]

    Integrating [tex]\sin (2\theta)d\theta[/tex] is trivial.

    Integrating [tex]\cot \theta[/tex] is easy because, being the ratio of cosine to sine, it is of the form [tex]f'(x)g(f(x))[/tex] giving the integral as [tex]\ln|\sin\theta|[/tex]

    Put everything together and you'll have your answer as an indefinite integral. Evaluate for the bounds for the final numerical answer.
     
    Last edited: Feb 7, 2005
  7. Feb 7, 2005 #6

    dextercioby

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    I don't know what mathematica did to get that answer. :rolleyes: Try the substitution
    [tex] x=\sin u [/tex]
    under which the limits of integration become
    [tex] u_{1}=0;u_{2}=\frac{\pi}{2} [/tex]

    The integral becomes
    [tex] J=\int_{0}^{\frac{+\pi}{2}} \sin^{2}u \ \ln\sin u \ du [/tex]

    Mathematica on the wolfram's site gives the antiderivative:
    (see attachement) which is pretty nasty.Not to mention the fact that applying the FTC will produce a terrible headache.

    Daniel.
     

    Attached Files:

    Last edited: Feb 7, 2005
  8. Feb 8, 2005 #7
    First of all, you folks are interpreting the integral incorrectly. I never said you could take the x inside!! The first integral must be evaluated first and then the second one:

    [tex]\int udv = uv - \int vdu[/tex]

    In my case v is a function defined by an integral (say [itex]v=\int f(x)dx[/itex]). So the above integral can be rewritten in terms of f(x) as:

    [tex]\int udv = u(\int f(x)dx) - \int (\int f(x)dx)du[/tex]

    It is obviously understood that the two integrals cannot be commuted. So I thought you would understand and hence did not place the parantheses. Sorry for the confusion though.

    Also when I entered the original integral in Mathematica with the limits, I got the answer mentioned in my second post. I wonder how you got a different answer dextercioby. Curious3141, I'm going to try out the approach you've suggested in a while. Thanks for your help mates :approve:

    I did make the obvious substitution but couldn't take it further by hand. (This was a question on a math test :grumpy:).

    Cheers
    Vivek
     
    Last edited: Feb 8, 2005
  9. Feb 8, 2005 #8
    To add to my last post (to clear your doubts Healey01 and Hallsofivy),

    [itex]u = lnx[/itex]
    [itex]dv = \frac{x^2}{\sqrt{1-x^2}}[/itex] or equivalently [itex]v = \int \frac{x^2}{\sqrt{1-x^2}}dx[/itex]

    And by the way hallsofivy,

    velocity = [itex]v = \frac{dx}{dt}[/tex]
    acceleration = [itex]a = \frac{dv}{dt} = \frac{d^2x}{dt^2}[/itex]
    position = [itex]x = \int v dt = \int (\int a dt)dt[/itex]

    Is that wrong? Just because you can find [itex]\int a dt[/itex] first and call it v and then put it inside the first integral doesn't mean you can't write it this way?!????
     
    Last edited: Feb 8, 2005
  10. Feb 8, 2005 #9
    let your [itex] u = \frac{x^2}{\sqrt{1-x^2}}[/itex], and [itex] dv = lnx dx [/itex] and do it by part....
     
  11. Feb 8, 2005 #10

    Curious3141

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    Vivek, I think I may have made a bad error in my computation in the later steps. I tried it again, and I get an integrand of [tex]\theta \cot \theta[/tex] which I cannot integrate easily. It still comes back to [tex]\int \ln|\sin \theta| d\theta[/tex] which I don't think has an elementary integral.

    Sorry I misled you. :( I think this isn't an elementary integral at all.
     
  12. Feb 8, 2005 #11
    Interestingly, I tried to do this problem in 3 ostensibly different ways on the test but each time I came up with [itex]\int \theta \cot\theta d\theta[/itex] and eventually [itex]\int ln \sin\theta d\theta[/itex]. And thats when I gave up so when I saw your first post, I thought I was indeed missing something....

    cheers
    vivek
     
  13. Feb 8, 2005 #12
    why don't you use my method posted above.....
     
  14. Feb 8, 2005 #13

    dextercioby

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    I don't know,because it lead...nowhere...? :confused:

    Da
     
  15. Mar 7, 2005 #14
    The definite integral to be done is given by
    [itex]
    I = \int^{1}_{0} \frac{x^{2} \ln x}{\sqrt{1-x^{2}}} \ dx \ .
    [/itex]

    The primary difficulty of this integral is that it contains a logarithmic function and a square root in the integrand. However, it can be simplified by an integration of parts.
    [itex]
    I
    = - \int^{1}_{0} x \ln x \cdot \frac{-x}{\sqrt{1-x^{2}}} \ dx
    = - \int^{1}_{0} x \ln x \cdot \frac{d}{dx}\sqrt{1-x^{2}} \ dx
    [/itex]
    [itex]
    = - \left[ x \ln x \sqrt{1-x^{2}} \right]^{1}_{0}
    + \int^{1}_{0} \left( 1 + \ln x \right) \sqrt{1-x^{2}} \ dx \ .
    [/itex]

    It can be easily determined that the first term on the right of the above equation is zero. Hence, we have
    [itex]
    I
    = \int^{1}_{0} \sqrt{1-x^{2}} \ dx + \int^{1}_{0} \ln x \sqrt{1-x^{2}} \ dx \ .
    [/itex]

    The first term on the RHS can be easily solved by using the substitution [itex]x=\sin\theta[/itex], which yields
    [itex]
    \int \sqrt{1-x^{2}} \ dx
    = \frac{1}{2} \left( \sin^{-1}x + x\sqrt{1-x^{2}} \right) \ .
    [/itex]

    Thus, we have
    [itex]
    I
    =
    \frac{1}{2} \left[ \sin^{-1}x + x\sqrt{1-x^{2}} \right]^{1}_{0} +
    \int^{1}_{0} \ln x \sqrt{1-x^{2}} \ dx
    =
    \frac{\pi}{4} + \int^{1}_{0} \ln x \sqrt{1-x^{2}} \ dx \ .
    [/itex]

    The second term can be simplified by applying another integration by part.
    [itex]
    I
    =
    \frac{\pi}{4} + \int^{1}_{0} \ln x \sqrt{1-x^{2}} \ dx
    \nonumber \\
    =
    \frac{\pi}{4} +
    \frac{1}{2} \int^{1}_{0} \ln x \cdot
    \frac{d}{dx}\left( \sin^{-1}x + x\sqrt{1-x^{2}} \right) \ dx
    \nonumber \\
    [/itex]
    [itex]
    =
    \frac{\pi}{4} +
    \frac{1}{2} \left[\left(\sin^{-1}x + x\sqrt{1-x^{2}}\right)\right]^{1}_{0} -
    \frac{1}{2} \int^{1}_{0} \frac{1}{x} \left(\sin^{-1}x + x\sqrt{1-x^{2}}\right) \ dx
    \nonumber \\
    =
    \frac{\pi}{8} - \frac{1}{2} \int^{1}_{0} \frac{\sin^{-1}x}{x} \ dx \ .
    [/itex]

    The integral [itex]I[/itex] is now more tractable now that we have got rid of the [itex]\log[/itex] function. The solution can be completed once we evaluate the integral term in the last eqation. Using the substitution [itex]x=\sin t[/itex] and applying an integration by part,
    [itex]
    \int^{1}_{0} \frac{\sin^{-1}x}{x} \ dx
    =
    \int^{\pi/2}_{0} t \cot t \ dt
    =
    \int^{\pi/2}_{0} t \frac{d}{dt}\ln(\sin t) \ dt
    [/itex]
    [itex]
    =
    \left[t \ln(\sin t)\right]^{\pi/2}_{0} -
    \int^{\pi/2}_{0} \ln(\sin t) \ dt
    =
    - \int^{\pi/2}_{0} \ln(\sin t) \ dt \ .
    [/itex]

    Thus, it simplifies to
    [itex]
    I
    = \frac{\pi}{8} + \frac{1}{2} \int^{\pi/2}_{0} \ln(\sin t) \ dt \ .
    [/itex]

    Fortunately, the definite integral term in the above equation can be readily evaluated.
    [itex]
    \int^{\pi/2}_{0} \ln(\sin t) \ dt
    =
    \frac{1}{2} \int^{\pi}_{0} \ln(\sin t) \ dt
    =
    \frac{1}{2} \int^{\pi}_{0} \ln\left(2\sin(t/2)\cos(t/2)\right) \ dt
    [/itex]
    [itex]
    =
    \frac{1}{2} \int^{\pi}_{0} \ln 2 \ dt +
    \frac{1}{2} \int^{\pi}_{0} \ln(\sin(t/2)) \ dt +
    \frac{1}{2} \int^{\pi}_{0} \ln(\cos(t/2)) \ dt
    =
    \frac{\pi}{2} \ln 2 + 2 \int^{\pi/2}_{0} \ln(\sin t) \ dt
    [/itex]

    since
    [itex]
    \frac{1}{2} \int^{\pi}_{0} \ln(\sin(t/2)) \ dt = \int^{\pi/2}_{0} \ln(\sin t) \ dt
    [/itex]
    and
    [itex]
    \frac{1}{2} \int^{\pi}_{0} \ln(\cos(t/2)) \ dt
    = \int^{\pi/2}_{0} \ln(\cos t) \ dt
    = \int^{\pi/2}_{0} \ln(\sin t) \ dt \ .
    [/itex]

    Therefore, it is obvious that
    [itex]
    \int^{\pi/2}_{0} \ln(\sin t) \ dt
    = -\frac{\pi}{2} \ln 2
    [/itex]

    and that the final answer is
    [itex]
    I = \frac{\pi}{8}\left(1-\ln 4\right) \ .
    [/itex]
     
    Last edited: Mar 7, 2005
  16. Mar 7, 2005 #15
    My answer is consistent with Mathematica's so it's alright!
     
  17. Mar 8, 2005 #16

    dextercioby

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    Yes,i didn't claim it could't be done (congratulations!!),just that it was very difficult to find the antiderivative and the apply the FTC...You made use however of the fact that it was a definite integral...

    Daniel.
     
  18. Mar 8, 2005 #17
    My first stab three weeks ago at evaluating that integral ended up horribly as I attempted to evaluate [itex]\int^{\pi}_{0} \ln(\sin t) \ dt[/itex] using complex integrals. Functions like [itex]\ln z[/itex] are neither very analytic nor friendly with branch cuts appearing in unwanted places. I gave up after trying for an hour. Then, I had to report to camp for reservist training for the next three weeks...

    In the course of doing the complex integrals, I fiddled around with the limits of the integral and discovered some nice symmetries in the definite integral. After that, it was all downhill.
     
  19. Mar 8, 2005 #18
    By the way, there's no closed form for the indefinite integral [itex]\int \ln(\sin t) \ dt[/itex]. You'll end up with a series of horrible beta functions.
     
  20. Mar 8, 2005 #19

    dextercioby

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    Okay.I don't know how u end up with beta functions,my Maple didn't...

    Daniel.
     

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  21. Mar 8, 2005 #20

    Zurtex

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    Can someone please explain to me how the original integral is valid at x=0?
     
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