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Recursive equation for an integral

  1. May 29, 2006 #1
    i have:
    [tex]I_n=\int_{0}^{1} \frac{x^{n+1/2}}{\sqrt{1-x^2}}dx[/tex]

    and i need to find a recursive equation for the above integral, and this is what i got so far:
    [tex]I_n=\left[x^{n+1/2}*arcsin(x)\right]_{0}^{1}+(n+1/2)\int_{0}^{1}x^{n-1/2}arcsin(x)dx[/tex]
     
    Last edited: May 29, 2006
  2. jcsd
  3. May 29, 2006 #2

    StatusX

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    Try multiplying the top and bottom of the original integrand by 1-x2.
     
  4. May 29, 2006 #3
    it still doesn't work, because i get an integral with x^(n+3/2)/(sqrt(1-x^2))^3, where i should get an integral of x^(n+3/2)/(sqrt(1-x^2)).
     
  5. May 29, 2006 #4

    VietDao29

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    Uhmm, I just wanna ask, is it (n + 1) / 2 or n + (1 / 2)?
    Is it:
    [tex]I_n=\int_{0}^{1} \frac{x^{n+ \frac{1}{2}}}{\sqrt{1-x^2}}dx[/tex]
    or
    [tex]I_n=\int_{0}^{1} \frac{x^{\frac{n + 1}{2}}}{\sqrt{1-x^2}}dx[/tex]?
     
  6. May 29, 2006 #5
    it's n+(1/2)
     
  7. May 29, 2006 #6
    HINT: Consider [tex]\frac{d}{dx}[x^{n+\frac{1}{2}}\sqrt{1-x^2}][/tex].

    Once you have obtained the recursive equation, you will just need to modify it slightly to obtain [tex]I_{n}[/tex] in terms of [tex]I_{n-2}[/tex].
     
    Last edited: May 29, 2006
  8. May 29, 2006 #7

    VietDao29

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    You can use StatusX's hint to solve your problem. And also you should choose u, and v wisely.
    [tex]I_n = \int_0 ^ 1 \frac{x ^ {n + \frac{1}{2}}}{\sqrt{1 - x ^ 2}} dx = \int_0 ^ 1 \frac{x ^ {n + \frac{1}{2}} (1 + x ^ 2)}{\sqrt{(1 - x ^ 2) ^ 3}} dx = - \frac{1}{2} \int_0 ^ 1 \frac{x ^ {n - \frac{1}{2}} (1 + x ^ 2)}{\sqrt{(1 - x ^ 2) ^ 3}} d(1 - x ^ 2)[/tex]
    [tex]= \int_0 ^ 1 x ^ {n - \frac{1}{2}} (1 + x ^ 2) d \left( \frac{1}{\sqrt{1 - x ^ 2}} \right)[/tex]
    Now let [tex]u = x ^ {n - \frac{1}{2}} (1 + x ^ 2)[/tex], and [tex]dv = d \left( \frac{1}{\sqrt{1 - x ^ 2}} \right)[/tex].
    Can you go from here? :)
    Recheck my maths if possible. I'm having a slight headache. :frown:
     
  9. May 30, 2006 #8

    StatusX

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    You should split the integral into the sum of two integrals, both of the form xsomething/(1-x2)3/2. These can both be integrated by parts to get two boundary terms (which must be combined before being evaluated), and multiples of Im for a couple values of m.
     
    Last edited: May 30, 2006
  10. May 30, 2006 #9

    benorin

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    Use x=sin(u)

    Use the substitution [tex]x=\sin u[/tex] so that [tex]dx=\cos u du[/tex] to get

    [tex]I_n=\int_{0}^{1} \frac{x^{n+1/2}}{\sqrt{1-x^2}}dx=\int_{0}^{\frac{\pi}{2}} \sin ^{n+1/2} u du [/tex]​

    The recurrence you are looking for is [tex]I_{n+2}=f(n)I_{n}[/tex]
    where f(n) is a rational function of n.

    P.S. (the one you are not looking for is [tex]I_{n}I_{n+1}=\frac{\pi}{2n+3}[/tex])
     
  11. May 30, 2006 #10
    benorin, i tried your approach but still didn't get a recursive eq, here what i got:
    [tex]I_n=\int_{0}^{1} \frac{x^{n+1/2}}{\sqrt{1-x^2}}dx=\int_{0}^{\frac{\pi}{2}} \sin ^{n+1/2} (t) dt=\left[tsin^{n+1/2}(t)\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}(n+1/2)sin^{n-1/2}(t)d(sin(t))[/tex]
     
    Last edited: May 30, 2006
  12. May 30, 2006 #11
    and statusx, here what i got from your appraoch:
    [tex]I_n=\int_{0}^{1} \frac{x^{n+1/2}}{\sqrt{1-x^2}}dx=\left[\frac{x^{n+3/2}}{(n+3/2)(\sqrt{1-x^2})^{3}}\right]_{0}^{1}-\int_{0}^{1}\frac{3x^{n+5/2}}{(n+3/2)(\sqrt{1-x^2})^{5}}dx-\left[\frac{x^{2n+2}}{(2n+2)(\sqrt{1-x^2})^{3}}\right]_{0}^{1}+\int_{0}^{1}3/(2n+2)\frac{x^{2n+3}}{(\sqrt{1-x^2})^{5}}dx[/tex]
     
    Last edited: May 30, 2006
  13. May 30, 2006 #12

    benorin

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    Use a trick

    [tex]I_{n+2}=\int_{0}^{\frac{\pi}{2}} \sin ^{n+2+1/2} u du [/tex]
    [tex]=\int_{0}^{\frac{\pi}{2}} \sin ^{n+1/2} u\sin ^{2} u du [/tex]
    [tex]= \int_{0}^{\frac{\pi}{2}} \sin ^{n+1/2} u(1-\cos ^{2} u) du [/tex]
    [tex]= \int_{0}^{\frac{\pi}{2}} \sin ^{n+1/2} u du -\int_{0}^{\frac{\pi}{2}} \sin ^{n+1/2} u\cos ^{2} u du [/tex]
    [tex]= I_{n}-\int_{0}^{\frac{\pi}{2}} \left( \sin ^{n+1/2} u\cos u\right) \cos u du[/tex]
    integration by parts (in the last integral): put

    [tex]U=\cos u, dU = -\sin u, dV=\sin ^{n+1/2} u\cos u du, V=\frac{1}{n+1+1/2}\sin ^{n+1+1/2} u[/tex]

    to get

    [tex]I_{n+2}= I_{n}-\int_{0}^{\frac{\pi}{2}} \left( \sin ^{n+1/2} u\cos u\right) \cos u du [/tex]
    [tex]= I_{n}-\left[ \frac{1}{n+2+1/2}\cos u\sin ^{n+2+1/2} u \right] _{u=0}^{\frac{\pi}{2}} - \frac{1}{n+1+1/2}\int_{0}^{\frac{\pi}{2}} \sin ^{n+2+1/2} u du [/tex]
    [tex]=I_{n}- \frac{1}{n+1+1/2}I_{n+2}[/tex]

    rewritten this becomes [tex]I_{n+2}=I_{n}- \frac{2}{2n+3}I_{n+2}[/tex]

    which gives what your after, namely [tex] I_{n+2}= \frac{2n+3}{2n+5}I_{n}[/tex]
     
    Last edited: May 30, 2006
  14. May 30, 2006 #13

    StatusX

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    I think you're taking the wrong u and dv. Remember:

    [tex]\int \frac{x}{(1-x^2)^{3/2}}dx = \frac{1}{\sqrt{1-x^2}}[/tex]
     
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