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Math Help - Integration by reduction formulae

  1. #1
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    Integration by reduction formulae

    Given that  I_n = \int_{0}^{1}\frac {1}{(1+x^2)^n} dx , show that, for n≥2,

    2(n-1)I_n = 2^{1-n} + (2n-3)I_{n-1}


    Please help me to solve this.
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  2. #2
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    Quote Originally Posted by geton View Post
    Given that  I_n = \int_{0}^{1}\frac {1}{(1+x^2)^n} dx , show that, for n≥2,

    2(n-1)I_n = 2^{1-n} + (2n-3)I_{n-1}


    Please help me to solve this.
    Its less messy to prove:

    \boxed{2nI_{n+1} = 2^{-n} + (2n-1)I_{n}}

    First step: Integration by parts-

    I_{n} =  \int_{0}^{1}\frac {1}{(1+x^2)^{n}} \, dx = \frac{x}{(1+x^2)^n}\bigg{|}_0^1 + 2n \int_0^1 \frac{x^2}{(1+x^2)^{n+1}} \, dx

    Second Step: Partial Fractions

    Now we observe that \int_0^1 \frac{x^2}{(1+x^2)^{n+1}} \,dx = \int_0^1 \frac{1}{(1+x^2)^n}\, dx - \int_0^1 \frac{1}{(1+x^2)^{n+1}}\, dx = I_{n} - I_{n+1}.

    Third Step: Rearranging

     I_{n} = 2^{-n} + 2n(I_{n} - I_{n+1})
     2n I_{n+1} = 2^{-n} + (2n-1)I_{n}
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  3. #3
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    Quote Originally Posted by geton View Post
    Given that  I_n = \int_{0}^{1}\frac {1}{(1+x^2)^n} dx , show that, for n≥2,

    2(n-1)I_n = 2^{1-n} + (2n-3)I_{n-1}


    Please help me to solve this.
    Use integration by parts: \int u \, dv = uv - \int v \, du.

    Let u = \frac{1}{(1 + x^2)^n} \Rightarrow du = - \frac{2n x}{(1 + x^2)^{n+1}} \, dx and dv = dx \Rightarrow v = x.

    Note that \frac{x^2}{(1 + x^2)^{n+1}} = \frac{(1 + x^2) - 1}{(1 + x^2)^{n+1}} = \frac{1}{(1 + x^2)^{n}} - \frac{1}{(1 + x^2)^{n+1}}.

    Then I_n = \frac{1}{2^n} + 2n (I_n - I_{n+1}).

    Therefore: 2n I_{n+1} = \frac{1}{2^n} + (2n-1) I_n.

    Re-label n \rightarrow n - 1 and solve for I_n.
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