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Math Help - a challenging integral (perhaps)

  1. #1
    Super Member Random Variable's Avatar
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    a challenging integral (perhaps)

     \int^{\pi}_{-\pi} \frac{\sin nx}{(1+2^{x}) \sin x} \ dx for n =0,1,2,...
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  2. #2
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    put x\mapsto-x and your integral equals \int_{-\pi }^{\pi }{\frac{2^{x}\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx}, thus \int_{-\pi }^{\pi }{\frac{\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx}\,+\,\int_{-\pi }^{\pi }{\frac{2^{x}\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx}=\int_{-\pi }^{\pi }{\frac{\sin nx}{\sin x}\,dx}=2\int_{0}^{\pi }{\frac{\sin nx}{\sin x}\,dx}.

    on the last integral let I_n=\int_0^\pi\frac{\sin nx}{\sin x}\,dx and get that I_n-I_{n-2}=0, thus I_n=\left\{\begin{array}{cl}\pi,&\text{if }n\text{ is odd.}\\<br />
0,&\text{if }n\text{ is even.}\end{array}\right.
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  3. #3
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Krizalid View Post
    put x\mapsto-x and your integral equals \int_{-\pi }^{\pi }{\frac{2^{x}\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx}, thus \int_{-\pi }^{\pi }{\frac{\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx}\,+\,\int_{-\pi }^{\pi }{\frac{2^{x}\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx}=\int_{-\pi }^{\pi }{\frac{\sin nx}{\sin x}\,dx}=2\int_{0}^{\pi }{\frac{\sin nx}{\sin x}\,dx}.

    on the last integral let I_n=\int_0^\pi\frac{\sin nx}{\sin x}\,dx and get that I_n-I_{n-2}=0, thus I_n=\left\{\begin{array}{cl}\pi,&\text{if }n\text{ is odd.}\\<br />
0,&\text{if }n\text{ is even.}\end{array}\right.
    There shouldn't be a two in front of  I_{n}

     \int^{\pi}_{-\pi} \frac{\sin nx}{(1+2^{x})\sin x} \ dx = \int_{-\pi}^{0} \frac{\sin nx}{(1+2^{x})\sin x} \ dx + \int^{\pi}_{0} \frac{\sin nx}{(1+2^{x})\sin x} \ dx

     = \int_{0}^{\pi} \frac{\sin nx}{(1+2^{-x})\sin x} \ dx + \int^{\pi}_{0}  \frac{\sin nx}{(1+2^{x})\sin x} \ dx

     = \int_{0}^{\pi} \frac{2^{x} \sin nx}{(1+2^{x})\sin x} \ dx + \int^{\pi}_{0}   \frac{\sin nx}{(1+2^{x})\sin x} \ dx

     = \int_{0}^{\pi} \frac{\sin nx}{\sin x} \ dx
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  4. #4
    Super Member Random Variable's Avatar
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    and for anyone who cares

     I_{n}-I_{n-2} = \int^{\pi}_{0} \frac{\sin nx}{\sin x} \ dx - \int^{\pi}_{0} \frac{\sin (n-2)x}{\sin x} \ dx

     = \int^{\pi}_{0} \frac{\sin nx - \sin (n-2)x}{\sin x} \ dx

     = 2 \int^{\pi}_{0} \frac{\sin x \cos (n-1)x}{\sin x} \ dx = 2 \int^{\pi}_{0} \cos (n-1)x \ dx

     = \frac{2}{n-1} \sin (n-1)x \Big|^{\pi}_{0} = 0


    and obviously  I_{0} = 0

    and  I_{1} = \int^{\pi}_{0} \frac{\sin x}{\sin x} \ dx = \int^{\pi}_{0} \ dx = \pi
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  5. #5
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    note that i never said that your integral equals 2I_n, the latter was because \frac{\sin nx}{\sin x} is even, what i actually did was

    \int_{-\pi }^{\pi }{\frac{\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx}=\frac{1}{2}\left( \int_{-\pi }^{\pi }{\frac{2^{x}\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx}+\int_{-\pi }^{\pi }{\frac{\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx} \right), and that yields the same you got.
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  6. #6
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Krizalid View Post
    note that i never said that your integral equals 2I_n, the latter was because \frac{\sin nx}{\sin x} is even, what i actually did was

    \int_{-\pi }^{\pi }{\frac{\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx}=\frac{1}{2}\left( \int_{-\pi }^{\pi }{\frac{2^{x}\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx}+\int_{-\pi }^{\pi }{\frac{\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx} \right), and that yields the same you got.
    OK. My bad.
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  7. #7
    Super Member PaulRS's Avatar
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    A general formula can be found for: I(n,m) = \int_{-\pi}^{\pi}\left(\frac{\sin(n\cdot x)}{\sin(x)}\right)^mdx

    Remember \sin(z)=\tfrac{\exp(z\cdot i)-\exp(-z\cdot i)}{2\cdot i} hence: I(n,m) = \int_{-\pi}^{\pi}\left(\frac{e^{n\cdot x\cdot i}-e^{-n\cdot x\cdot i}}{e^{x\cdot i}-e^{-x\cdot i}}\right)^mdx

    Note that: \tfrac{b^n-a^n}{b-a}=\sum_{k=0}^{n-1}a^k\cdot b^{n-1-k} let a=e^{-x\cdot i}; b=e^{x\cdot i} then : \frac{e^{n\cdot x\cdot i}-e^{-n\cdot x\cdot i}}{e^{x\cdot i}-e^{-x\cdot i}} = e^{(n-1)\cdot x\cdot i}\cdot \sum_{k=0}^{n-1}e^{-2 k\cdot x\cdot i }

    Thus: I(n,m) = \int_{-\pi}^{\pi}e^{(n-1)\cdot m\cdot x\cdot i}\cdot \left(\sum_{k=0}^{n-1}e^{-2k\cdot x\cdot i }\right)^mdx = 2\pi \cdot [x^{(n-1)\cdot m}]\left\{ \left(\sum_{k=0}^{n-1}x^{2k}\right)^m \right\} - that is to say 2\pi multiplied by the coefficient of x^{(n-1)\cdot m} of that polynomial in there - (*)

    (*) Because \int_{-\pi}^{\pi}e^{n\cdot x\cdot i}dx is 2\cdot\pi for n=0 and 0 for all other integer value of n.

    In particular then:

    • I(2n+1, 1) = 2\pi
    • I(n, 2) = 2 \pi \cdot n, since our answer is the number of pairs of integers (x, y) with 0\leq x,y\leq n-1 such that x+y = n -1, fix a valid x and that determines y.
    Last edited by PaulRS; March 12th 2010 at 01:43 PM. Reason: a little typo at the end.
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