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Thread: lebesgue integration

  1. #1
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    lebesgue integration

    i have exams on this course coming up in two days (i should probably write this post earlier). i was hoping to read more and try to find a solution myself but unfortunately failed.
    let f be a positive function, integrable over R. prove that


    <br />
\sum_{n \in Z} \int_0^1 f(x+n) d \mu = \int_{-\infty}^{\infty} f(x) d \mu <br />
    and deduce that \sum_{n \in Z} f(x+n) converges

    prove that



    this is based on the assumption that for all x in 0..pi/2 ( i have done this part)

    also one mroe if p >-1 and q belongs to the natural number. prove that

    (i have done this part)

    deduce that



    any thoughts would be appreciated! thank you!
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  2. #2
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    Quote Originally Posted by alexthepenguin View Post
    i have exams on this course coming up in two days (i should probably write this post earlier). i was hoping to read more and try to find a solution myself but unfortunately failed.
    let f be a positive function, integrable over R. prove that


    <br />
\sum_{n \in Z} \int_0^1 f(x+n) d \mu = \int_{-\infty}^{\infty} f(x) d \mu <br />
    and deduce that \sum_{n \in Z} f(x+n) converges

    prove that



    this is based on the assumption that for all x in 0..pi/2 ( i have done this part)

    also one mroe if p >-1 and q belongs to the natural number. prove that

    (i have done this part)

    deduce that



    any thoughts would be appreciated! thank you!

    Okay,

    Ive done the first one,

    LHS = sum ( int(f(x+n) x = 0 to 1), n = -inf, ..., inf)

    consider the integral component,

    inf(f(x+n) , x = 0 to 1)
    here let t = x + n --> dt/dx = 1 or dt = dx
    Change Limits: at x = 0 t = n, at x = 1 t = n +1

    Thus,

    inf(f(x+n), x = 0 to 1) becomes int( f(t) t = n to n+1) or int(f(x), x = n to n+1)

    Plug back into the sum,

    sum ( int(f(x), x = n to n + 1) , n = -inf,...,inf)

    which as Im sure you can easily see is
    sum(int(f(x), x = -inf...inf))

    Hope this helps,

    I'll keep plugging away on the rest,

    Regards,

    David
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  3. #3
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    Quote Originally Posted by alexthepenguin View Post
    prove that



    this is based on the assumption that for all x in 0..pi/2 ( i have done this part)
    Substitute y = x√n to get \sqrt n\int_0^{\pi/2}\cos^nx\,dx = \int_0^{\pi\sqrt n/2}\cos^n(y/\sqrt n)\,dy.

    Use good ol' l'H˘pital to check that \lim_{x\to0}\frac{\ln(\cos x)}{x^2} = \frac12. Put x=y/\sqrt n in that, to see that \lim_{n\to\infty}n\ln\bigl(\cos(y/\sqrt n)\bigr) = -\frac{y^2}2. Taking exponentials, \lim_{n\to\infty}\cos^n(y/\sqrt n) = e^{-y^2/2}. Also, the inequality \ln(\cos x)\leqslant-x^2/2 shows that \cos^n(y/\sqrt n) \leqslant e^{-y^2/2}. You can then apply the Dominated Convergence Theorem to get \lim_{n\to\infty} \int_0^{\pi\sqrt n/2}\cos^n(y/\sqrt n)\,dy = \int_0^\infty e^{-y^2/2}dy.
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