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Math Help - Integral limit

  1. #1
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    Integral limit

    \lim_{n\to \infty} \int_{0}^{\frac{\pi}{2}} \frac{1}{tg^nx+ctg^nx} dx
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi,
    Quote Originally Posted by petter View Post
    \lim_{n\to \infty} \int_{0}^{\frac{\pi}{2}} \frac{1}{tg^nx+ctg^nx} dx
    Let u=\tan^n x ( \implies x=\arctan(\sqrt[n]{u}) and \mathrm{d}x=\frac{1}{n}\cdot \frac{u^{\frac{1-n}{n}}}{1+u^2}\,\mathrm{d}u)

    \int_{0}^{\frac{\pi}{2}} \frac{1}{\tan^nx +\mathrm{cotan}^n x} \,\mathrm{d}x=\frac{1}{n} \int_0^\infty \frac{1}{u+\frac{1}{u}}\cdot \frac{u^{\frac{1-n}{n}}}{1+u^2}\,\mathrm{d}u=\frac{1}{n} \int_0^\infty \frac{\sqrt[n]{u}}{(1+u^2)^2}\,\mathrm{d}u

    Now it must be clear what the limit is...
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  3. #3
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    Quote Originally Posted by flyingsquirrel View Post

    Let u=\tan^n x ( \implies x=\arctan(\sqrt[n]{u}) and \mathrm{d}x=\frac{1}{n}\cdot \frac{u^{\frac{1-n}{n}}}{1+u^2}\,\mathrm{d}u)

    i don't think what you've got for \text{dx} is correct.
    finding a good upper bound for the integral looks more interesting to me than the limit, which is 0:

    let \tan^{2n}x=t. then: 0 \leq I_n=\int_0^{\frac{\pi}{2}} \frac{dx}{\tan^n x + \cot^n x}=\frac{1}{2n} \int_0^{\infty} \frac{t^{\frac{1}{2n} - \frac{1}{2}}}{(1+t)(1+\sqrt[n]{t})} \ dt \leq \frac{1}{2n} \int_0^{\infty} \frac{t^{\frac{1}{2n} - \frac{1}{2}}}{1+t} \ dt. but: \int_0^{\infty} \frac{t^{\frac{1}{2n} - \frac{1}{2}}}{1+t} \ dt=\frac{\pi}{\cos \left(\frac{\pi}{2n} \right)}. thus: 0 \leq I_n \leq \frac{\pi}{2n\cos \left(\frac{\pi}{2n} \right)}, and hence: \lim_{n\to\infty}I_n = 0.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    I posted an answer to this question here

    http://www.mathhelpforum.com/math-he...ral-limit.html

    This back when I was less careful, so you may want to wait for NCA or another senior member to verify it.

    The basic gist is that since \left|\frac{1}{\tan^n(x)+\cot^n(x)}\right|\leqslan  t 1 on \left[0,\frac{\pi}{2}\right] and \frac{1}{\tan^n(x)+\cot^n(x)}\to 0 almost everywhere (pointwise) then by Lebesgue's Dominated Convergence Theorem \lim_{n\to\infty}\int_0^{\frac{\pi}{2}}\frac{dx}{\  tan^n(x)+\cot^n(x)}=\int_0^{\frac{\pi}{2}}\lim_{n\  to\infty}\frac{dx}{\tan^n(x)+\cot^n(x)}dx and using the fact that \lim_{n\to\infty}\frac{dx}{\tan^n(x)+\cot^n(x)}=\l  eft\{\begin{array}{rcl} \frac{1}{2}& \mbox{if} & x=\frac{\pi}{4}\\ 0 & \mbox{if} & \text{otherwise} \end{array}\right..


    NOTE: Once again the original post should not have probably been done because I was using machinery a bit beyond my capabilities of yet so you should wait for confirmation...but I thought it may give another perspective to this problem.
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    I posted an answer to this question here

    http://www.mathhelpforum.com/math-he...ral-limit.html

    This back when I was less careful, so you may want to wait for NCA or another senior member to verify it.

    The basic gist is that since \left|\frac{1}{\tan^n(x)+\cot^n(x)}\right|\leqslan  t 1 on \left[0,\frac{\pi}{2}\right] and \frac{1}{\tan^n(x)+\cot^n(x)}\to 0 almost everywhere (pointwise) then by Lebesgue's Dominated Convergence Theorem \lim_{n\to\infty}\int_0^{\frac{\pi}{2}}\frac{dx}{\  tan^n(x)+\cot^n(x)}=\int_0^{\frac{\pi}{2}}\lim_{n\  to\infty}\frac{dx}{\tan^n(x)+\cot^n(x)}dx and using the fact that \lim_{n\to\infty}\frac{dx}{\tan^n(x)+\cot^n(x)}=\l  eft\{\begin{array}{rcl} \frac{1}{2}& \mbox{if} & x=\frac{\pi}{4}\\ 0 & \mbox{if} & \text{otherwise} \end{array}\right..


    NOTE: Once again the original post should not have probably been done because I was using machinery a bit beyond my capabilities of yet so you should wait for confirmation...but I thought it may give another perspective to this problem.
    looks good to me, although the theorem is an overkill for the problem and it doesn't give us any decent bound for the integral!
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    looks good to me, although the theorem is an overkill for the problem and it doesn't give us any decent bound for the integral!
    Thanks NCA

    To the OP: When I edited my response I seem to cut off the ending...it should have been something to the effect that

    "So we can see by this that the integral is zero"

    Im sure you could have surmised that...but just not to leave any threads unraveled
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