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Find:

$$\lim_{n \to \infty} n \int_0^1 (\cos x - \sin x)^n dx$$

This is one of the problems i have to solve so that i could join college. I tried using integration by parts, i tried using notations but nothing works. If someone could please help me i would deeply appreciate it. ! thanks in advance ! I know the answer to the limit is 1. But i need help proving it.

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  • $\begingroup$ L'hopitals may help. $\endgroup$ – Ahmed S. Attaalla Jul 21 '16 at 10:42
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    $\begingroup$ hmm, you mean to use L'hospitals like this ? I transform that n into 1/n and i put it under the integral then i apply L'hospital? @AhmedS.Attaalla gyazo.com/6576519a9581adab111bb180e67eac83 like this? $\endgroup$ – Razvan Jul 21 '16 at 10:50
  • $\begingroup$ At first glance it seemed to be a standard L'hopital rule problem but now I'm not sure. What exact section are you working so we can get an idea of what method is involved? @Razvan $\endgroup$ – Ahmed S. Attaalla Jul 21 '16 at 11:19
  • $\begingroup$ Im not sure to what do you reffer by "section". I just finished highschool and im trying to go to university. I studied mathematics-informatics in school. Im not sure i understood your question though. @AhmedS.Attaalla $\endgroup$ – Razvan Jul 21 '16 at 11:27
  • $\begingroup$ Okay just don't mind what I've said. @Razvan $\endgroup$ – Ahmed S. Attaalla Jul 21 '16 at 11:37
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Write the integrand as $(\cos(x)-\sin(x))^n=e^{n\log(\cos(x)-\sin(x))}$. Looking at a few plots it becomes immediatly clear the integral will be dominated by a small region around the origin with width $\epsilon\sim1/n$ as $n \rightarrow\infty$. The more formal reason for that is, that the exponent is nearly zero around the origin

$$ I_n=\int_0^1(\cos(x)-\sin(x))^n\sim \int_0^{\epsilon}e^{n\log(\cos(x)-\sin(x))} $$

Taylor expansion of the exponent yields

$$ I_n\sim\int_0^{\epsilon}e^{-x n} $$

by pushing $\epsilon$ to infinity we introduce only an exponentially small error so

$$ I_n \sim\int_0^{\infty}e^{-x n}=\frac{1}{n} $$

which yields

$$ \lim_{n \rightarrow \infty} n I_n=1 $$

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    $\begingroup$ thanks, but this is not necessary anymore...i finished my phd in theoretical physics yesterday ;) $\endgroup$ – tired Jul 21 '16 at 11:35
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    $\begingroup$ @tired, your proof is not rigorous enough. $\endgroup$ – xpaul Jul 22 '16 at 13:40
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    $\begingroup$ @ClaudeLeibovici "I have been amazed by the fact that ∫(cos(x)−sin(x))ndx has a "rather simple" expression" Hmm, has it? Not that I know. Would you be confusing asymptotics when $n\to\infty$ with exact formulas for some given $n$? $\endgroup$ – Did Jan 15 '17 at 11:48
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    $\begingroup$ To turn this post into a mathematical answer, one could mention that this is a simple case of Laplace's method and one should state precisely the general theorem one wishes to apply and check that its hypotheses hold in the present case. In particular, the fact that the function $\cos x-\sin x$ has a single maximum at $x=0$ on the interval $[0,1]$ and its behaviour near $x=0$ should be made explicit. $\endgroup$ – Did Jan 15 '17 at 11:53
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    $\begingroup$ @Did my post outlines the procedure to obtain the given limit and also presents the correct results. i don't think it is my job here to fill in all the steps to make this completly rigorous . This is something the op might want to do if it seems necessary (and he can also ask politily if he needs help with that). $\endgroup$ – tired Jan 16 '17 at 8:31
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The following common inequalities suffice here: $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

  • $1+x+\lfrac12 x^2 \le \exp(x) \le 1 + x + x^2$ for every real $x \in [0,\ln(2)]$.

  • $1 - \lfrac12 x^2 \le \cos(x) \le 1 - \lfrac12 x^2 + \lfrac1{24} x^4$ for every real $x$.

  • $x - \lfrac16 x^3 \le \sin(x) \le x$ for every real $x \ge 0$.

They can be proven elementarily by recursively comparing derivatives.


As $n \to \infty$:

  Let $r = n^{2/3}$.

  Given any $x \in [0,\lfrac1r]$:

    $\cos(x) - \sin(x) \le 1 - x + \lfrac16 x^3 \le \exp(-x)$.

    $\cos(x) - \sin(x) \ge 1 - x - \lfrac12 x^2 \ge \exp(-x-2x^2)$ since $x \to 0$.

  Thus $(\cos(x)-\sin(x))^n \in \exp(-x) ^ n \cdot [\exp(-\frac{2}{r^2}),1]^n$.

  Thus ${\displaystyle\int}_0^{\lfrac1r} n ( \cos(x) - \sin(x) )^n\ dx \in [\exp(-\frac{2n}{r^2}),1] \cdot {\displaystyle\int}_0^{\lfrac1r} n \exp(-nx)\ dx$.

  Also ${\displaystyle\int}_0^{\lfrac1r} n \exp(-nx)\ dx = 1 - \exp(-\lfrac{n}{r}) \to 1$.

  Given any $x \in [\lfrac1r,1]$:

    $\cos(x) - \sin(x) \le 1 - x - x^2 ( \lfrac12 - \lfrac16 x - \lfrac1{24} x^2 ) \le 1 - \lfrac1r$.

    $\cos(x) - \sin(x) \ge 1 - x - \lfrac12 x^2 \ge -\frac12$.

    Thus $| n ( \cos(x) - \sin(x) )^n | \le n ( 1 - \lfrac1r )^n \to 0$.

  Thus ${\displaystyle\int}_{\lfrac1r}^1 n ( \cos(x) - \sin(x) )^n\ dx \to 0$.

  Therefore ${\displaystyle\int}_0^1 n ( \cos(x) - \sin(x) )^n\ dx \to 1$.

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  • $\begingroup$ @ClaudeLeibovici: You may be interested in my answer. I cut the integral into two pieces at $n^{-2/3}$ so that both parts can be easily proven to converge, rigorously. $\endgroup$ – user21820 Jan 15 '17 at 17:10

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