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I have seen this problem somewhere on the internet but I could not prove it.

Let $$I_{0}=\int^{\infty}_{0}\frac{\sin x}{x}dx$$ and then define $$I_{n+1}=\int^{I_{n}}_{0}\frac{\sin x}{x}dx.$$

Show that $$\lim_{n\rightarrow\infty}\sqrt{n}\ I_{n}=3.$$

MZperX
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  • Where on the internet did you find it? Can you bring some more context? Have you atleast tried to show that $\lim\limits_{n\to \infty} I_n\to 0$? – Zacky Feb 02 '20 at 23:58
  • Possible duplicate: https://math.stackexchange.com/questions/3215/convergence-of-sqrtnx-n-where-x-n1-sinx-n // Exactly same skill can be applied with only replacing $sin(x)$ by $\int sin(x)/x = x - x^3/18 +x^5/ 600 +...$ – Needmoremath Feb 03 '20 at 00:23

2 Answers2

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It is easy to see that $I_n\to0$ as $n\to\infty$. So by Stolze's Theorem, one has \begin{eqnarray} \lim_{n\rightarrow\infty}n I^2_{n}&=&\lim_{n\rightarrow\infty}\frac{n}{I^{-2}_{n}}\\ &=&\lim_{n\rightarrow\infty}\frac{1}{I^{-2}_{n+1}-I^{-2}_{n}}\\ &=&\lim_{n\rightarrow\infty}-\frac{I^{2}_{n+1}I^{2}_{n}}{I^{2}_{n+1}-I^{2}_{n}}\\ &=&\lim_{x\to0}-\frac{x^2\left(\int_0^x\frac{\sin t}{t}dt\right)^2}{\left(\int_0^x\frac{\sin t}{t}dt\right)^2-x^2}\\ &=&\lim_{x\to0}-\frac{x^2\left(\int_0^x(1-\frac16t^2+O(t)^5)dt\right)^2}{\left(\int_0^x(1-\frac16t^2+O(t)^5)dt\right)^2-x^2}\\&=&9. \end{eqnarray}

xpaul
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Denote the integral by $S(I_n)$

$$ I_{n+1}+I_n=S_{n+1}+S_n\\ I_{n+1}-I_n=S_{n+1}-S_n \\ \rightarrow 2 I_{n}= 2 S_{n} $$ writing (Taylor expansion) $$ S_{n}=I_{n-1}-\frac1{18}I_{n-1}^3+O(I_{n-1}^5) $$

in the limit of big $n$ this leads to a "continuum approximation" (justification follows)

$$ 18\,\partial_nI_n\sim-I^3_n+O(I_{n}^5) \quad (\star) $$

or (the constant of Integration can be neglected in the limit $n\rightarrow\infty$)

$$ I_n \sim\frac{3}{\sqrt{n}} \quad (\star \star) $$

Justification of $(\star)$:

We assume that $I_n$ is in the form of $(\star \star)$:

$$ I^{(k)}_n\ll I^{(m)}_n\quad (1) $$ for $m<k$, so we can approximate the finite difference by a continious derivative.

$$ I_n-I_{n-1}\ll 1/\sqrt{n}\quad (2) $$ so we can replace $I_{n-1}$ by $I_n$.

$$ I^{k}_n\ll I^{m}_n\quad (3) $$ for $m<k$, so we can neglect higher order Terms in the Taylor approximation to $S(I_n)$.

Combining $(1)$ & $(2)$ & $(3)$ shows that our approximation $(\star)$ is consistent.

approxolotl
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