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I would like to estimate $\int^{1}_{-1} \left(\frac{\sin{x}}{x}\right)^{300} dx$ to $1$ significant figure. (This question is taken from a quant exam).

My (vague) idea is to use Taylor series expansion and to estimate the remainder term. But then I run into problems immediately as I don't see a straightforward way to compute the first few terms of Taylor series for $\left(\frac{\sin{x}}{x}\right)^{300}$...

Any ideas?

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    $\begingroup$ Did you mean -1? In any case you can use the Laplace method, after rewriting $(\sin(x)/x)^{300}=e^{300 \log(\sin(x)/x))}$. $\endgroup$
    – Ian
    Oct 18, 2017 at 23:55
  • $\begingroup$ Yes, I just corrected it @Ian $\endgroup$
    – Alex
    Oct 18, 2017 at 23:57
  • $\begingroup$ Thanks - I totally forgot about Laplace method! @Ian $\endgroup$
    – Alex
    Oct 19, 2017 at 0:00

3 Answers 3

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The saddle point approximation for \begin{align} \int_{-1}^1\frac{\sin^{n}x}{x^{n}}dx&=\int_{-1}^1 e^{n\ln\frac{\sin x}{x}}dx=\int_{-1}^1e^{-n(\frac{x^2}{6}+\frac{x^4}{180}+\frac{x^6}{2835}+\cdots)}dx\\ &\approx\int_{-\infty}^\infty e^{-\frac{n}{6}x^2}dx=\sqrt{\frac{6\pi}{n}},\quad n\rightarrow\infty. \end{align} It does not matter whether $n$ is even or odd, because $\frac{\sin x}{x}$ is even and the negative parts haven't been reached by the integration bounds yet, or even if they are reached, they become too small in the $n\rightarrow\infty$ limit.

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Approximation by exponential

Approximating $\frac{\sin(x)}x\approx1-\frac{x^2}6$, $$ \begin{align} \int_{-1}^1\left(\frac{\sin(x)}x\right)^{300}\,\mathrm{d}x &\approx\int_{-\infty}^\infty e^{-\frac{300}6x^2}\,\mathrm{d}x\\ &=\frac{\sqrt{2\pi}}{10}\\[9pt] &=0.25066 \end{align} $$ where we can compute $\sqrt{2\pi}$ by hand using $\pi=3.1416$ and the scaffold method for square roots: $$ \begin{align} \sqrt{2\pi} &=\sqrt{6.2832}\\ &=2\sqrt{1.5708}\\ &=2(1.2533)\\ &=2.5066 \end{align} $$

For comparison, the original integral is approximately $0.250537$.


Contour Integration

There are no singularities so we can offset the contour by $-i$. $$\require{cancel} \begin{align} \int_{-\infty}^\infty\left(\frac{\sin(x)}x\right)^{300}\,\mathrm{d}x &=\frac1{2^{300}}\int_{-\infty-i}^{\infty-i}\frac{\left(e^{ix}-e^{-ix}\right)^{300}}{x^{300}}\,\mathrm{d}x\\ &=\frac1{2^{300}}\sum_{k=0}^{149}\int_{\gamma^+}(-1)^k\binom{300}{k}\frac{e^{i(300-2k)x}}{x^{300}}\,\mathrm{d}x\\ &+\cancel{\frac1{2^{300}}\sum_{k=151}^{300}\int_{\gamma^-}(-1)^k\binom{300}{k}\frac{e^{i(300-2k)x}}{x^{300}}\,\mathrm{d}x}\\ &=\frac{2\pi i}{2^{300}}\sum_{k=0}^{149}(-1)^k\binom{300}{k}\frac{-i(300-2k)^{299}}{299!}\\ &=\frac\pi{299!}\sum_{k=0}^{149}(-1)^k\binom{300}{k}(150-k)^{299}\\[9pt] &=0.25053746380056856955 \end{align} $$ where $$ \gamma^+=[-R-i,R-i]\cup Re^{i[0,\pi]}-i $$ and $$ \gamma^-=[-R-i,R-i]\cup Re^{-i[0,\pi]}-i $$ Note that $\gamma^-$ does not contain the origin.

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  • $\begingroup$ Hmm... I thought the original question had an exponent of $100$. I see that it was $300$. I guess it is clear enough how to adjust my answers to work. If desired, I can fix my answers. $\endgroup$
    – robjohn
    Oct 19, 2017 at 15:02
  • $\begingroup$ Amazing ! was my first reaction when reading your answer. The next and final one is Beautiful !. Many thanks for providing us so elegant solutions. Cheers. $\endgroup$ Oct 19, 2017 at 15:03
  • $\begingroup$ I have adjusted the exponent to $300$. $\endgroup$
    – robjohn
    Oct 19, 2017 at 15:15
  • $\begingroup$ @ClaudeLeibovici: Thanks! However, my first approach is just a simplification of Zhuoran He's answer. For such a large exponent, $\left(1-\frac{x^2}6\right)^{300}=e^{-50x^2}$ is pretty good. By the time the $5x^4/3$ term becomes big enough to matter, $e^{-50x^2}$ is very small. $\endgroup$
    – robjohn
    Oct 19, 2017 at 15:28
  • $\begingroup$ I was mainly speaking about the contour Integration : this is very elegant. $\endgroup$ Oct 19, 2017 at 15:54
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More than inspired by Zhuoran He's answer and assuming that we could require more accuracy $$I_n=\int_{-1}^1\frac{\sin^{n}x}{x^{n}}\,dx\approx \int_{-\infty}^\infty e^{-\frac{n}{6}x^2-\frac{n}{180}x^4}\,dx=\sqrt{\frac{15}{2}} e^{5 n/8} K_{\frac{1}{4}}\left(\frac{5 n}{8}\right)$$ where appears the modified Bessel function of the second kind.

Expanding for infinitely large values of $n$, we have $$I_n\approx \sqrt{\frac{6\pi}{n}}-\frac{3}{10} \sqrt{\frac{3 \pi }{2n^3}}+\frac{21}{80} \sqrt{\frac{3 \pi }{2n^5}}+O\left(\frac{1}{n^{7/2}}\right)$$ which leads to more than acceptable results when compared to numerical integration $$\left( \begin{array}{ccc} n & \text{exact} & \text{approximation} \\ 100 & 0.4335090102 & 0.4335152100 \\ 200 & 0.3067676765 & 0.3067687712 \\ 300 & 0.2505374649 & 0.2505378616 \\ 400 & 0.2169989558 & 0.2169991493 \\ 500 & 0.1941043334 & 0.1941044444 \\ 600 & 0.1772010667 & 0.1772011384 \\ 700 & 0.1640621723 & 0.1640622203 \\ 800 & 0.1534702230 & 0.1534702566 \\ 900 & 0.1446979461 & 0.1446961543 \\ 1000 & 0.1372730901 & 0.1372731089 \end{array} \right)$$

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