How to estimate $\int^{1}_{-1} \left(\frac{\sin{x}}{x}\right)^{300} dx$ to 1 significant figure? 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?
 A: 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.
A: 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.
