一定积分题

青青子衿

$\int_0^{\pi}e^{\sin x}dx$[阴险]

青青子衿

$\int_0^{\pi}e^{\sin x}dx$
青青子衿 发表于 2013-11-2 15:36

$\int_0^{\pi}e^{\sin x}dx=\pi  (L_0(1)+I_0(1)) = 6.208758036$

Czhang271828

$\int_0^{\pi}e^{\sin x}dx=\pi  (L_0(1)+I_0(1)) = 6.208758036$
青青子衿 发表于 2014-1-20 13:11

See from math.stackexchange.com/questions/1128367/inte … help-question-e-sinx

hbghlyj

We have:
$$I=\int_{0}^{\pi/2}e^{\sin x}\,dx = \int_{0}^{1}\frac{e^{t}}{\sqrt{1-t^2}}\,dt$$
and since:
$$e^{t}=\sum_{k\geq 0}\frac{ t^k}{k!},\qquad \int_{0}^{1}\frac{t^k}{\sqrt{1-t^2}}\,dt =\int_{0}^{\pi/2}\sin^k\theta\,d\theta=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{k+1}{2}\right)}{2\,\Gamma\left(\frac{k}{2}+1\right)}$$
(see Wallis' integrals for more information), it follows that:
$$ I = \sum_{k\geq 0}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{k+1}{2}\right)}{2\,\Gamma\left(\frac{k}{2}+1\right)\Gamma(k+1)}=\frac{\pi}{2}\sum_{k\geq 0}\frac1{2^k \Gamma\left(\frac{k}{2}+1\right)^2}=\color{red}{\frac{\pi}{2}\left(I_0(1)+L_0(1)\right)},$$
where $I_0$ and $L_0$ are a Bessel and a Struve function.

hbghlyj

使用Laplace's method估计 $\displaystyle \int _{a}^{b}e^{Mf(x)}\rmd x$ 型积分.
$$\sin x=1-\frac{1}{2} \left(x-\frac{\pi }{2}\right)^2+O\left(\left(x-\frac{\pi }{2}\right)^3\right)$$代入$\int_0^{\pi}e^{\sin x}\rmd x$,$$e\int_0^{\pi}e^{-\frac12\left(x-\frac{\pi }{2}\right)^2}\rmd x$$将$x$换成$x+\pi/2$得$$e\int_{-\pi/2}^{\pi/2}e^{-\frac12x^2}\rmd x=e\sqrt{2\pi}\left(2\Phi(\frac\pi2)-1\right)\approx6.02176$$

hbghlyj

使用WolframAlpha计算$e\sqrt{2\pi}\left(2\Phi(\frac\pi2)-1\right)$
(N[ReplaceAll[CDF[NormalDistribution[0, 1], x], {x -> Pi/2}]]*2-1)*E*sqrt(2pi)

如果把ReplaceAll改成Replace就识别不了
(N[Replace[CDF[NormalDistribution[0, 1], x], {x -> Pi/2}]]*2-1)*E*sqrt(2pi)