Integration help - question: $e^{-\sin(x)}$

Question

I would really like some help with the integration of $e^{-\sin(x)}$. Thanks to anyone who will help :)

Given that $\sin(x) > \frac{2x}{\pi}$ for $0 < x < \frac{\pi}{2}$, where $$\int_0^{\pi/2}e^{-\sin x}\,dx<\int_0^{\pi/2}e^{-2x/\pi}\,dx$$

RTS:

$$\int_0^{\pi/2}e^{-\sin x}\,dx=\int_{\pi/2}^{\pi}e^{-\sin x}\,dx$$

i think that its cannot be made in terms of elementary functions — cand, Feb 01 '15 at 00:17
i mean that probably that the primitive cannot be given in terms of elementary functions. — cand, Feb 01 '15 at 00:22
Do you mean that we are integrating over the interval $[0, \pi/2]$? — user24142, Feb 01 '15 at 00:29
See here: http://www.wolframalpha.com/input/?i=integral+exp%28-sin%28x%29%29+from+x%3D0..pi%2F2 — Cameron Williams, Feb 01 '15 at 00:42

score 6 · Accepted Answer · edited Feb 01 '15 at 05:01

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

As to why the integrand does not possess an anti-derivative expressible in terms of elementary functions, see Liouville's theorem and the Risch algorithm.

even though this is a great solution, we have not learnt this approach in our lessons — betty_carr, Feb 01 '15 at 00:54

score 1 · Answer 2 · answered Feb 01 '15 at 19:58

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{}$ \begin{align}&\overbrace{\color{#66f}{\int_{0}^{\pi/2}\expo{-\sin\pars{x}}\,\dd x}} ^{\ds{\dsc{\sin\pars{x}}=\dsc{t}\ \imp\ \dsc{x}=\dsc{\arcsin\pars{t}}}}\ =\ \int_{0}^{1}\frac{\expo{-t}}{\root{1 - t^{2}}}\,\dd t =\color{#66f}{\large -\,\frac{\pi}{2}\,{\rm M}_{0}\pars{1}} \approx{\tt 0.8731}\tag{1} \end{align} $\ds{\,{\rm M}_{\nu}\pars{z}}$ is a Modified Struve Function.

The result $\ds{\pars{~\mbox{given in expression}\ \pars{1}~}}$ corresponds to $11.5.4$ in this link.

Integration help - question: $e^{-\sin(x)}$

2 Answers2