Integral $\int-\sin x\mathrm e^{-\sin x}$

Question

I want to integrate the function $$-\sin x\mathrm{e}^{-\sin x}.$$ I tried integration by parts with $u=-\sin x$ but I got another difficult integration which is $$\int (\cos x)^{2}\mathrm{e}^{-\sin x} \,\mathrm{d}x.$$ Could someone help me out?

Answer 1

I do not think that you could find the antiderivative.

Set $\sin(x)=t$ to make the integrand $-t\, e^{-t}$ and then $$-t e^{-t}=-\sum_{n=0}^\infty (-1)^n \frac{t^{n+1}}{n!}$$ making $$\int-\sin (x)\,\mathrm e^{-\sin (x)}\,dx=\sum_{n=0}^\infty \frac{(-1)^{n+1}}{n!}\int \sin^{n+1}(x)\,dx$$

Edit

Since, in comments, the problem is said for the integral between $0$ and $2\pi$, let us use $$\int_0^{2\pi} \sin^{n+1}(x)\,dx=-\sqrt{\pi }\frac{ \left((-1)^n-1\right) \Gamma \left(\frac{n+2}{2}\right)}{\Gamma \left(\frac{n+3}{2}\right)}$$ which are equal to $0$ if $n$ is even.

This makes by the end$$-\int_0^{2\pi}\sin (x)\,\mathrm e^{-\sin (x)}\,dx=\sum_{n=0}^\infty \frac{ 2^{-2 n}\, \pi}{\Gamma (n+1) \Gamma (n+2)} =2 \pi I_1(1)$$ where appears the modified Bessel function of the first kind.

Looking for explicit results, using the same approach as above for $\int_0^{\frac{n\pi}2} \sin^{n+1}(x)\,dx$, I only found some for $$I_n=-\int_0^{\frac{n\pi}2} \sin(x)\,\mathrm e^{-\sin (x)}\,dx$$ the key ones being $$\left( \begin{array}{cc} n & I_n \\ 1 & \frac{1}{2} \pi (I_1(1)-\pmb{L}_{-1}(1)) \\ 2 & \pi (I_1(1)-\pmb{L}_{-1}(1)) \\ 3 & \frac{1}{2} \pi (3 I_1(1)-\pmb{L}_{-1}(1)) \\ 4 & 2 \pi I_1(1) \end{array} \right)$$ where also appears the modified Struve function.

Answer 2

$$\int\cos x\,e^{\sin x}dx$$ is elementary. Using the chain rule, it is easy to see that the antiderivative is $$e^{\sin x}.$$

On the opposite,

$$\int\sin x\,e^{\sin x}dx$$ has no closed-form expression.

Answer 3

$$ \begin{align} -\int_0^{2\pi}\sin(x)\,e^{-\sin(x)}\,\mathrm{d}x &=\int_0^{2\pi}\sin(x)\,e^{\sin(x)}\,\mathrm{d}x\\ &=\int_0^{2\pi}\sin(x)\,\sinh(\sin(x))\,\mathrm{d}x\\ &=\sum_{k=0}^\infty\int_0^{2\pi}\frac{\sin^{2k+2}(x)}{(2k+1)!}\,\mathrm{d}x\\ &=2\pi\sum_{k=0}^\infty\frac1{4^{k+1}}\binom{2k+2}{k+1}\frac1{(2k+1)!}\\ &=\pi\sum_{k=0}^\infty\frac1{4^kk!(k+1)!}\\[6pt] &=2\pi\operatorname{I}_1(1)\\[15pt] &=3.5509993784243618938 \end{align} $$ $\operatorname{I}_1$ is a modified Bessel function of the first kind. In any case, the sum converges quite quickly.

Answer 4

$-\int\mathrm e^{-\sin x}\sin x~dx=\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+2}x}{(2n+1)!}dx-\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}x}{(2n)!}dx$

For $n$ is any non-negative integer,

$\int\sin^{2n+2}x~dx=\dfrac{(2n+2)!x}{4^{n+1}((n+1)!)^2}-\sum\limits_{k=0}^n\dfrac{(2n+2)!(k!)^2\sin^{2k+1}x\cos x}{4^{n-k+1}((n+1)!)^2(2k+1)!}+C$

This result can be done by successive integration by parts.

$\int\sin^{2n+1}x~dx$

$=-\int\sin^{2n}x~d(\cos x)$

$=-\int(1-\cos^2x)^n~d(\cos x)$

$=-\int\sum\limits_{k=0}^nC_k^n(-1)^k\cos^{2k}x~d(\cos x)$

$=-\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}x}{k!(n-k)!(2k+1)}+C$

$\therefore\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+2}x}{(2n+1)!}dx-\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}x}{(2n)!}dx$

$=\sum\limits_{n=0}^\infty\dfrac{(2n+2)!x}{4^{n+1}(2n+1)!((n+1)!)^2}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(2n+2)!(k!)^2\sin^{2k+1}x\cos x}{4^{n-k+1}(2n+1)!((n+1)!)^2(2k+1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}x}{(2n)!k!(n-k)!(2k+1)}+C$

$=\sum\limits_{n=0}^\infty\dfrac{2(n+1)x}{4^{n+1}((n+1)!)^2}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{2(n+1)(k!)^2\sin^{2k+1}x\cos x}{4^{n-k+1}((n+1)!)^2(2k+1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}x}{(2n)!k!(n-k)!(2k+1)}+C$

$=\sum\limits_{n=0}^\infty\dfrac{x}{2^{2n+1}n!(n+1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(k!)^2\sin^{2k+1}x\cos x}{2^{2n-2k+1}n!(n+1)!(2k+1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}x}{(2n)!k!(n-k)!(2k+1)}+C$