0
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ If you do the variable substitution $u = - \sin x$, you should get a different integral from what you write -- something more like $- \int (u/\sqrt{1-u^2})\, e^u du$. $\endgroup$ – Stijn Feb 22 '18 at 9:30
  • $\begingroup$ @Stijn Isn't the integral just $ue^u$? The original integral is $-\sin x e^{- \sin x}$, and the second integral is just what the OP tried. $\endgroup$ – Toby Mak Feb 22 '18 at 9:33
  • $\begingroup$ @TobyMak This integral is not $\int ue^u\;du$. $du = -\cos(x) dx \neq dx$. $\endgroup$ – bames Feb 22 '18 at 9:34
  • $\begingroup$ It is cos^2(x) in the second one and not just cos (x) but don't know how to edit that $\endgroup$ – F.O Feb 22 '18 at 9:39
  • $\begingroup$ Just to add, to do integration by parts (with "$u$" and "$v$" in whatever implicit conventions you have), since you would need $du/dx$ to be something like $-\sin x$, $u$ would have to be $\cos x$, not $-\sin x$. I do not think this will really help either way. $\endgroup$ – Stijn Feb 22 '18 at 9:42
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @Claude Leibovici, thanks for the answer actually in the origin problem it was possible to go from line integral to double integral by Green therom or to triple integral by Gauss theorem which was easy but I wanted to evaluate the line integral and check if it was possible and will give the same answer but unfortunately I got that integral in my way. $\endgroup$ – F.O Feb 22 '18 at 10:50
  • 1
    $\begingroup$ @klirk. I am not computing the integral. Just representing the integrand as an infinite sum of powers of $\sin(x)$. $\endgroup$ – Claude Leibovici Feb 22 '18 at 10:54
  • $\begingroup$ Yes, I got it now...thats why I deleted my comments. You only use the series expansion of the exponential function. That was not clear (at least to me) from your answer. $\endgroup$ – klirk Feb 22 '18 at 10:58
  • 1
    $\begingroup$ @klirk. This is correct. Sorry for not being clear ! Cheers. $\endgroup$ – Claude Leibovici Feb 22 '18 at 11:01
1
$\begingroup$

$$\int\cos x\,e^{\sin x}dx$$ 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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.