Prove $$\large\int_{-\pi}^{\pi}\sin (\sin x) \,dx =0$$ without using the fact that $\sin(x)$ is odd.

Computing this in wolfram says that it is uncomputable, which leads me to believe that the only way to find this would be methods for solving definite integrals. I am wondering if it is possible with any other techniques such as DUIS or residues?

  • $\begingroup$ May be we can use the fact that For $0\le x\le\pi,$ $$\sin\sin x\le\sin x$$ $\endgroup$ – Bumblebee Jan 19 '15 at 7:11
  • $\begingroup$ @Nilan yes, if you can include a proof $\endgroup$ – Teoc Jan 19 '15 at 7:12

Here is a very weird solution. Notice that for $s \in \Bbb{R}$,

$$ \int_{-\pi}^{\pi} \sin(s\sin x) \, dx = \Im \int_{-\pi}^{\pi} \exp(is\sin x) \, dx. $$

Now it follows that

\begin{align*} \int_{-\pi}^{\pi} \exp(is\sin x) \, dx &= \int_{-\pi}^{\pi} \exp(se^{ix}/2) \exp(-se^{-ix}/2) \, dx \\ &= \int_{-\pi}^{\pi} \exp(se^{ix}/2) \cdot \overline{\exp(-se^{ix}/2)} \, dx \\ &= 2\pi \sum_{n=0}^{\infty} \left\{ \frac{1}{n!}\left(\frac{s}{2}\right)^{n} \right\}\left\{ \frac{(-1)^{n}}{n!}\left(\frac{s}{2}\right)^{n} \right\} \\ &= 2\pi \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(n!)^{2}}\left(\frac{s}{2}\right)^{2n} \\ &= 2\pi J_{0}(s), \end{align*}

where $J_{0}$ is the Bessel function of the first kind. Here we only need the fact that $J_{0}(s) \in \Bbb{R}$ if $s \in \Bbb{R}$. Consequently we get

$$ \int_{-\pi}^{\pi} \sin(s\sin x) \, dx = 0 $$


$$ \int_{-\pi}^{\pi} \cos(s\sin x) \, dx = 2\pi J_{0}(s). $$

(Of course, now both two identities extend to all of $s \in \Bbb{C}$ by the principle of analytic continuation.)

  • 1
    $\begingroup$ you must know your bessel function good. $\endgroup$ – abel Jan 19 '15 at 8:19
  • $\begingroup$ I think this is the answer that hides the use of oddness of sine best, so far. $\endgroup$ – mickep Jan 19 '15 at 8:31
  • $\begingroup$ @mickep on the contrary, it uses it in the very first line. $\endgroup$ – Start wearing purple Jan 20 '15 at 10:03
  • $\begingroup$ @O.L. I agree (and this I also write about in my answer). $\endgroup$ – mickep Jan 20 '15 at 10:23

$$\int_{-\pi}^\pi \sin(\sin(x)) \; dx = \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!}\int_{-\pi}^\pi \sin^{2n+1}(x)\; dx$$ (the interchange of sum and integral justified by uniform absolute convergence) so it suffices to show that $\int_{-\pi}^\pi \sin^{2n+1}\; dx = 0$ for nonnegative integers $n$. Now using the substitution $u = \cos(x)$, $$\int_{-\pi}^\pi \sin^{2n+1}(x)\; dx = \int_{-\pi}^\pi (1 - \cos^2(x))^n \sin(x)\; dx = -\int_{-1}^{-1} (1-u^2)^n \; du = 0 $$

  • $\begingroup$ The "change of variables" $u=\cos x$ in the last integral looks bad to me. It looks better if you split the integral in two pieces. $\endgroup$ – mickep Jan 19 '15 at 10:40
  • 1
    $\begingroup$ Why? It's a perfectly good substitution. $\endgroup$ – Robert Israel Jan 19 '15 at 17:31
  • $\begingroup$ You are of course correct, Robert. I just don't like changing variables so that the new integral collapses to be over one point. I should have stated that differently. $\endgroup$ – mickep Jan 20 '15 at 10:25

I think this question is a bit funny (and in principal, I think that any calculation leading to the desired conclusion uses that sine is odd in one or another way).

Are we allowed to use $\sin(t\pm\pi)=-\sin(t)$ and that $\sin t=\frac{\exp(it)-\exp(-it)}{2i}$ (I agree that it follows from the second identity directly that sine is odd. But in some sense it is not worse than saying that $\sin t=\text{Im}\,\exp(it)$ and then use that $\overline{\exp(it)}=\exp(-it)$)?

Nevertheless, here are some calculations using these formulas, leading to the desired result:

We divide the integral in two pieces $$ I=\int_{-\pi}^\pi \sin(\sin x)\,dx = \int_{-\pi}^0\sin(\sin x)\,dx+\int_0^{\pi}\sin(\sin x)\,dx. $$ Performing the change of variables $t=x+\pi$ and $t=x-\pi$ respectively in these integrals give $$ \begin{align} I&=\int_0^{\pi}\sin(\sin(t-\pi))\,dt+\int_{-\pi}^0 \sin(\sin(t+\pi))\,dt\\ &=\int_{-\pi}^{\pi} \sin(-\sin t)\,dt. \end{align} $$ Next, we use that $\sin z=\frac{\exp(iz)-\exp(-iz)}{2i}$, $$ \begin{align} I & = \int_{-\pi}^\pi \sin(\sin x)\,dx\\ & = \int_{-\pi}^{\pi}\frac{\exp(i\sin x)-\exp(-i\sin x)}{2i}\,dx\\ & = -\int_{-\pi}^{\pi} \frac{\exp(-i\sin x)-\exp(i\sin x)}{2i}\,dx\\ & = -\int_{-\pi}^{\pi} \sin(-\sin x)\,dx\\ & = -I. \end{align} $$ Thus $I=0$.


Another silly answer, using complex analytic methods (similar to sos440's answer, but avoiding use of Bessel functions):

Rewrite the integrand using Euler's formulas and put $z = e^{ix}$, thus mapping $[-\pi,\pi]$ to the unit circle (some algebra omitted): $$ \int_{-\pi}^\pi \sin \sin x \, dx = -\frac1{2} \int_{|z|=1} \frac{\exp\Big( \frac12 ( z - \frac1z ) \Big) - \exp\Big( \frac12 ( -z + \frac1z ) \Big)}{z}\,dz. $$

The integrand has an essential singularity at $z=0$, but we can still compute the relevant residue. Thanks to the $z$ in the denominator, we only have to compute the $0$:th terms of the Laurent series for the numerator.

We have \begin{align} \exp\Big( \frac12 ( z - \frac1z ) \Big) &= e^{z/2} \cdot e^{-1/(2z)} \\ &= \Big( 1 + \frac{1}{1!} \big( \frac{z}{2} \big) + \frac{1}{2!} \big( \frac{z}{2} \big)^2 + \cdots \Big) \Big( 1 - \frac{1}{1!} \big( \frac{1}{2z} \big) + \frac{1}{2!} \big( \frac{z}{2} \big)^2 - \cdots \Big) \end{align} Hence, the $0$:th term will be $$ 1 - \frac{1}{1!} \frac{1}{2^2} + \frac{1}{2!} \frac{1}{2^4} - \frac{1}{3!} \frac{1}{2^6} + \cdots $$

Similarly \begin{align} \exp\Big( \frac12 (-z + \frac1z ) \Big) &= e^{-z/2} \cdot e^{1/(2z)} \\ &= \Big( 1 - \frac{1}{1!} \big( \frac{z}{2} \big) + \frac{1}{2!} \big( \frac{z}{2} \big)^2 - \cdots \Big) \Big( 1 + \frac{1}{1!} \big( \frac{1}{2z} \big) + \frac{1}{2!} \big( \frac{z}{2} \big)^2 + \cdots \Big) \end{align} And again, the $0$:th term will be $$ 1 - \frac{1}{1!} \frac{1}{2^2} + \frac{1}{2!} \frac{1}{2^4} - \frac{1}{3!} \frac{1}{2^6} + \cdots $$

Summing up, the $0$:th term in the Laurent series for the numerator vanishes, and by the residue theorem, the integral, unsurprisingly is $0$.

Of course, this approach also uses, albeit implicitly, since I bothered to write out a number of unnecessary calculations, that the integrand is odd.


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.