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

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?

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

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

and

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

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

• you must know your bessel function good. – abel Jan 19 '15 at 8:19
• I think this is the answer that hides the use of oddness of sine best, so far. – mickep Jan 19 '15 at 8:31
• @mickep on the contrary, it uses it in the very first line. – Start wearing purple Jan 20 '15 at 10:03
• @O.L. I agree (and this I also write about in my answer). – 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 $u = \cos(x)$, $$\int_{-\pi}^\pi \sin^{2n+1}(x)\; dx = \int_{-\pi}^\pi (1 - \cos^2(x))^n \sin(x)\; dx = -\int_{-1}^{-1} (1-u^2)^n \; du = 0$$

• 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. – mickep Jan 19 '15 at 10:40
• Why? It's a perfectly good substitution. – Robert Israel Jan 19 '15 at 17:31
• 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. – 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.