Convergence of $\int\limits_{-\infty}^{\infty}\frac{\sin{x^2}}{x}dx$

I had to prove that the integral $\int\limits_{-\infty}^{\infty}\frac{\sin{(x^2)}}{x}dx$ converges. I thought splitting it to $$\int_{-\infty}^{-1}\frac{\sin{x^2}}{x}dx+\int_{-1}^{0}\frac{\sin{x^2}}{x}dx+\int_0^1\frac{\sin{x^2}}{x}dx+\int_1^\infty\frac{\sin{x^2}}{x}dx$$ and then to activative on the first and last dirichle test but I'm sure that $\int_a^bsinx^2$ is bounded... Furthermore what can be done with the integrals in the middle. Mupad claims the integral converges to 0. How can I apply convergence tests to these integrals?

• If we let $f(0)=0$, the function $f(x)=\frac{\sin x^2}{x}$ becomes continuous, hence $\int_{-1}^1 f(x)\,\mathrm dx$ poses no problem. (Also, $f$ is an odd function, which explains the result $0$). – Hagen von Eitzen Jul 6 '13 at 11:15
• I can't let $f(0)=0$. About odd functions: It doesn't enough: take $g(x)=\frac 1 x$ which is also odd but $\int_{-\infty}^{\infty}g\ne 0$ but $\infty$ since $\int_0^{\infty}gdx$ diverges. – user65985 Jul 6 '13 at 11:20
• $\int_{-1}^1\sin(x^2)dx/x$ isn't an improper integral! We should note that whatever $f(0)$ is, $f(x)$ is integrable on $[-1,1]$ and the value doesn't change! – Yai0Phah Jul 6 '13 at 11:38
• @FrankScience this $\int_{-1}^1$ is formally an improper integral, only by the existence of a continuous extension into $0$ it is essentially no longer improper. – Hagen von Eitzen Jul 6 '13 at 11:45
• @FrankScience As you refer to Wikipedia: They say "Integrals are also improper if the integrand is undefined at an interior point of the domain of integration" This is the case here, so Coargu's doubts about $x=0$ are a priori justified. – Hagen von Eitzen Jul 6 '13 at 11:50

I should say something more: we can determine the value of the integral (from $0$ to $\infty$)!

It's easy, however, if we apply the substitution $x=\sqrt u$ where $u\ge 0$. Since $u=x^2$ is (strictly) monotone, the substitution is justified.

$$\int_0^\infty\frac{\sin x^2}xdx=\frac12\int_0^\infty\frac{\sin u}udu=\frac\pi4$$

The last one is due to Dirichlet.

EDIT: We should note that there's something schematic: $$\frac{f(x^m)}xdx=\alpha\frac{f(u)}udu$$ where $u=x^m$ and $\alpha$ is a constant.

Yes, splitting it is a good idea.

According to the comments, by the fact that $\lim_{x\to 0} x\cdot\displaystyle\frac{\sin(x^2)}{x^2} = 0\cdot 1=0$, we can conclude that $\sin(x^2)/x$ can be made continuous, hence its integral must exist ocer the compact integral $[-1,1]$. (And, anyway it is $0$ because the function is odd.)

For the other two parts, consider the zeroes: $x=\pm\sqrt{k\pi},\ k\in\Bbb N$, and let $$A_k:=\int_{\sqrt{k\pi}}^{\sqrt{(k+1)\pi}} \frac{\sin(x^2)}{x}dx$$ Then we have that the sequence $(A_k)$ is alternating, and as $$|A_k|\le \frac{\sqrt{(k+1)\pi}-\sqrt{k\pi}}{\sqrt{k\pi}}=\frac{\pi}{\sqrt{k\pi}\cdot(\sqrt{(k+1)\pi}+\sqrt{k\pi})}\ \to 0$$ as $k\to\infty$, we have that $(A_k)$ is a Leibniz sequence, hence converges.

The function $$f: \mathbb{R} \to \mathbb{R},\ f(x)=\begin{cases} \frac{\sin(x^2)}{x} & \text{ for } x \ne 0\\ 0 & \text{ for } x=0\end{cases}$$ is continuous and theorefore integrable on every compact subset of $\mathbb{R}$.

For every $a,b \in \mathbb{R}$ with $-a,b>1$ we have \begin{eqnarray} \int_a^bf&=&\int_{-1}^1f+\int_a^{-1}f+\int_1^bf=\int_1^bf-\int_1^{-a}f=\int_{-a}^b f\\ &=&\frac12\int_{a^2}^{b^2}\frac{\sin(x)}{x}\,dx =-\frac12\left[\frac{\cos(x)}{x}\right]_{a^2}^{b^2}-\int_{a^2}^{r^2}\frac{\cos(x)}{x^2}\,dx\\ &=&\frac12\left(\frac{\cos(b^2)}{b^2}-\frac{\cos(a^2)}{a^2}\right)-\int_{a^2}^{b^2}\frac{\cos(x)}{x^2}\,dx. \end{eqnarray} It follows that \begin{eqnarray} \left|\int_a^bf(x)\,dx\right|&\le& \frac{1}{2a^2}+\frac{1}{2b^2}+\int_{\min\{a^2,b^2\}}^{\max\{a^2,b^2\}}\frac{1}{x^2}\,dx\\ &=&\frac{1}{2a^2}+\frac{1}{2b^2}-\frac{1}{3(\max\{a^2,b^2\})^6}+\frac{1}{3(\min\{a^2,b^2\})^6}, \end{eqnarray} and hence $$\left|\int_{-\infty}^\infty f\right|=\lim_{a\to -\infty}\lim_{b\to \infty}\left|\int_a^b f(x)\,dx\right|=0,$$ i.e. $$\int_{-\infty}^\infty f=0.$$