Compare the Integral

We can use the following formula: $\sin(a)\sin(b) = \dfrac{\cos(a-b) - \cos(a+b)}{2}$ :

$\begin{aligned} \int_{0}^{\infty} \sin(x)\sin(x^2) dx & = \dfrac{1}{2} \int_{0}^{\infty} \left(\cos(x^2-x) - \cos(x^2+x)\right) dx \\ & = \lim_{r \to \infty} \dfrac{1}{2}\left(\int_{0}^{r} \cos(x^2-x) dx - \int_{0}^{r} \cos(x^2+x) dx \right) \\ & = \dfrac{1}{2} \lim_{r \to \infty} \left(\int_{0}^{r} \cos(x^2-x) dx - \int_{1}^{r+1} \cos(x^2-x) dx \right) \\ & = \dfrac{1}{2} \lim_{r \to \infty} \left(\int_{0}^{1} \cos(x^2-x) dx - \int_{r}^{r+1} \cos(x^2-x) dx \right) \\ & = \dfrac{1}{2}\int_{0}^{1} \cos(x^2-x) dx - \dfrac{1}{2}\lim_{r \to \infty} \int_{r}^{r+1} \cos(x^2-x) dx \end{aligned}$

For $0 \le x \le 1$ we have $-\dfrac{1}{4} \le x^2-x \le 0$ , so $\cos(x^2-x) \ge 0$ and thus $\displaystyle \int_{0}^{1} \cos(x^2-x) dx > 0$ .

Fix $r > 3\pi$ and let $I(r) =\displaystyle\int_{r}^{r+1} \cos(x^2-x) dx$ . Now we just need to prove that $\displaystyle\lim_{r \to \infty} I(r) = 0$ :

First let $y = x^2 - x$ . Then $x = \dfrac{1}{2} + \sqrt{y+\dfrac{1}{4}}$ . Then, using integration by substitution, we get: $I(r) = \int_{r^2-r}^{r^2+r} \dfrac{\cos(y)}{\sqrt{4y+1}} dy$

Define the sequence $(a_n)$ as follows. Let $a_1 = p$ be the smallest number in the interval $[r^2-r, r^2+r]$ of the form $2\pi k-\dfrac{\pi}{2}$ for $k \in \mathbb{Z}$ , and let $a_{2m-1} = q$ be the largest number in that interval of that form (where $m = \dfrac{q-p}{2\pi} + 1$ ). And for $2 \le n \le 2m - 2$ let $a_n = a_{n-1} + \pi$ . Note that $a_{2m-1} = a_{2m-2} + \pi$ as well. Also let $a_0 = r^2 - r$ and $a_{2m} = r^2 + r$ .

Now for $0 \le n \le 2m - 1$ , let $f(n) = \displaystyle\int_{a_n}^{a_{n+1}} \dfrac{\cos(y)}{\sqrt{4y+1}} dy$ .

When $n = 0$ we have $a_1 - a_0 < \pi$ . Therefore: $\begin{aligned} \left|f(0) \right| & = \left| \displaystyle\int_{a_0}^{a_1} \dfrac{\cos(y)}{\sqrt{4y+1}} dy \right| \\ & \le \displaystyle\int_{a_0}^{a_1} \dfrac{\left|\cos(y)\right|}{\sqrt{4y + 1}} dy \\ & \le \displaystyle\int_{a_0}^{a_1} \dfrac{1}{\sqrt{4a_0+1}} dy \\ & = \dfrac{a_1 - a_0}{\sqrt{4a_0+1}} \\ & < \dfrac{\pi}{\sqrt{4a_0+1}} \\ |f(0)| & < \dfrac{\pi}{2r-1} ... \to (A) \end{aligned}$

Similarly, $|f(2m-1)| < \dfrac{\pi}{\sqrt{4a_{2m-1}+1}} < \dfrac{\pi}{\sqrt{4(r^2+r-\pi)+1}} < \dfrac{\pi}{2r} ... \to (B)$ .

Now let $k$ be an integer $1 \le k \le m - 1$ . For $a_{2k-1} \le y \le a_{2k}$ we have $\cos(y) \ge 0$ . Then, $\begin{array}{rcccl} \displaystyle\int_{a_{2k-1}}^{a_{2k}} \dfrac{\cos(y)}{\sqrt{4a_{2k}+1}} dy & \le & \displaystyle\int_{a_{2k-1}}^{a_{2k}} \dfrac{\cos(y)}{\sqrt{4y+1}} dy & \le & \displaystyle\int_{a_{2k-1}}^{a_{2k}} \dfrac{\cos(y)}{\sqrt{4a_{2k-1}+1}}dy \\ \dfrac{\sin(a_{2k}) - \sin(a_{2k-1})}{\sqrt{4a_{2k}+1}} & \le & & \le & \dfrac{\sin(a_{2k}) - \sin(a_{2k-1})}{\sqrt{4a_{2k-1}+1}} \\ \dfrac{2}{\sqrt{4a_{2k}+1}} & \le & & \le & \dfrac{2}{\sqrt{4a_{2k-1}+1}} \\ \dfrac{2}{\sqrt{4(r^2+r) + 1}} & < & & < & \dfrac{2}{\sqrt{4(r^2-r) + 1}} \\ \dfrac{2}{2r+1} & < & f(2k-1) & < & \dfrac{2}{2r-1} \end{array}$

Similarly we can show that: $\dfrac{-2}{2r-1} < f(2k) < \dfrac{-2}{2r+1}$

Therefore, combining all these equations gives: $\left|\sum_{j = 1}^{2m-2} f(j)\right| \le (m-1)\left(\dfrac{2}{2r-1} - \dfrac{2}{2r+1} \right) ... \to (C)$

Now since $a_{2m-1} = a_1 + (2m-2)\pi$ , then we have: $\begin{aligned} 2r & = a_{2m} - a_0 \\ & = (a_{2m} - a_{2m-1}) + (a_{2m-1} - a_1) + (a_1 - a_0) \\ & \ge a_{2m-1}-a_1 \\ & = 2(m-1)\pi \end{aligned}$ Therefore $m \le 1+ \dfrac{r}{\pi}$ . Substituting this into $(C)$ gives:

$\left|\sum_{j = 1}^{2m-2} f(j)\right| \le \left(\dfrac{r}{\pi}\right)\left(\dfrac{2}{2r-1} - \dfrac{2}{2r+1} \right) = \dfrac{4r}{\pi(4r^2 - 1)} ... \to (D)$

Therefore, combining $(A), (B)$ and $(D)$ gives:

$\begin{aligned} \left|I(r) \right| & = \left| \displaystyle\int_{r^2 - r}^{r^2+r} \dfrac{\cos(y)}{\sqrt{4y+1}}dy \right| \\ & = \left|\sum_{j = 0}^{2m-1} f(j)\right| \\ & \le \left|f(0)\right| + \left|\sum_{j = 1}^{2m-2} f(j)\right| + |f(2m-1)| \\ |I(r)| & <\dfrac{\pi}{2r-1} + \dfrac{4r}{\pi(4r^2 - 1)} + \dfrac{\pi}{2r} \end{aligned}$

As $r$ approaches infinity, this expression approaches $0$ . Therefore by the Squeeze Theorem, $\displaystyle\lim_{r \to \infty} I(r) = 0$ .

Therefore, $\displaystyle\int_{0}^{\infty} \sin(x)\sin(x^2) dx = \displaystyle\dfrac{1}{2}\int_{0}^{1} \cos(x^2-x) dx$ , which as we proved earlier, converges to a positive value.