Does the improper integral $\int_0^\infty\sin(x)\sin(x^2)\,\mathrm dx$ converge

Does the following improper integral converge?

$$\lim_{B \to \infty}\int_0^B\sin(x)\sin(x^2)\,\mathrm dx$$

Solutions Collecting From Web of "Does the improper integral $\int_0^\infty\sin(x)\sin(x^2)\,\mathrm dx$ converge"

Fix $A<B$; then
$$I_{A,B}:=\int_{A^2}^{B^2}\sin t\sin t^2dt=\frac 12\int_A^B\frac{\sin(\sqrt s)}{\sqrt s}\sin sds.$$
Integrating by parts, this gives
$$I_{A,B}=\frac 12\left[-\cos s\frac{\sin(\sqrt s)}{\sqrt s}\right]_A^B+\frac 12\int_A^B \cos s\cdot \left(\frac{\cos(\sqrt s)}{2s}-\frac{\sin(\sqrt s)}{2s^{3/2}}\right),$$
so the problem reduces to the convergence of $\int_1^{\infty}\cos s\cos(\sqrt s)/sds$, which can be tackled similarly.

Notice $x^2 + x = (x+1)x = (x+1)^2 – (x+1)$, we have:
$$\begin{align}\int_0^B \sin(x) \sin(x^2) dx
&= \frac12\int_0^B(\cos(x^2-x)-\cos(x^2+x)) dx\\
&= \frac12\left(\int_0^B – \int_1^{B+1}\right)\cos(x^2-x)dx\\
&= \frac12\left(\int_0^1 – \int_B^{B+1}\right)\cos(x^2-x)dx
\end{align}$$
Integrate by parts. For large $B$, we have:
$$\begin{align}
&\int_B^{B+1} \cos(x^2 – x)dx = \left[ \frac{\sin(x^2 – x)}{2x – 1} \right]_B^{B+1} + \int_B^{B+1}\frac{2\sin(x^2 – x)dx}{(2x – 1)^2}\\
\implies & \left|\int_B^{B+1} \cos(x^2 – x)dx\right| \le \frac{2}{2B-1} + \frac{2}{(2B-1)^2}\\
\implies & \lim_{B\to\infty} \int_B^{B+1} \cos(x^2 – x)dx = 0\\
\implies & \lim_{B\to\infty} \int_0^B \sin(x) \sin(x^2) dx = \frac12 \int_0^1 \cos(x^2 – x) dx.
\end{align}$$

When we express the above in terms of complex exponentials, we get integrals of the form

$$\int_0^{B} dx \, e^{i (x^2 \pm x)}$$

Therefore, for convergence purposes, consider the following integral in the complex plane:

$$\oint_C dz \, e^{i z^2} e^{i z}$$

where $C$ is a wedge-shaped contour that has a segment along the positive real axis between $[0,B]$, a circular arc of 45 degrees and of radius $B$, and a line joining the end of the arc and the origin.

By Cauchy’s integral theorem, the above contour integral is zero. On the other hand, it may be expressed as integrals over each segment:

$$\int_0^B dx \, e^{i x^2} e^{i x} + i B \int_0^{\pi/4} d\theta \, e^{i \theta} e^{i B e^{i \theta}} e^{i b^2 e^{i 2 \theta}} + e^{i \pi/4} \int_B^0 dt \, e^{-t^2} e^{e^{i \pi/4} t} = 0$$

The first integral is the integral which we seek. The second integral may be bounded by

$$B \int_0^{\pi/4} d\theta e^{-B \sin{\theta}} e^{-B^2 \sin{2 \theta}} $$

which goes to zero as $B \to \infty$ because $\sin{\theta} \ge (2/\pi) \theta$ when $\theta \in [0,\pi/2]$. The above integral is then bounded by

$$B \int_0^{\pi/4} d\theta e^{-B (2/\pi) \theta} e^{-B^2 (2/\pi) 2 \theta} = \frac{\pi}{4 B + 2} $$

The integral then may be written as

$$\int_0^B dx \, e^{i x^2} e^{i x} = e^{i \pi/4} \int_0^B dt \, e^{-t^2} e^{-t/\sqrt{2}} e^{i t\sqrt{2}}$$

Because the integral on the RHS converges in the limit of $B \to \infty$, the integral on the LHS does. The same goes for the minus sign as well. Thus, the stated integral converges.

hint
Notice that $$\sin (x^2) = (\cos (x^2))’ \frac{1}{2x}$$
Which gives the idea if you manage to do this thing twice (or enough times) and you reach something like
$\frac{g(x)}{x^n}$ where $n>1$ and $g(x)$ is something bounded you can use comparison test to conclude convergence.

One thing that bugs us is the $\frac{1}{2x}$ is undbounded at zero but this does not really matter because if instead we study the convergence of $\lim _{B \rightarrow \infty}\int _{0}^{B} \sin (x) \sin (x^2) \mathrm{d}x$ is the same as the convergence of $\lim _{B \rightarrow \infty}\int _{1}^{B} \sin (x) \sin (x^2)\mathrm{d}x$.

$\int _{0}^{B}\!\sin \left( x \right) \sin \left( {x}^{2} \right) {dx}=
-\int _{0}^{B}\!1/2\,\cos \left( x \left( 1+x \right) \right) {dx}+
\int _{0}^{B}\!1/2\,\cos \left( x \left( -1+x \right) \right) {dx}$

Change variable on the first integral on the LHS $x \Rightarrow x-1$:

$\int _{0}^{B}\!\sin \left( x \right) \sin \left( {x}^{2} \right) {dx}=
-\int _{1}^{1+B}\!1/2\,\cos \left( x \left( -1+x \right) \right) {dx}
+\int _{0}^{B}\!1/2\,\cos \left( x \left( -1+x \right) \right) {dx}$

separate a finite and an infinite integration from the first integral:

$=
\int _{0}^{1}\!1/2\,\cos \left( x \left( -1+x \right) \right) {dx}-
\int _{0}^{1+B}\!1/2\,\cos \left( x \left( -1+x \right) \right) {dx}+
\int _{0}^{B}\!1/2\,\cos \left( x \left( -1+x \right) \right) {dx}
$

observe that:

$\lim _{B\rightarrow \infty } \left( -\int _{0}^{1+B}\!1/2\,\cos
\left( x \left( -1+x \right) \right) {dx}+\int _{0}^{B}\!1/2\,\cos
\left( x \left( -1+x \right) \right) {dx} \right) =0$

and thus:

$\lim _{B\rightarrow \infty } \left( \int _{0}^{B}\!\sin \left( x
\right) \sin \left( {x}^{2} \right) {dx} \right) =\int _{0}^{1}\!1/2
\,\cos \left( x \left( -1+x \right) \right) {dx}$

A good student of infi should be able to tell more or less right away that yes, the integral converges, and that it converges conditionally but not absolutely.

Why not absolutely? Because absolute value of the integrand does not decrease with x -> infinity.

Why yes conditionally? Because sin(x^2) oscillates around zero faster and faster, as x grows, like it is mentioned here. To convert this qualitative claim into actual proof, one can do an estimate like this.

Note, there is no need to actually evaluate the integral; it does not matter whether the first factor in sin(x)sin(x^2) is sin(x) or cos(x) or cos^3(x) etc. The defining moment here is the behavior of sin(x^2).