A calculus problem by คลุง แจ็ค

$\begin{aligned} I & = \int_0^\infty \sin \left(\frac 1{x^2} \right) dx & \small \color{#3D99F6} \text{By }\int_0^\infty f(x)\ dx = \int_0^\infty \frac {f\left(\frac 1x \right)}{x^2}\ dx \\ & = \int_0^\infty \frac {\sin \left(x^2 \right)}{x^2} dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = - \frac {\sin (x^2)}x \ \bigg|_0^\infty + \color{#3D99F6} \int_0^\infty 2\cos (x^2)\ dx & \small \color{#3D99F6} \text{Fresnel integral } C(u) = \int_0 ^u \cos \left(2\pi x^2 \right) dx \\ & = - 0 + {\color{#D61F06} \lim_{x \to 0} \frac {\sin (x^2)}x} + \color{#3D99F6} \sqrt{2\pi} C (\infty) & \small \color{#D61F06} \text{By L'Hôpital's rule} \\ & = {\color{#D61F06} \lim_{x \to 0} \frac {2x\cos (x^2)}1} + \color{#3D99F6} \sqrt{\frac \pi 2} & \small \color{#3D99F6} \text{Note that }C (\infty) = \frac 12 \\ & = \sqrt{\frac \pi 2} \end{aligned}$

Therefore $A = \boxed 2$ .

My solution probably is not fully correct, but here I go. Define for $t < 0$ $I(t) = \int_0^{\infty} e^{t/x^2}\sin\left(\frac{1}{x^2}\right)\mathrm{d}x,$ then $I'(t) = \int_0^{\infty} \frac{e^{t/x^2}\sin\left(\frac{1}{x^2}\right)}{x^2}\mathrm{d}x.$ Substitute $u = \frac{1}{x}$ , then $\mathrm{d}u = -\frac{1}{x^2}\mathrm{d}x,$ so that $I'(t) = -\int_{\infty}^0 e^{tu^2}\sin\left(u^2\right)\mathrm{d}u = \int_0^\infty e^{tu^2}\sin\left(u^2\right)\mathrm{d}u.$ Use that $\sin(z) = \frac{e^{zi} - e^{-zi}}{2i},$ for $z \in \mathbb{C}$ , to get that $I'(t) = \frac{1}{2i}\int_0^\infty e^{tu^2} \cdot (e^{u^2i} - e^{-u^2i}) \mathrm{d}u = \frac{1}{2i}\int_0^\infty e^{(t+i)u^2} - e^{(t-i)u^2}) \mathrm{d}u.$ Now, use that $\int_0^{\infty} e^{-zx^2} \mathrm{d}x = \frac{\sqrt{\pi}}{2\sqrt{z}}$ for any complex number $z$ , such that $\mathrm{Re}(z) > 0$ , to notice that $I'(t) = \frac{\sqrt{\pi}}{4i}\left(\frac{1}{\sqrt{-(t+i)}} - \frac{1}{\sqrt{i-t}}\right).$ Now, note that $I(t) = \frac{\sqrt{\pi}}{4i} \int \left(\frac{1}{\sqrt{-(t+i)}} - \frac{1}{\sqrt{i-t}}\right) \mathrm{d}t = \frac{\sqrt{\pi}}{2i}\left(\sqrt{i-t} - \sqrt{-(t+i)}\right) + C.$ Now, use that if $t \to -\infty$ , then $I(t) \to 0$ , from which follows that $C = 0$ after doing some work, so that $I(t) = \frac{\sqrt{\pi}}{2i}\left(\sqrt{i-t} - \sqrt{-(t+i)}\right)$ We can now conclude that $\int_0^\infty \sin\left(\frac{1}{x^2}\right)\mathrm{d}x = \lim_{t \, \uparrow \, 0} I(t) = \frac{\sqrt{\pi}}{2i}\left(\sqrt{i} - \sqrt{-i}\right) = \sqrt{\pi} \cdot \frac{e^{\frac{i\pi}{4}}-e^{\frac{-i\pi}{4}}}{2i} = \sqrt{\pi} \cdot \sin\left(\frac{\pi}{4}\right) = \sqrt{\frac{\pi}{2}}, \text{ so that } \boxed{A = 2}.$

This is a "sort-of" solution.

$\int \sin \left(\frac{1}{x^2}\right) \, dx \Rightarrow x \sin \left(\frac{1}{x^2}\right)-\sqrt{2 \pi } C\left(\frac{\sqrt{\frac{2}{\pi }}}{x}\right)$

$\underset{x\to 0}{\text{lim}}\,x \sin \left(\frac{1}{x^2}\right)\Rightarrow 0$ and $\underset{x\to \infty }{\text{lim}}\,x \sin \left(\frac{1}{x^2}\right)\Rightarrow 0$

$\underset{x\to 0}{\text{lim}}\,C(x)\Rightarrow 0$ and $\underset{x\to \infty }{\text{lim}}\,C(x)\Rightarrow \frac12$ assuming that $\frac10=\infty$ when, in fact, its sign is dependent on the approach direction.

$-(-\sqrt{2\pi})\frac12\Rightarrow \sqrt{\frac{\pi}{2}}$