How prove this $\int_{0}^{\infty}\sin{x}\sin{\sqrt{x}}\,dx=\frac{\sqrt{\pi}}{2}\sin{\left(\frac{3\pi-1}{4}\right)}$ Prove that
$$\int_{0}^{\infty}\sin{x}\sin{\sqrt{x}}\,dx=\frac{\sqrt{\pi}}{2}\sin{\left(\frac{3\pi-1}{4}\right)}$$
I have some question. Using this, find this integral is not converge, I'm wrong?
Thank you everyone

Numerical calculation shows that the graph of
$$ y = \int_{0}^{x} \sin t \sin \sqrt{t} \, dt $$
is given by

Though a graph cannot constitute a proof, it strongly suggests that $y$ cannot converge as $x \to \infty$. Indeed, we can show that
$$ y(x) = -\cos x \sin \sqrt{x} + \frac{\sqrt{\pi}}{2} \sin \left( \frac{3\pi - 1}{4} \right) + o(1) $$
as $x \to \infty$. Thus in ordinary sense, the integral
$$ \int_{0}^{\infty} \sin x \sin \sqrt{x} \, dx $$
diverges. But if we understand the integral in Abel summation senas as follows
$$ \int_{0}^{\infty} \sin x \sin \sqrt{x} \, dx := \lim_{s \to 0^{+}} \int_{(0, \infty)} \sin x \sin \sqrt{x} \; e^{-sx} \, dx, $$
then the oscillating part vanishes and we obtain the identity. In fact, it converges to the proposed limit as weak as in Cesaro summation sense.
Now let us show that the integral converges in Able summation sense. By some calculation, we have
\begin{align*}
\int_{0}^{\infty} \sin x \sin \sqrt{x} \; e^{-sx} \, dx
&= \int_{0}^{\infty} 2x \sin x \sin (x^2) \; e^{-sx^{2}} \, dx \\
&= \int_{0}^{\infty} x \Re \left( e^{-i(x^2-x)} - e^{-i(x^2+x)} \right) e^{-sx^{2}} \, dx \\
&= \Re \int_{-\infty}^{\infty} x e^{-(s+i)x^2 + ix} \, dx.
\end{align*}
By noting that
$$ z e^{-(s+i)z^2 + iz} = z \exp \left\{ -(s+i) \left( z - \tfrac{i}{2(s+i)} \right)^2 + \tfrac{i}{4(1-is)} \right\} $$
is an entire function with a nice vanishing speed as $ \left| \Re z \right| \to \infty$, we find that we can shift the contour of integration so that
\begin{align*}
\int_{0}^{\infty} \sin x \sin \sqrt{x} \; e^{-sx} \, dx
&= \Re \left[ e^{\frac{i}{4(1-is)}} \int_{-\infty}^{\infty} \left(x + \frac{i}{2(s+i)} \right) e^{-(s+i)x^2} \, dx \right] \\
&= \Re \left[ e^{\frac{i}{4(1-is)}} \frac{i}{2(s+i)} \int_{-\infty}^{\infty} e^{-(s+i)x^2} \, dx \right] \\
&= \Re \left[ e^{\frac{i}{4(1-is)}} \frac{i}{2(s+i)} \frac{\sqrt{\pi}}{\sqrt{s + i}} \right].
\end{align*}
Taking $s \to 0^{+}$, we find that in Abel summation sense,
\begin{align*}
\int_{0}^{\infty} \sin x \sin \sqrt{x} \, dx
&= \frac{\sqrt{\pi}}{2} \Re \left( e^{\frac{i-i\pi}{4}} \right)
 = \frac{\sqrt{\pi}}{2} \cos \left( \frac{1-\pi}{4} \right)
 = \frac{\sqrt{\pi}}{2} \sin \left( \frac{3\pi - 1}{4} \right)
\end{align*}
as desired.
