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

First make the substitution $x=u^2$ to get:
$\displaystyle \int _{0}^{\infty }\!\sin \left( x \right) \sin \left( \sqrt {x}
 \right) {dx}=\int _{0}^{\infty }\!2\,\sin \left( {u}^{2} \right) \sin
 \left( u \right) u{du}$,
$\displaystyle=-\int _{0}^{\infty }\!u\cos \left( u \left( u+1 \right)  \right) {du}+
\int _{0}^{\infty }\!u\cos \left( u \left( u-1 \right)  \right) {du}$,
and changing variable again in the second integral on the R.H.S such that $u\rightarrow u+1$ this becomes:
$=\displaystyle\int _{0}^{\infty }\!-u\cos \left( u \left( u+1 \right)  \right) {du}+
\int _{-1}^{\infty }\!\left(u+1\right)\cos \left( u \left( u+1 \right)  \right) {du}
$,
$\displaystyle=\int _{0}^{\infty }\!\cos \left( u \left( u+1 \right)  \right) {du}+
\int _{-1}^{0}\! \left( u+1 \right) \cos \left( u \left( u+1 \right) 
 \right) {du}
$.
Now we write $u=v-1/2$ and this becomes:
$\displaystyle\int _{1/2}^{\infty }\!\cos \left( {v}^{2}-1/4 \right) {dv}+\int _{-1/
2}^{1/2}\! \left( v+1/2 \right) \cos \left( {v}^{2}-1/4 \right) {dv}=$
$\displaystyle \left\{\int _{0}^{\infty }\!\cos \left( {v}^{2}-1/4 \right) {dv}\right\}$
$\displaystyle  +\left\{\int _{-1/2}
^{1/2}\!v\cos \left( {v}^{2}-1/4 \right) {dv}+\int _{-1/2}^{0}\!1/2\,
\cos \left( {v}^{2}-1/4 \right) {dv}-1/2\,\int _{0}^{1/2}\!\cos
 \left( {v}^{2}-1/4 \right) {dv}\right\},$
but the second curly bracket is zero by symmetry and so:
$\displaystyle \int _{0}^{\infty }\!\sin \left( x \right) \sin \left( \sqrt {x}
 \right) {dx}=\displaystyle \int _{0}^{\infty }\!\cos \left( {v}^{2}-1/4 \right) {dv}$,
$\displaystyle =\int _{0}^{
\infty }\!\cos \left( {v}^{2} \right) {dv}\cos \left( 1/4 \right) +
\int _{0}^{\infty }\!\sin \left( {v}^{2} \right) {dv}\sin \left( 1/4
 \right) 
$.
We now quote the limit of Fresnel integrals:
$\displaystyle\int _{0}^{\infty }\!\cos \left( {v}^{2} \right) {dv}=\int _{0}^{
\infty }\!\sin \left( {v}^{2} \right) {dv}=\dfrac{\sqrt{2\pi}}{4}$,
to obtain:
$\displaystyle \int _{0}^{\infty }\!\sin \left( x \right) \sin \left( \sqrt {x}
 \right) {dx}=\dfrac{\sqrt{2\pi}}{4}\left(\cos\left(\dfrac{1}{4}\right)+\sin\left(\dfrac{1}{4}\right)\right)=\dfrac{\sqrt{\pi}}{2}\sin{\left(\dfrac{3\pi-1}{4}\right)}$.
