Various evalutions of $\int_0^\infty \sin x \sin \sqrt{x} \,dx$ I'm looking for various ways to evaluate the integral:
$$\int_0^\infty \sin x\sin \sqrt{x}\,dx$$
I'm mainly interested in complex analysis. I can think of a wedge -shaped contour of angle $\pi/4$ but I'm having trouble constructing the integrand properly. Perhaps we have to take into account that the root here may cause some trouble and define a function having a branch and this complicates things. 
I know two solutions using real analysis. One uses Laplace transformations and the other using only elementary tools plus the known results of the Fresnel Integrals.
Can someone help me with the contour integration?
Thank you!
 A: I'm posting an asnwer (of the $2$ I have) using real analysis methods:
$$\begin{aligned} 
\int_{0}^{\infty}\sin x \sin \sqrt{x}\,dx &\overset{\sqrt{x}=u}{=\! =\! =\!}2\int_{0}^{\infty}u\sin u \sin u^2 \,du \\  
 &=-\int_{0}^{\infty}u\cos \left ( u^2+u \right )\,du+\int_{0}^{\infty}u\cos(u^2-u)\,du \\  
 &\overset{u \mapsto u+1}{=\! =\! =\! =\!}-\int_{0}^{\infty}u\cos(u^2+u)\,du+\int_{-1}^{\infty}\left ( u+1 \right )\cos\left ( u^2+u \right )\,du \\  
 &= \int_{0}^{\infty}\cos\left ( u^2+u \right )\,du+\int_{-1}^{0}\left ( u+1 \right )\cos\left ( u^2+u \right )\,du\\  
 &\overset{u={\rm v}-\frac{1}{2}}{=\! =\! =\! =\!}\int_{1/2}^{\infty}\cos\left ( {\rm v}^2-\frac{1}{4} \right )\,d{\rm v}+\int_{-1/2}^{1/2}\left ( {\rm v}+\frac{1}{2} \right )\cos \left ( {\rm v}^2-\frac{1}{4} \right )\,d{\rm v} \\  
 &= \int_{0}^{\infty}\cos \left ( {\rm v}^2-\frac{1}{4} \right )\,d{\rm v}+\\ 
 &  \left [ \int_{-1/2}^{1/2}{\rm v}\cos \left ( {\rm v}^2-\frac{1}{4} \right )\,d{\rm v}+\frac{1}{2}\int_{-1/2}^{0}\cos\left ( {\rm v}^2-\frac{1}{2} \right )\,d{\rm v}- \frac{1}{2}\int_{0}^{1/2}\cos\left ( {\rm v}^2-\frac{1}{2} \right )\,d{\rm v} \right ] 
\end{aligned}$$
However, the equation in the bracket equals zero due to symmetry.
Hence:
$$\begin{aligned}\int_{0}^{\infty}\sin x \sin x^2\,dx&=\int_{0}^{\infty}\cos\left ( {\rm v}^2-\frac{1}{2} \right )\,d{\rm v}\\ 
 &=\cos \frac{1}{4}\int_{0}^{\infty}\cos {\rm v}^2\,d{\rm v}+\sin \frac{1}{4}\int_{0}^{\infty}\sin {\rm v}^2\,d{\rm v}\\ 
&\overset{(*)}{=}\int_{0}^{\infty}\sin {\rm v}^2\,d{\rm v}\left ( \cos \frac{1}{4}+\sin \frac{1}{4} \right )\\ 
 &=\frac{\sqrt{\pi}}{2}\sin \left ( \frac{3\pi-1}{4} \right )\;\; \;\;\;\; \square 
\end{aligned}$$
$(*)$ We used the Frensel integrals stating that $\displaystyle \int_{0}^{\infty}\cos x^2 \,dx=\int_{0}^{\infty}\sin x^2 \,dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}$. 
A: This is not an answer since it does not use complex analysis.
Using a CAS, what was found is that if $$I=\int_0^a \sin x\sin \sqrt{x}\,dx$$ $$I=\frac{1}{2} \sqrt{\frac{\pi }{2}} \left(\cos \left(\frac{1}{4}\right) (C(\alpha
   )+C(\beta ))+\sin \left(\frac{1}{4}\right) (S(\alpha )+S(\beta ))\right)-\sin
   \left(\sqrt{a}\right) \cos (a)$$ where appear Fresnel integrals using $\alpha=\frac{2 \sqrt{a}-1}{\sqrt{2 \pi }}$, $\beta=\frac{2 \sqrt{a}+1}{\sqrt{2 \pi }}$ which oscillates for ever as commented by Lucian.
