# 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!

• The integral is undefined. Mar 8, 2015 at 21:42
• I agree with @science, as written the integrand function is not Riemann integrable over $\mathbb{R}$^+. Are you sure it is $dx$ and not $\frac{dx}{\sqrt{x}}$ or $\frac{dx}{x}$? Mar 8, 2015 at 21:47
• What do you mean undefined? Yep, I'm completely sure about that. Mar 8, 2015 at 21:49
• The result of the above integral is $\displaystyle \frac{\sqrt{\pi}}{2}\sin\bigl({\tfrac{3\pi-1}{4}}\bigl)$. And I can assure you it converges. Mar 8, 2015 at 21:54
• I can assure you it converges. - And I can assure you that it does not, but rather oscillates indefinitely within the limits of a specified interval of the form $(a-1,a+1)$, where $$a=\dfrac{\sqrt{2\pi}}4\bigg(\cos\dfrac14+\sin\dfrac14\bigg).$$ Mar 8, 2015 at 23:04

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 $\displaystyle \int_{0}^{\infty}\cos x^2 \,dx=\int_{0}^{\infty}\sin x^2 \,dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}$.

• Thnaks for the nice answer. $+1$ Mar 10, 2015 at 11:10
• You begin by writing the original integral as the difference of two others, both of which diverge. So I assume that the integral converges by means of the Cauchy principal value of its particular alternate expression. Mar 11, 2015 at 20:26

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.