Does the following integral converge?

$$\int_1^\infty \sin^2 (x^2) \, dx$$

I tried $$\int_1^\infty \sin^2(x^2) \, dx=\int_1^\infty \frac{1-\cos(2x^2)}{2} \, dx = \frac{\sqrt{2}}{4} \int_1^\infty\frac{1-\cos(u)}{2\sqrt{u}} \, du$$

The idea is that, I want to compare the original integral to a divergent $p$-integral. But I am not sure how to proceed from here.

  • $\begingroup$ Does $\int_1^{\infty} \cos (2x^2)\,dx$ converge? $\endgroup$ – Daniel Fischer Oct 19 '16 at 17:50
  • $\begingroup$ $\int_{k\pi+\pi/2}^{k\pi+3\pi/2} \frac{1-\cos(u)}{\sqrt{u}} du \geq \int_{k\pi+\pi/2}^{k\pi+3\pi/2} \frac{1-\cos(u)}{\sqrt{k\pi+3\pi/2}} du = \dots$ $\endgroup$ – Ian Oct 19 '16 at 17:50
  • $\begingroup$ sum of two integrals.div $+$ conv. $\endgroup$ – hamam_Abdallah Oct 19 '16 at 17:51
  • $\begingroup$ Here's Ian's comment written more legibly:$$\int_{k\pi+\pi/2}^{k\pi+3\pi/2} \frac{1-\cos(u)}{\sqrt u} \, du \geq \int_{k\pi+\pi/2}^{k\pi+3\pi/2} \frac{1-\cos(u)}{\sqrt{k\pi+3\pi/2}} \, du = \cdots$$ $\endgroup$ – Michael Hardy Oct 19 '16 at 17:55
  • 1
    $\begingroup$ Why not just write$$ \int_1^\infty \sin^2 x^2\,dx = \int_1^\infty \sin^2 u \, \frac{du}{2\sqrt u} $$instead of also using the half-angle formula? What is gained by the latter? $\qquad$ $\endgroup$ – Michael Hardy Oct 19 '16 at 17:58

No, it does not. We have, due to $\sin^2(z)=\frac{1-\cos(2z)}{2}$, $$ \int_{1}^{M}\sin^2(x^2)\,dx \stackrel{x^2\to z}{=} \int_{1}^{M^2}\frac{\sin^2(z)}{2\sqrt{z}}\,dz =\frac{M-1}{2}-\int_{1}^{M^2}\frac{\cos(2z)}{4\sqrt{z}}\,dz$$ where $\int_{1}^{+\infty}\frac{\cos(2z)}{4\sqrt{z}}\,dz$ is finite by Dirichlet's test/integration by parts, but $\frac{M-1}{2}$ grows unbounded.
In particular, $$\int_{1}^{M^2}\frac{\cos(2z)}{4\sqrt{z}}\,dz = \left.\frac{\sin(2z)}{8\sqrt{z}}\right|_{1}^{M^2}+\int_{1}^{M^2}\frac{\sin(2z)}{16 z\sqrt{z}}\,dz$$ is bounded in absolute value by $\frac{3}{8}$ (by the trivial inequality $\sin\leq 1$), hence $$\boxed{ \int_{1}^{M}\sin^2(x^2)\,dx \color{red}{\geq \frac{M}{2}-\frac{7}{8}}. }$$


We know that $\sin^2(x)\geq\dfrac{1}{2}$ $\text{whenever}\,\,\dfrac{\pi}{4}+n\pi\leq x\leq\dfrac{3\pi}{4}+n\pi, n\in\mathbb{N}$. Hence $$\sqrt{\dfrac{\pi}{4}+n\pi}\leq x\leq\sqrt{\dfrac{3\pi}{4}+n\pi}\Longrightarrow \sin^2(x^2)\geq\dfrac{1}{2}$$ Let $I_n=\left[\sqrt{\dfrac{\pi}{4}+n\pi},\sqrt{\dfrac{3\pi}{4}+n\pi}\right]$. Then the length of $I_n$, $n\geq 1$, is more than $\dfrac{C}{\sqrt{n}}$ for some constant $C>0$ independent of $n$. Hence: $$\int_1^{\infty}\sin^2(x^2)dx\geq\sum_{n\geq 1}^{\infty}\int_{I_n}\sin^2(x^2)dx\geq \sum_{n\geq 1}^{\infty}\dfrac{1}{2}\dfrac{C}{\sqrt{n}}$$ which is infinite.


$$\int\sin^2x^2\,dx=\frac12\int dx-\frac12\int\cos2x^2\,dx.$$ The second term is a Fresnel cosine integral, which is known to converge.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.