I am trying to prove that the improper integral $$ I=\int_0^\infty \sin x\sin(x^2)\mathrm{d}x$$ converges.

Here's my work:

It suffices to show that $$\int_\frac{\pi}{2}^\infty \sin (x) \sin(x^2)\mathrm{d}x$$ converges. Using integration by parts,

\begin{align*} \int_\frac{\pi}{2}^\infty \sin (t) \sin(t^2)\mathrm{d}t&=\int_\frac{\pi}{2}^\infty \frac{\sin (t)}{2t}\cdot 2t\sin(t^2)\mathrm{d}t\\ &=\underline{\bigg[ -\frac{\sin (t)}{2t}\cos(t^2)\bigg]_\frac{\pi}{2}^\infty}+{\int_\frac{\pi}{2}^\infty \left(\frac{\sin (t)}{2t}\right)'\cos(t^2)\mathrm{d}t} \end{align*} The underlined part is a constant...

Then I got stuck. I'd like to use "sandwich rule" using the fact that $-1\leq \cos(t^2)\leq 1$, but I can't find a way to apply it properly.

How can I proceed from here? Any correction and/or help would be appreciated. :)

  • 1
    $\begingroup$ Do you know about the Fresnel Integral value $\int_{0}^{\infty} \sin(x^2) \ dx = \sqrt{\frac{\pi}{8}}$ ? If so you can use the comparison test to prove convergence ? $\endgroup$ May 15, 2017 at 2:47
  • $\begingroup$ @VivekKaushik I have to solve it in high school level, so I can't use such techniques. Good idea though. $\endgroup$ May 15, 2017 at 2:53
  • $\begingroup$ Just for fun, the value of $I$ is given by $$I = \sqrt{\frac{\pi }{2}} \left(C\left(\frac{1}{\sqrt{2 \pi }}\right) \cos \left(\frac{1}{4}\right)+S\left(\frac{1}{\sqrt{2 \pi }}\right) \sin \left(\frac{1}{4}\right)\right) \approx 0.4917$$ Nice result! $\endgroup$
    – Dmoreno
    May 15, 2017 at 16:34
  • $\begingroup$ I would like to see a proof that splits the integral into the regions where $sin(x^2)$ has constant sign. This might generalize to s proof that $\int \sin(x)\sin(f(x))dx$ convergences whenever$f$ grows fast enough. For example, would $f(x)=x^{1+c}$ converge for all $c > 0$. How about $x\ln(1+x)$? $\endgroup$ Oct 31, 2019 at 13:12

4 Answers 4


We shall use only substitution and integration by parts to show that the integral of interest, $\int_1^L \sin(x)\sin(x^2)\,dx$, is convergent.

First, enforcing the substitution $x\to \sqrt{x}$ reveals

$$\int_1^L \sin(x)\sin(x^2)\,dx=\frac12\int_1^L \frac{\sin(\sqrt{x})\sin(x)}{\sqrt{x}}\,dx \tag 1$$

Second, integrating by parts the integral on the right-hand side of $(1)$ with $u=\frac{\sin(\sqrt{x})}{\sqrt{x}}$ and $v=-\cos(x)$ yields

$$\begin{align} \int_1^L \sin(x)\sin(x^2)\,dx&=\frac12\left.\left(-\frac{\sin(\sqrt{x})\cos(x)}{\sqrt{x}}\right)\right|_{x=1}^{x=L}\\\\ &+\frac14 \int_1^L \left(\frac{\cos(\sqrt{x})\cos(x)}{x}-\frac{\sin(\sqrt{x})\cos(x)}{x^{3/2}}\right)\,dx\tag2 \end{align}$$

Third, integrating by parts the first term in the integral on the right-hand side of $(2)$ with $u=\frac{\cos(\sqrt{x})}{x}$ and $v=\sin(x)$, we obtain

$$\begin{align}\int_1^L \frac{\cos(\sqrt{x})\cos(x)}{x}\,&=\left.\left(\frac{\cos(\sqrt{x})\sin(x)}{x}\right)\right|_{x=1}^{x=L}\\\\ &- \int_1^L \left(\frac{\sin(\sqrt{x})\sin(x)}{2x^{3/2}}+\frac{\cos(\sqrt{x})\sin(x)}{x^2}\right)\,dx\tag 3 \end{align}$$

Substituting $(3)$ into $(2)$ shows that all integrals involved are of the forms

$$I_1=\int_1^L \frac{\sin(\sqrt{x})\cos(x)}{x^{3/2}}\,dx$$


$$I_2=\int_1^L \frac{\cos(\sqrt{x})\sin(x)}{x^2}\,dx$$

Both $I_1$ and $I_2$ are absolutely convergent as $L\to\infty$ since $\int_1^\infty \frac{1}{x^{3/2}}\,dx<\infty$ and $\int_1^L \frac{1}{x^2}\,dx<\infty$.

Therefore, the integral of interest converges as was to be shown!.

  • $\begingroup$ Wow, thank you! Can it be done without using the absolute convergence test and/or p-value test? (I'm trying to find the simplest solution to this problem...) $\endgroup$ May 15, 2017 at 3:09
  • $\begingroup$ You're welcome. My pleasure. Substitution and two integration by parts is about as simple as this one gets. $\endgroup$
    – Mark Viola
    May 15, 2017 at 3:12
  • $\begingroup$ @MarkViola The lower limit of Integration in given problem is $0$, don't it matters? $\endgroup$ Jul 25, 2017 at 14:55
  • $\begingroup$ @ArpitYadav The integral from $0$ to $1$ is a proper integral. $\endgroup$
    – Mark Viola
    Oct 31, 2019 at 14:31

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\Lambda > 0}$:

\begin{align} \int_{0}^{\Lambda}\sin\pars{x}\sin\pars{x^{2}}\,\dd x & = {1 \over 2}\int_{0}^{\Lambda}\cos\pars{x^{2} - x}\,\dd x - {1 \over 2}\int_{0}^{\Lambda}\cos\pars{x^{2} + x}\,\dd x \\[5mm] & = {1 \over 2}\int_{1/2}^{\Lambda + 1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x - {1 \over 2}\int_{-1/2}^{\Lambda - 1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x \\[1cm] & = {1 \over 2}\int_{1/2}^{\Lambda - 1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x + {1 \over 2}\int_{\Lambda - 1/2}^{\Lambda + 1/2} \cos\pars{x^{2} - {1 \over 4}}\,\dd x \\[2mm] & - {1 \over 2}\int_{-1/2}^{1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x - {1 \over 2}\int_{1/2}^{\Lambda - 1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x \\[1cm] & = {1 \over 2}\int_{-1/2}^{1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x \\[2mm] & + \bracks{% {1 \over 2}\int_{1/2}^{\Lambda + 1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x - {1 \over 2}\int_{1/2}^{\Lambda - 1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x} \end{align}

As $\ds{\Lambda \to \infty}$, the last two integrals converge since they can be reduced to convergent Fresnel Integrals.

$$ \int_{0}^{\infty}\sin\pars{x}\sin\pars{x^{2}}\,\dd x = \bbx{\int_{0}^{1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x} \quad\mbox{which is clearly}\ convergent. $$


You could just keep going by explicitly taking the derivative:

$$\left(\sin t\over2t\right)'={\cos t\over 2t}-{\sin t\over2t^2}$$

which now gives two integrals to be shown convergent:

$$\int_{\pi/2}^\infty{\cos t\cos t^2\over2t}dt\qquad\text{and}\qquad\int_{\pi/2}^\infty{\sin t\cos t^2\over2t^2}dt$$

The second of these is clearly convergent, since $\left|\sin t\cos t^2\over2t^2\right|\le{1\over2t^2}$ and $\int_1^\infty{dx\over x^2}=1$. For the first integral, do integration by parts again, using

$$u={\cos t\over4t^2}\qquad\text{and}\qquad dv=2t\cos t^2\,dt$$

so that

$$\int_{\pi/2}^\infty{\cos t\cos t^2\over2t}dt={\cos t\sin t^2\over4t^2}\Big|_{\pi/2}^\infty+\int_{\pi/2}^\infty\left({\sin\over4t^2}+{\cos t\over8t^3}\right)\sin t^2\,dt$$

and the remaining integral is again clearly convergent.


It is enough to show that both the integrals $$ I_1 = \int_{0}^{+\infty}\cos(x+x^2)\,dx,\qquad I_2=\int_{0}^{+\infty}\cos(x^2-x)\,dx $$ are (conditionally) convergent. $f(x)=x+x^2$ is an increasing function on $\mathbb{R}^+$ and by setting $x=f^{-1}(t)$ we get: $$ I_1 = \int_{0}^{+\infty}\frac{\cos t}{\sqrt{1+4t}}\,dt $$ that clearly is a convergent integral by Dirichlet's test: $\cos t$ has a bounded primitive and $\frac{1}{\sqrt{1+4t}}$ is decreasing towards zero on $\mathbb{R}^+$. A similar argument proves the convergence of $I_2$: $g(x)=x^2-x$ is increasing on $\left(\frac{1}{2},+\infty\right)$ and the integral $\int_{0}^{1/2}\cos(x^2-x)\,dx$ is clearly bounded by $\frac{1}{2}$ in absolute value.

The Laplace transform gives a way for computing a good numerical approximation of $I_1$. We have: $$ 0\leq I_1 = \frac{1}{2\sqrt{\pi}}\int_{0}^{+\infty}\frac{\sqrt{s}e^{-s/4}}{1+s^2}\,ds \stackrel{\text{Cauchy-Schwarz}}{\leq}\frac{1}{2\sqrt{\pi}}\sqrt{\int_{0}^{+\infty}e^{-s/2}\,ds\int_{0}^{+\infty}\frac{s\,ds}{(1+s^2)^2}}$$ hence $0\leq I_1\leq \frac{1}{2\sqrt{\pi}}$.


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.