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.

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


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 $0\leq I_1\leq \frac{1}{2\sqrt{\pi}}$.


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.


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