Find the Fourier transform of $\sin x^2$.

Question

I've tried it by applying integratrion by parts, but I'm not getting the answer correct. Its answer is $$\frac{1}{\sqrt{2}}\,\sin\left(\frac{k^2}{4} +\frac{\pi}{4}\right).$$ Please help in this.

Answer 1

$$I(\xi)=\int_{\mathbb{R}}\sin(x^2)\cos(\xi x)\,dx = \int_{\mathbb{R}}\sin(x^2)\,dx - \int_{0}^{+\infty}\frac{1-\cos(\xi\sqrt{x})}{\sqrt{x}}\sin(x)\,dx $$ is a convergent integral if $\int$ is intended as an improper Riemann integral, by Dirichlet's criterion.

Moreover, $\sin(x^2)\cos(\xi x) = \frac{1}{2}\left(\sin(x^2+\xi x)+\sin(x^2-\xi x)\right)$, and:

$$ \int_{\mathbb{R}}\sin(x^2+\xi x)\,dx = \int_{\mathbb{R}}\sin\left(\left(x+\frac{\xi}{2}\right)^2-\frac{\xi^2}{4}\right)\,dx = \int_\mathbb{R}\sin(x^2-\xi^2/4)\,dx$$ just depends on $\sin(\xi^2/4)$ and $\cos(\xi^2/4)$ by the sine addition formulas, since: $$ \int_{\mathbb{R}}\sin(x^2)\,dx = \int_{\mathbb{R}}\cos(x^2)\,dx = \sqrt{\frac{\pi}{2}}$$ are Fresnel's integrals. Put everything together and you will get your Fourier transform defined by an improper Riemann integral.

Answer 2

There is not enough information in the question to allow us writing a full answer. However, we can formally derive the sine form.

If $u(x):=\sin x^2$, then one can check that u solves the $2$nd order ODE $$xu''-u'+4x^3u=0.$$ If $\hat{u}$ is the unitary Fourier transform of $u$, then $\hat{u}$ must solve the $3$rd order ODE $$\hat{u}'''+\frac{k^2}{4}\hat{u}'+\frac{3k}{4}\hat{u}=0$$ because of the formula (see e.g. https://en.wikipedia.org/wiki/Fourier_transform#Applications) $$\widehat{x^n u}(k)=\mathrm{i}^n\frac{\mathrm{d}^n}{\mathrm{d}k^n}\hat{u}(k),\quad\quad \widehat{\left(\frac{\mathrm{d}^n}{\mathrm{d}x^n}u\right)}(k)=\mathrm{i}^nk^n\hat{u}(k).\quad\quad\quad\quad(*)$$ Now define $v(k):=A\sin\left(\frac{k^2}{4}+K\right)$ for some real constants $A,K$. One computes : \begin{align} v'(k)&=\frac{Ak}{2}\cos\left(\frac{k^2}{4}+K\right),\\ v''(k)&=\frac{A}{2}\cos\left(\frac{k^2}{4}+K\right)-\frac{Ak^2}{4}\sin\left(\frac{k^2}{4}+K\right),\\ v'''(k)&=-\frac{3Ak}{4}\sin\left(\frac{k^2}{4}+K\right)-\frac{Ak^3}{8}\cos\left(\frac{k^2}{4}+K\right).\\ \end{align} One can check that $v$ solves $(*)$. Now, it remains to show that $v=\hat{u}$ (by invoking an argument of unicity of the solution of $(*)$) and to find the constants $A,K$, but one needs more information.

Answer 3

Let be $$ \sin(ax^2)=\frac{\mathrm e^{iax^2}-\mathrm e^{iax^2}}{2i}\qquad\text{and}\qquad\cos(ax^2)=\frac{\mathrm e^{iax^2}+\mathrm e^{iax^2}}{2} $$ anf use the Fourier transform defined as $$\mathcal F\left\{f(x)\right\}=F(k)=\int_{-\infty}^{\infty} f(x)\mathrm e^{-i2\pi k x}\,\mathrm d x$$

The Fourier transform can be found by using the Fourier Transform of the Gaussian. Observing that $$\mathcal F\left\{\mathrm e^{-\beta x^2}\right\}=\sqrt{\frac{\pi}{\beta}}\mathrm e^{-\frac{pi^2 k^2}{\beta}}$$ for $\beta=ia$ for $a>0$ $$\mathcal F\left\{\mathrm e^{-i ax^2}\right\}=\sqrt{\frac{\pi}{ai}}\mathrm e^{-\frac{\pi^2 k^2}{ia}}=\sqrt{\frac{\pi}{a}}\mathrm e^{-i\frac{\pi}{4}}\mathrm e^{-\frac{\pi^2 k^2}{ia}}=\sqrt{\frac{\pi}{a}}\mathrm{exp}\left(i\left[\frac{\pi^2 k^2}{a}-\frac{\pi}{4}\right]\right)$$ for $a<0$, $\beta=-|a|i$

$$ \mathcal F\left\{\mathrm e^{-i ax^2}\right\}= \sqrt{\frac{\pi}{-|a|i}}\mathrm e^{-\frac{\pi^2 k^2}{i|a|}} =\sqrt{\frac{\pi}{|a|}}\mathrm e^{i\frac{\pi}{4}}\mathrm e^{-i\frac{\pi^2 k^2}{|a|}} =\sqrt{\frac{\pi}{|a|}}\mathrm{exp}\left(-i\left[\frac{\pi^2 k^2}{|a|}-\frac{\pi}{4}\right]\right) $$

that is $$ \mathcal F\left\{\mathrm e^{-i ax^2}\right\}=\begin{cases} \sqrt{\frac{\pi}{a}}\mathrm{exp}\left(i\left[\frac{\pi^2 k^2}{|a|}-\frac{\pi}{4}\right]\right) & \text{for }a>0\\ \sqrt{\frac{\pi}{|a|}}\mathrm{exp}\left(-i\left[\frac{\pi^2 k^2}{|a|}-\frac{\pi}{4}\right]\right) & \text{for }a<0 \end{cases} $$

So we have, for positive $a$, $$ \mathcal F\left\{\mathrm e^{-iax^2}\right\}=\sqrt{\frac{\pi}{a}}\mathrm{exp}\left({i\left[\frac{\pi^2k^2}{a}-\frac{\pi}{4}\right]}\right) $$ and $$ \mathcal F\left\{\mathrm e^{iax^2}\right\}=\mathcal F\left\{\mathrm e^{-i(-a)x^2}\right\}=\sqrt{\tfrac{\pi}{|-a|}}\mathrm{exp}\left({-i\left[\tfrac{\pi^2k^2}{|a|}-\tfrac{\pi}{4}\right]}\right)=\sqrt{\frac{\pi}{a}} \mathrm{exp}\left({-i\left[\frac{\pi^2k^2}{a}-\frac{\pi}{4}\right]}\right) $$ So we have, for positive $a$, \begin{align} \mathcal F\left\{\sin(ax^2)\right\}&=\frac{\mathcal F\left\{\mathrm e^{iax^2}\right\}-\mathcal F\left\{\mathrm e^{-iax^2}\right\}}{2i}\\ &=\frac{1}{2i}\left[\sqrt{\frac{\pi}{a}}\mathrm e^{-i\left(\frac{\pi^2k^2}{a}-\frac{\pi}{4}\right)}-\sqrt{\frac{\pi}{a}}\mathrm e^{i\left(\frac{\pi^2k^2}{a}-\frac{\pi}{4}\right)}\right]\\ &=-\sqrt{\frac{\pi}{a}}\sin\left(\frac{\pi^2k^2}{a}-\frac{\pi}{4}\right) \end{align} and observing that $\sin(-t)=-\sin(t)$, we have $$ \mathcal F\left\{\sin(ax^2)\right\}=\begin{cases} -\sqrt{\frac{\pi}{a}}\sin\left(\frac{\pi^2k^2}{a}-\frac{\pi}{4}\right) & \text{for }a>0\\ \sqrt{\frac{\pi}{|a|}}\sin\left(\frac{\pi^2k^2}{|a|}-\frac{\pi}{4}\right) & \text{for }a<0 \end{cases} $$ In the same way we have $$ \mathcal F\left\{\cos(ax^2)\right\}=\begin{cases} \sqrt{\frac{\pi}{a}}\cos\left(\frac{\pi^2k^2}{a}-\frac{\pi}{4}\right) & \text{for }a>0\\ \sqrt{\frac{\pi}{|a|}}\cos\left(\frac{\pi^2k^2}{|a|}-\frac{\pi}{4}\right) & \text{for }a<0 \end{cases} $$ Thus \begin{align} \mathcal F\left\{\sin(x^2)\right\}=-\sqrt{\pi}\sin\left(\pi^2k^2-\frac{\pi}{4}\right) \end{align}

Using the Fourier transform defined as $$\mathcal F\left\{f(x)\right\}=\hat f(\xi)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x)\mathrm e^{-i\xi x}\,\mathrm d x=\frac{1}{\sqrt{2\pi}}F\left(\frac{\xi}{2\pi}\right)$$ we find $$ \mathcal F\left\{\sin(x^2)\right\}=-\frac{1}{\sqrt{2}}\sin\left(\frac{\xi^2}{4}-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\cos\left(\frac{\xi^2}{4}+\frac{\pi}{4}\right) $$