Fourier transform with $\sin(t^2)$

This exercise gave me nightmares this night. I have $$x(t)=\sin(t^2)e^{-2|t-2|}$$ to Fourier transform.

First I though about solving the integral. (should I divide the signal in $2$, first for $t-2<0$ then for $t-2\ge 0$ and then study then separately?) I then express the sin in exponentials with Euler and write up the Fourier integral (edit: which is also wrong). Well I'm not able to solve this integral, so I guess there is a smarter way to go about it, any hint?

• Where problem does this come from? You can certainly use the convolution theorem to take Fourier transforms of $\sin(t^2)$ and $\exp(-2|t-2|)$, and convolve the results to achieve a solution. The problem is that I'm not sure this convolution integral simplifies much afterwards. Are you looking for completely closed-form solution? – Chester Jun 28 '15 at 1:47
• It's an exam question for a telecommunications course. What do you mean by closed-form solution? – Crysis85 Jun 28 '15 at 4:16
• You have a product of two functions. The convolution theorem says that the Fourier transform of the product is equal to the convolution of the transforms of the two functions (up to scale depending on your def of the transform). It is mostly straight forward to compute each transform and write the convolution of the two. What I mean by closed-form is that this convolutional integral can be expressed in terms of simple functions. I'm not sure if this is the case. – Chester Jun 28 '15 at 4:33

My convention for the Fourier transform is $$\widehat{f}(\xi)=\int_{\mathbb{R}}f(x)e^{-2\pi i x\xi}\mathrm{d}x,\qquad \xi\in\mathbb{R} \tag{1}$$ You can adjust to your own convention using the scaling properties of the Fourier transform.

As Chester noted in the comments, you have a product of two functions $\sin t^{2}$ and $e^{-2\left|t-2\right|}$. However, the convolution theorem $\widehat{fg}=\widehat{f}\ast\widehat{g}$ only applies for suitable functions $f$ and $g$ or a distribution $f$ and a suitable function $g$. In any case, the Fourier transform of $\sin t^{2}$ only exists in the sense of distributions: $$\langle{\sin(\cdot)^{2},\widehat{\phi}}\rangle=\int_{\mathbb{R}}\sin(t^{2})\widehat{\phi}(t)\mathrm{d}t,\qquad\forall\phi\in\mathcal{S}(\mathbb{R})$$ As I don't know how familiar you are with distribution theory, I am going to try to avoid referring to it.

First, observe that by writing

$$\sin(t^{2})=\dfrac{e^{it^{2}}-e^{-it^{2}}}{2i}, \tag{2}$$ it suffices to replace $\sin t^{2}$ by $e^{\pm it^{2}}$ and then use linearity.

By the dominated convergence theorem, we have

$$\lim_{\delta\rightarrow 0^{+}}\int_{\mathbb{R}}e^{-\delta x^{2}}e^{\pm ix^{2}}e^{-2\left|x-2\right|}e^{-2\pi i x\xi}\mathrm{d}x=\int_{\mathbb{R}}e^{-ix^{2}}e^{-2\left|x-2\right|}\mathrm{d}x$$ For any $\delta>0$, the LHS is the Fourier transform of two $L^{1}$ functions $e^{-\delta x^{2}}e^{-ix^{2}}$ and $e^{-2\left|x-2\right|}$, which is given by the convolution of their Fourier transforms. Alternatively, we have a product of two square-integrable functions, so we can use Plancherel's theorem.

Lemma 1. If $f_{\delta}(x)=e^{-(\delta +i\sigma)x^{2}}$, for $\delta>0$ and $\sigma\in\mathbb{R}$, then $$\widehat{f_{\delta}}(\xi)=\int_{\mathbb{R}}e^{-(\delta+i\sigma)x^{2}}e^{-2\pi i x\xi}\mathrm{d}x=\left(\dfrac{\pi}{\delta+i\sigma}\right)^{1/2}e^{-(\pi\xi^{2})/(\delta+i\sigma)}, \tag{3}$$ where we take the principal branch of $\arg(z)$.

This result is typically proven using Cauchy's theorem from complex analysis, and a proof can be found in a typical Fourier analysis book. Since we will need it later below, we note that

$$\lim_{\delta\rightarrow 0^{+}}\widehat{f_{\delta}}(\xi)=\left(\dfrac{\pi}{\left|\sigma\right|}\right)^{1/2}e^{-\text{sgn}(\sigma)i\pi/4}e^{i\pi\xi^{2}/\sigma},\qquad\forall\xi\in\mathbb{R} \tag{4}$$

Lemma 2. If $g(x)=e^{-2\left|x-2\right|}$, then $$\widehat{g}(\xi)=\dfrac{e^{-4\pi i\xi}}{1+\pi^{2}\xi^{2}} \tag{5}$$

Proof. It suffices by the translation and scaling properties to compute the Fourier transform of the function $e^{-\left|t\right|}$l. \begin{align*} \int_{\mathbb{R}}e^{-\left|x\right|}e^{-2\pi i x\xi}\mathrm{d}x=\int_{0}^{\infty}e^{-x}2\cos(2\pi x\xi)\mathrm{d}x \end{align*} Integrating by parts twice, we obtain that the RHS is equal to $$-(2\pi^{2}\xi^{2})^{-1}\int_{0}^{\infty}e^{-x}\cos(2\pi x\xi)\mathrm{d}x,$$ which with some algebra yields $$\int_{\mathbb{R}}e^{-\left|x\right|}e^{-2\pi i x\xi}\mathrm{d}x=\dfrac{2}{1+4\pi^{2}\xi^{2}}$$ $\Box$

Applying Lemma 1 with $\sigma=\pm 1$ and Lemma 2, we obtain that

\begin{align*} \lim_{\delta\rightarrow 0^{+}}\int_{\mathbb{R}}e^{-(\delta\pm i) x^{2}}e^{-2\left|x-2\right|}e^{-2\pi i x\xi}\mathrm{d}x=\lim_{\delta\rightarrow 0^{+}}\int_{\mathbb{R}}\left(\dfrac{\pi}{\delta\pm i}\right)^{1/2}e^{-(\pi y^{2})/(\delta\pm i)}\dfrac{e^{-4\pi i(\xi-y)}}{1+\pi^{2}(\xi-y)^{2}}\mathrm{d}y \end{align*} The integrand on the RHS above is dominated by an $L^{1}$ function for all $\delta>0$, so by dominated convergence, we conclude that

\begin{align*} \int_{\mathbb{R}}e^{\mp ix^{2}}e^{-2\left|x-2\right|}e^{-2\pi i x\xi}\mathrm{d}x=\sqrt{\pi}e^{\mp i\pi/4}\int_{\mathbb{R}}e^{\pm i(\pi y^{2})}\dfrac{e^{-4\pi i(\xi-y)}}{1+\pi^{2}(\xi-y)^{2}}\mathrm{d}y \tag{6} \end{align*}

Whence, \begin{align*} \int_{\mathbb{R}}\sin(x^{2})e^{-2\left|x-2\right|}e^{-2\pi i x\xi}\mathrm{d}x&=\dfrac{\sqrt{\pi}}{2i}\int_{\mathbb{R}}\left[e^{-i(\pi y^{2}-\pi/4)}-e^{i(\pi y^{2}-\pi/4)}\right]\dfrac{e^{-4\pi i(\xi-y)}}{1+\pi^{2}(\xi-y)^{2}}\mathrm{d}y\\ &=-\sqrt{\pi}\int_{\mathbb{R}}\sin(\pi y^{2}-\pi/4)\dfrac{e^{-4\pi i(\xi-y)}}{1+\pi^{2}(\xi-y)^{2}}\mathrm{d}y \tag{7} \end{align*}

At the moment, I don't know how to obtain a closed form expression for the integral on the right. Perhaps another user could chime in on this point.