Integral of sinc function multiplied by Gaussian

Question

I am wondering whether the following integral

$$\int_{-\infty}^{\infty} \frac{\exp( - a x^2 ) \sin( bx )}{x} \,\mathrm{d}x$$

exists in closed form. I would like to use it for numerical calculation and find an efficient way to evaluate it. If analytical form does not exist, I really appreciate any alternative means for evaluating the integral. One method would be numerical quadrature including Gaussian quadrature, but it may be inefficient when the parameters $a$ and $b$ are very different in scale.

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EDIT : In view of this discussion, we have decided to add OP's self-answer to the end of the question, for it does not qualify as an answer yet contains vital details. The copying is unabridged.

Thanks very much for your comments, and the following result was obtained including the case for $x_0 \ne 0$: $$ \int_{-\infty}^{\infty} dx \exp[-a(x-x_0)^2] \frac{ \sin(bx) }{ x } = \pi \exp(-a x_0^2) \mathrm{Re}\left(\mathrm{erf}\left[\frac{b+2iax_0}{2\sqrt{a}}\right] - \mathrm{erf}\left[\frac{2iax_0}{2\sqrt{a}}\right]\right) $$ where $a\gt0, b, x_0$ are assumed to be all real. (note: coefficients etc may be still wrong...)

This integral appears in a type of electronic structure calculation based on a grid representation (sinc-function basis). I believe the above result should be definitely useful.

Thanks much!! --jaian

Answer 1

We assume $a,b>0$. Then $$\begin{eqnarray*} \int_{-\infty}^\infty dx\, e^{-a x^2}\frac{\sin b x}{x} &=& \int_0^b d\beta \, \int_{-\infty}^\infty dx\, e^{-a x^2} \cos \beta x \\ &=& \int_0^b d\beta \, \mathrm{Re} \int_{-\infty}^\infty dx\, e^{-a x^2+i \beta x} \\ &=& \int_0^b d\beta \, \mathrm{Re}\, \sqrt{\frac{\pi}{a}} e^{-{\beta}^2/(4a)} \\ &=& \pi \, \mathrm{erf}\left(\frac{b}{2\sqrt{a}}\right). \end{eqnarray*}$$ This approach can be generalized to $x_0\ne0$.

Answer 2

We can consider the general type of integral, means

$$\int_{-\infty}^{\infty}e^{-a(x-x_{0})^{2}}\frac{\sin(bx)}{x}dx$$

Case 1. If $b=0$, the function identically 0, so the integral converges and equals to 0.

Remark. The function is an odd function of $b$ (we consider $b$ as a variable), so we can only consider cases of $b>0$ in the following.

Case 2. If $a<0$, the integral divergent and didn't exist.

Case 3. If $a>0$, we can calculate as following:

$$\begin{align*} \int_{-\infty}^{\infty}e^{-a(x-x_{0})^{2}}\frac{\sin(bx)}{x}dx &=\int_{-\infty}^{\infty}e^{-a(x-x_{0})^{2}}(\int_{0}^{b}\cos(xy)dy)dx\\ &=\int_{0}^{b}(\int_{-\infty}^{\infty}e^{-a(x-x_{0})^{2}}\cos(xy)dx)dy \end{align*}$$

$$\begin{align*} \int_{-\infty}^{\infty}e^{-a(x-x_{0})^{2}}\cos(xy)dx &=\int_{-\infty}^{\infty}e^{-ax^{2}}\cos((x+x_{0})y)dx\\ &=\cos(x_{0}y)\int_{-\infty}^{\infty}e^{-ax^{2}}\cos(xy)dx -\sin(x_{0}y)\int_{-\infty}^{\infty}e^{-ax^{2}}\sin(xy)dx\\ &=\cos(x_{0}y)\int_{-\infty}^{\infty}e^{-ax^{2}}\cos(xy)dx \end{align*}$$

We denote $\int_{-\infty}^{\infty}e^{-ax^{2}}\cos(xy)dx$ by $F(y)$, then

$$\begin{align*} F^{\prime}(y)&=-\int_{-\infty}^{\infty}xe^{-ax^{2}}\sin(xy)dx =\frac{1}{2a}\int_{-\infty}^{\infty}\sin(xy)de^{-ax^{2}}\\ &=-\frac{y}{2a}\int_{-\infty}^{\infty}e^{-ax^{2}}\cos(xy)dx =-\frac{y}{2a}F(y) \end{align*}$$

By calculation, we can obtain that $F(0)=\int_{-\infty}^{\infty}e^{-ax^{2}}dx =2\int_{0}^{\infty}e^{-ax^{2}}dx =\frac{1}{\sqrt{a}}\Gamma(\frac{1}{2}) =\frac{\sqrt{\pi}}{\sqrt{a}}$.

Then solve the ordinary differential equation with initial value, we can get:

$$F(y)=F(0)e^{-\frac{y^{2}}{4a}}=\frac{\sqrt{\pi}}{\sqrt{a}}e^{-\frac{y^{2}}{4a}}$$

So

$$\begin{align*} \int_{-\infty}^{\infty}e^{-a(x-x_{0})^{2}}\frac{\sin(bx)}{x}dx &=\int_{0}^{b}\cos(x_{0}y)F(y)dy\\ &=\frac{\sqrt{\pi}}{\sqrt{a}}\int_{0}^{b}e^{-\frac{y^{2}}{4a}}\cos(x_{0}y)dy\\ &=\sqrt{\pi}\int_{0}^{\frac{b}{\sqrt{a}}}e^{-\frac{t^{2}}{4}}\cos(\sqrt{a}x_{0}t)dt \end{align*}$$

Case 4. If $a=0$, then the integral becomes $\int_{-\infty}^{\infty}\frac{\sin(bx)}{x}dx$, by the criterion of sigular integral, we know that the integral converges.

In particular,

$$\int_{-\infty}^{\infty}\frac{\sin(bx)}{x}dx =\lim_{a\rightarrow0^{+}}\int_{-\infty}^{\infty}e^{-a(x-x_{0})^{2}}\frac{\sin(bx)}{x}dx$$

Set $a\rightarrow0^{+}$ in case 3, we can obtain that $$\int_{-\infty}^{\infty}\frac{\sin(bx)}{x}dx =\sqrt{\pi}\int_{0}^{\infty}e^{-\frac{t^{2}}{4}}dt =\sqrt{\pi}\Gamma(\frac{1}{2})=\pi$$

Answer 3

Another proof using Parseval's theorem. $\int f \cdot g^* \mathrm{d}t = \int F G^*\mathrm{d}\omega$

\begin{equation} \begin{aligned} \int_\mathbb{R} \frac{\sin(b x)}{b x} \cdot e^{-a x^2} \mathrm{d}x =& \int_\mathbb{R} \mathcal{F}\left[\frac{\sin(b x)}{b x}\right](\omega) \cdot \left(\mathcal{F}[e^{-a x^2}](\omega)\right)^* \mathrm{d}\omega \\ =& \int_\mathbb{R} \frac{\pi}{b} \mathrm{rect}\left(\frac{\pi}{b}\omega \right) \sqrt{\frac{\pi}{a}}e^{-\frac{\pi^2}{a}\omega^2} \mathrm{d}\omega \\ =& \frac{ \pi^{\frac{3}{2}} }{ b\sqrt{a}} \int_{-\frac{b}{2\pi}}^{\frac{b}{2\pi}} e^{-\frac{\pi^2}{a}\omega^2} \mathrm{d}\omega \\ =& \frac{\pi}{b}\cdot \frac{1}{\sqrt{\pi}} \int_{-\frac{b}{2\sqrt{a}}}^{\frac{b}{2\sqrt{a}}} e^{-\xi^2} \mathrm{d}\xi = \frac{\pi}{b}\mathrm{erf}\left(\frac{b}{2\sqrt{a}}\right) \end{aligned} \end{equation}