# Evaluating the integral $\int_{-\infty}^\infty \frac{\sin^2(x)}{x^2}e^{i t x} dx$

I want to evaluate the integral

$$\int_{-\infty}^\infty \frac{\sin^2(x)}{x^2}e^{i t x} dx$$

for all $t \in \mathbb{R}$. I would preferably do it using the tools of complex analysis, but since I haven't found any other thread dealing with this integral at all so far, I'd be open to any approach.

I must admit that I haven't really found a way to start. The function $f(z) := \frac{\sin^2(z)}{z^2}e^{i t z}$ has an isolated singularity (a removable singularity, with continuation $f(0) = 1$, to be precice) at $z = 0$. I think a clever way would be to integrate over the edge of rectangle, semicircle or "slice of cake" (the edges of which we then would let go against $-∞$ and $∞$), then maybe somehow apply the Cauchy theorem or Residue theorem. But I don't know how to concretely do that, or how to even get started with that.

On a more general note, I'm having trouble understanding how to transform a "normal" integral over a real interval into one where we integrate over the edges of a geometric shape; and I also don't know, based on which pattern one can see with geometric shape fits best. Are there any tips or hints on how to get more intuition for that? (Apart from, obviously, dealing with these integrals regularly, what I'm trying to do but where I'm still failing a lot of times.)

I also noticed that the above integrals looks strikingly similar to the Fourier transformed of $f$ (or, even more similar to the formula for the inverse transformed). So maybe this could help us with this integral? I'm not really sure though.

• Unless I am overlooking something, $f$ has a removable singularity at $z=0$, not a pole. – Martin R Jan 13 '16 at 19:03
• @MartinR Thanks, you're right. Corrected it. – moran Jan 13 '16 at 19:08
• I think you may find this useful math.stackexchange.com/questions/861407/… – Brightsun Jan 13 '16 at 19:08
• Note that this is the inverse Fourier transform of the square of the sinc function, which can be written as a convolution of step functions. – user296602 Jan 13 '16 at 19:14

It is well-known (and not difficult to check) that the Fourier transform $\hat{f}$ of the function $f(t) := 1_{[-1,1]}(t)$ equals

$$\hat{f}(x) = \frac{\sin x}{x}$$

(up to a multiplicative constant depending on the definition of the Fourier transform). Since the Fourier transform of a convolution of two functions equals the product of the Fourier transforms, we have

$$\widehat{f \ast f}(x) = \left( \frac{\sin x}{x} \right)^2.$$

Now, finally, the integral from your problem is essentially an inverse Fourier transform, so I would expect that the integral equals

$$c (f \ast f)(t) = c (1_{[-1,1]} \ast 1_{[-1,1]})(t) = c \int 1_{[-1,1]}(t-s) 1_{[-1,1]}(s) \, ds$$

for a suitable constant $c$ (which, as mentioned above, depends on the definition of the Fourier transform you are using). The latter integral can be calculated explicitly.

We have $$I(t)= \int_{R} \frac{\sin^2x}{x^2} e^{itx} dx = \int_{R} \sin^2x\int_0^\infty e^{-sx^2} ds e^{itx} dx = \int_0^\infty \int_R e^{-sx^2 +itx} \sin^2 x dx ds=\int_0^\infty e^{-t^2/4s} \left(\int_R e^{-s (x-it/2s)^2} \sin^2 x dx \right)ds.$$ Using residue theorem we have $$\int_R e^{-s (x-it/2s)^2} \sin^2 x dx =\int_R e^{-s x^2} \sin^2(x+it/2s) dx= -\int_R e^{-sx^2} \frac{\left(e^{2ix -t/s} + e^{-2ix +t/s} -2\right)}4dx.$$ We have $$\int_R e^{-sx^2+2ix -t/s} dx = e^{-(t+1)/s}\int_R e^{-s(x-i/s)^2 }dx =e^{-(t+1)/s}\int_R e^{-sx^2}dx= \sqrt{\frac\pi{s}}e^{-(t+1)/s},$$ and $$\int_Re^{-sx^2-2ix +t/s} dx =\sqrt{\frac\pi{s}}e^{(t-1)/s}.$$ Hence $$I(t) = -\frac{\sqrt{\pi}}4\int_0^\infty\frac{\left(e^{-(t+2)^2/4s} + e^{-(t-2)^2/4s} -2e^{-t^2/4s}\right)}{\sqrt{s}} ds\quad \quad (*)$$ If $|t|\not =2$, then $$\frac2{\sqrt\pi}I(t) = -\int_0^\infty\frac{\left(e^{-(t+2)^2/4s}-1\right)}{2\sqrt{s}}ds-\int_0^\infty\frac{\left(e^{-(t-2)^2/4s}-1\right)}{2\sqrt{s}}ds+2\int_0^\infty\frac{e^{-t^2/4s}-1}{2\sqrt{s}}ds.$$ If $a >0$, we have by changing the variable and integration by parts $$\int_0^\infty \frac{e^{-a/s}-1}{2\sqrt{s}}ds = \int_0^\infty \frac{e^{-as}-1}{2s^{3/2}} ds=-a\int_0^\infty e^{-as} s^{-1/2}ds= -\sqrt{\pi a}\quad \quad (1).$$ By $(1)$, if $|t|\not =2$ then $$\frac2{\sqrt\pi}I(t) = \sqrt{\pi} \left(\frac{|t+2|}2 +\frac {|t-2|}2 -|t|\right).$$

If $|t|=2$ then by using $(1)$ in $(*)$ we obtain $I(t) =0$.

Finally, $I(t) = 0$ if $|t|\geq 2$, and $I(t) = \frac{\pi}2 (2 -|t|)$ if $|t|<2$.

• Smooth trick! +1! – Brightsun Jan 13 '16 at 23:42

In the spirit of one that knows the answer, you can proceed as follows: let $$f(x) = \begin{cases} 0\ \ \qquad\text{ if }|x|\ge 2\\ 2-|x|\quad\text{ if }|x|<2. \end{cases}$$ Defining the Fourier transform by $$\hat f(\xi)\equiv \int_{-\infty}^{+\infty}f(x)e^{-i\xi x},$$ one can directly compute it: $$\hat f(\xi)=\int_{-2}^{2}(2-|x|)e^{-i\xi x}dx=2\int_{-2}^2e^{-i\xi x}dx+\int_{-2}^2te^{-i\xi x}dx+\int_0^2te^{-itx}dx\\ =\frac{2}{i\xi}\left( e^{2i\xi }-e^{-i2\xi}\right)+i\frac{\partial}{\partial\xi}\int_{-2}^0e^{i\xi x}dx -i\frac{\partial}{\partial \xi}\int_0^2e^{-i\xi x}dx\\ =\frac{2-e^{-i2\xi }-e^{i2\xi}}{\xi^2}=4\frac{\sin^2\xi}{\xi^2},$$ where some algebra has been omitted in the next to last step, and the last step follows by the Euler-De Moivre formula. Finally, one uses the inverse Fourier transform tool $$f(x) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}\hat f(\xi)e^{+i\xi x}d\xi,$$ so that $$\int_{-\infty}^{+\infty}\frac{\sin^2\xi}{\xi^2}e^{+i\xi t}d\xi = \frac{\pi}{2}f(t),$$ for $f$ given above. Notice that, having computed this integral one can give a distributional meaning to otherwise ill-defined integrals such as $$\text{PV}\int_{-\infty}^{+\infty}\frac{\sin^2\xi}{\xi}e^{+i\xi t}d\xi=\frac{\pi}{2i}f'(t)\\ \int_{-\infty}^{+\infty}{\sin^2\xi}e^{+i\xi t}d\xi=-\frac{\pi}{2}f''(t)$$ since $$f'(t) = 2\chi_{[-2,0]}(t)-2\chi_{[0,2]}(t)\\ f''(t)= 2\delta_{-2}(t)-4 \delta_0(t)+2\delta_2(t),$$ where $\delta_a$ is the Dirac distribution concentrated at the point $a$ and $\chi_B$ is the characteristic function of the set $B$.

$\newcommand{\angles}{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}{\overline{#1}} \newcommand{\pars}{\left(\, #1 \,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}{\left\vert\, #1 \,\right\vert}$ With the identity $\ds{{\sin\pars{x} \over x} = \half\int_{-1}^{1}\expo{\pm\ic k x}\,\dd k}$:

\begin{align} \color{#f00}{% \int_{-\infty}^{\infty}{\sin^{2}\pars{x} \over x^{2}}\,\expo{\ic tx}\,\dd x} & = \int_{-\infty}^{\infty}\expo{\ic tx} \pars{\half\int_{-1}^{1}\expo{\ic k x}\,\dd k} \pars{\half\int_{-1}^{1}\expo{-\ic q x}\,\dd q}\dd x \\[3mm] & = {\pi \over 2}\int_{-1}^{1}\int_{-1}^{1}\ \overbrace{% \int_{-\infty}^{\infty}\expo{\ic\pars{t + k - q}x}\,{\dd x \over 2\pi}} ^{\ds{=\ \delta\pars{t + k - q}}}\ \,\dd k\,\dd q \\[3mm] & = {\pi \over 2}\int_{-1}^{1}\int_{-1}^{1}\delta\pars{k - \bracks{q - t}} \,\dd k\,\dd q = \left.{\pi \over 2}\int_{-1}^{1}\dd q\,\right\vert_{\,-1\ <\ q - t\ <\ 1} \\[3mm] & = \left.{\pi \over 2}\int_{-1}^{1}\dd q\,\right\vert_{\,t - 1\ <\ q\ <\ t + 1} = \left\lbrace\begin{array}{lcl} \ds{0} & \mbox{if} & \ds{t < -2} \\[2mm] \ds{{\pi \over 2}\int_{-1}^{t + 1}\dd q} & \mbox{if} & \ds{-1 < t + 1 < 1} \\[2mm] \ds{{\pi \over 2}\int_{t - 1}^{1}\dd q} & \mbox{if} & \ds{-1 < t - 1 < 1} \\[2mm] \ds{0} & \mbox{if} & \ds{1 < t - 1} \end{array}\right. \\[3mm] & = \left\lbrace\begin{array}{lcl} \ds{0} & \mbox{if} & \ds{t < -2} \\[2mm] \ds{{\pi \over 2}\pars{t + 2}} & \mbox{if} & \ds{-2 < t < 0} \\[2mm] \ds{{\pi \over 2}\pars{2 - t}} & \mbox{if} & \ds{0 < t < 2} \\[2mm] \ds{0} & \mbox{if} & \ds{t > 2} \end{array}\right. \\[3mm] & = \color{#f00}{% \Theta\pars{\vphantom{\Large A}2 - \verts{t}}{\pi \over 2} \pars{\vphantom{\Large A}2 - \verts{t}}} \end{align}

$\ds{\Theta}$ is the Heaviside Step function.

Two more solutions. One in the spirit of this answer:

\begin{align} \int_{-\infty}^\infty\frac{\sin^2x}{x^2}\mathrm e^{\mathrm itx}\,\mathrm dx &= \int_{-\infty}^\infty\frac{\sin^2x\cos tx}{x^2}\,\mathrm dx \\ &= \frac14\int_{-\infty}^\infty\frac{\cos(t+2)x-2\cos tx+\cos(t-2)x}{x^2}\,\mathrm dx \\ &= \frac14\int_{-\infty}^\infty\frac{(t+2)\sin(t+2)x-2t\sin tx+(t-2)\sin(t-2)x}x\,\mathrm dx \\ &= \frac\pi4\left(|t+2|-2|t|+|t-2|\right) \\ &= \frac\pi2(2-|t|)\,\Theta(2-|t|)\;. \end{align}

And one in the spirit of this answer:

\begin{align} \int_{-\infty}^\infty \frac{1-\cos ax}{x^2}\mathrm dx &= \int_{-\infty}^\infty\Re\frac{1-\mathrm e^{\mathrm iax}}{x^2}\mathrm dx \\ &= \int_{-\infty}^\infty\Re\frac{1-\mathrm e^{\mathrm iax}+\mathrm iax/(1+x^2)}{x^2}\mathrm dx \\ &= \Re\int_{-\infty}^\infty \frac{1-\mathrm e^{\mathrm iax}+\mathrm iax/(1+x^2)}{x^2}\mathrm dx \\ &=|a|\pi\;, \end{align}

where the last line follows by the residue theorem. Applying this to the second line of the first solution again yields the fourth line of the first solution.