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How do I show that $$\int_{-\infty}^\infty \frac{\sin^2(\pi s)}{(\pi s)^2} e^{2\pi isx} \, ds = \begin{cases} 1+x & \text{if }-1 \le x \le 0 \\ 1-x & \text{if }0 \le x \le 1 \\ 0 & \text{otherwise} \end{cases}$$

I know that $\sin^2(\pi s)=\frac{1-\cos(2\pi s)}{2}=\frac{1-(e^{2\pi i s}-e^{-2\pi i s}))/2}{2}$, so $$\int_{-\infty}^\infty \frac{\sin^2(\pi s)}{(\pi s)^2} e^{2\pi isx} \, ds=2\int_0^\infty \frac{2e^{2\pi isx}-(e^{2\pi is(1+x)}+e^{2\pi i s(-1+x)})}{4\pi^2s^2} \, ds$$

I am also allowed to use the known identity $$\int_{-\infty}^\infty \frac{1-\cos(a \pi x)}{(\pi x)^2} \, dx = |a|$$ for some real number $a$.

user321401
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  • Probably easiest to go in the other direction, and then invoke Fourier inversion. – paul garrett Apr 03 '16 at 18:15
  • Too bad you can't just take the route of convolving two rectangular pulses in the frequency domain… – MonadBoy Apr 03 '16 at 18:18
  • I presume it's homework, then? Ok, your rearrangement is very close to allowing invocation of the "known identity": you can adjust by constants to get a linear combination of cosines, as you might anticipate. – paul garrett Apr 03 '16 at 18:18
  • As @A.Sh has said in a very laconic style, I think that this issue should be taken with much less calculations : we are faced to the Fourier transform of a product. We know that the F.T. of the Cardinal Sine is the characteristic function $\mathbb{1}([-a,a])$ of an interval [-a,a] (or "rectangular pulse"). Thus, the result is the convolution of $\mathbb{1}([-a,a])$ with itself, which is known to be triangular function. – Jean Marie Apr 03 '16 at 19:11

2 Answers2

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Note that we can write

$$\begin{align} \int_{-\infty}^\infty\frac{\sin^2(\pi s)}{(\pi s)^2}e^{i2\pi sx}\,ds&=\frac12\int_{-\infty}^\infty\frac{1-\cos(2\pi s)}{(\pi s)^2}e^{i2\pi sx}\,ds\\\\ &=\int_0^\infty \frac{1-\cos(2\pi s)}{(\pi s)^2}\,\cos(2\pi sx)\,ds\\\\ &=\int_0^\infty \frac{\cos(2\pi sx)-\frac12\left(\cos(2\pi s(x+1))+\cos(2\pi s(x-1))\right)}{(\pi s)^2}\,ds\\\\ &=\int_0^\infty \frac{\cos(2\pi sx)-1}{(\pi s)^2}\,ds\\\\ &+\frac12\int_0^\infty \frac{1-\cos(2\pi s(x+1))}{(\pi s)^2}\,ds\\\\ &+\frac12\int_0^\infty \frac{1-\cos(2\pi s(x-1))}{(\pi s)^2}\,ds\\\\ &=-|x|+\frac12|x+1|+\frac12|x-1| \end{align}$$

Mark Viola
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  • In your second step, how did you justify that $\frac 12 e^{2\pi i sx}=\cos(2\pi sx)$? – user321401 Apr 03 '16 at 18:34
  • We used Euler's Formula $e^{i2\pi sx}=\cos(2\pi sx)+i\sin(2\pi sx)$ and exploited the fact that the integral of an odd function over symmetric limits is $0$. – Mark Viola Apr 03 '16 at 18:36
  • My bad, the "known identity" is the integral over $(-\infty,\infty)$, but I made a typo by printing the integral over $(0,\infty)$ in my question. (I have now fixed this.) I think this error made its way into your work, so perhaps the $\frac 12$'s don't appear after all. (btw, I haven't finished my work yet, so I don't know if the $\frac 12$'s are still there or not; I was trying to look at your answer as little as possible.) – user321401 Apr 03 '16 at 20:47
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    This answer is actually correct using $$\int_0^\infty \frac{1-\cos(2\pi a x)}{(\pi x)^2},dx=|a|$$ – Mark Viola Apr 03 '16 at 21:36
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We can prove this formally without using the given identity; just using the various Fourier transform identities.

First consider, \begin{align*} \mathcal F^{-1}(\sin(\pi s)) &= \frac{1}{2i} \int_{\mathbb R} (e^{i\pi s} - e^{-i\pi s})e^{2\pi i xs} ds \\ &= \frac{1}{2i} \int_{\mathbb R} e^{2i\pi s\left(x + \tfrac 1 2 \right)} + e^{2i\pi s\left(x - \tfrac 1 2 \right)} \\ &= \frac{1}{2i}\left[\delta\left(x + \tfrac 1 2 \right) - \delta\left(x-\tfrac 1 2 \right) \right]. \end{align*} Then \begin{align*}\mathcal F^{-1}(\sin (\pi s)) &= \mathcal \pi F^{-1}\left(s \frac 1 {\pi s} \sin(\pi s) \right) \\ &= -\frac{1}{2 i} \frac{d}{dx} \mathcal F^{-1}\left(\frac{\sin(\pi s)}{\pi s}\right). \end{align*} Then \begin{align*}F^{-1}\left(\frac{\sin(\pi s)}{\pi s} \right) &= -2i \int^x_{-\infty}[F^{-1}(\sin (\pi s))](y) dy \\ &=\int^x_{-\infty} \left[\delta\left( y - \tfrac 1 2 \right) - \delta\left(y+\tfrac 1 2 \right) \right] dy = \left\{ \begin{matrix} 1, & x \in \left(-\tfrac 1 2, \tfrac 1 2\right) \\ 0, & \text{otherwise}\end{matrix} \right. \end{align*} Finally, \begin{align*}\mathcal F^{-1}\left( \frac{\sin^2(\pi s)}{(\pi s)^2} \right) &= F^{-1}\left(\frac{\sin(\pi s)}{\pi s} \right) * F^{-1}\left(\frac{\sin(\pi s)}{\pi s} \right) \\ &= \int_{\mathbb R} \chi_{(-1/2,1/2)}(y) \chi_{(-1/2,1/2)}(x-y) dy \\ &= \int^{1/2}_{-1/2} \chi_{(x-1/2, x+1/2)}(y) dy. \end{align*} where $\chi_A$ denotes the indicator function on the set $A$. If $\lvert x \rvert > 1$, then the function is zero over the range of integration, so the integral is zero. For $x \in (-1,0)$, we have $$\int^{1/2}_{-1/2} \chi_{(x-1/2, x+1/2)}(y) dy = \int^{1/2 + x}_{-1/2} dy = 1+x$$ and for $x \in (0,1)$, we get $$\int^{1/2}_{-1/2} \chi_{(x-1/2, x+1/2)}(y) dy = \int^{1/2}_{-1/2+x} dx = 1-x.$$ Thus $$\mathcal F^{-1}\left( \frac{\sin^2(\pi s)}{(\pi s)^2} \right) = \left \{\begin{matrix} 0, & \lvert x \rvert > 1 \\ 1+x, & x \in (-1,0), \\ 1-x, & x \in (0,1). \end{matrix}\right.$$

User8128
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