Cosine transform of sinc cubed

Some dude asked to evaluate the following integral:

$$\int_0^{\infty} dt \, \frac{\sin^3{\pi t}}{(\pi t)^3} \cos{u t}$$

I propose to perform a direct computation using Cauchy’s theorem, i.e., extension into the complex plane. I will then verify the solution using the convolution theorem.

DIRECT EVALUATION

Rewrite the integral as

$$\frac12 \int_{-\infty}^{\infty} dt \, \frac{\sin^3{\pi t}}{(\pi t)^3} e^{i u t}$$

Now use $\sin{\pi t} = (e^{i \pi t}-e^{-i \pi t})/(2 i)$ to obtain, as the value of the original integral:

$$\frac{i}{16 \pi^3} \int_{-\infty}^{\infty} dt \, \frac{e^{i (u+3 \pi)t}}{t^3}- i \frac{3}{16 \pi^3}\int_{-\infty}^{\infty} dt \, \frac{e^{i (u+ \pi)t}}{t^3} +\\ i \frac{3}{16 \pi^3}\int_{-\infty}^{\infty} dt \, \frac{e^{i (u- \pi)t}}{t^3} – \frac{i}{16 \pi^3} \int_{-\infty}^{\infty} dt \, \frac{e^{i (u-3 \pi)t}}{t^3} $$

Now evaluate these integrals by considering contour integrals over the complex variable $z$ in the complex plane:

$$\frac{i}{16 \pi^3} \oint_C dz \frac{e^{i (u+3 \pi)z}}{z^3} – i \frac{3}{16 \pi^3}\oint_C dz \frac{e^{i (u+ \pi)z}}{z^3} +\\ i \frac{3}{16 \pi^3}\oint_C dz \frac{e^{i (u- \pi)z}}{z^3} – \frac{i}{16 \pi^3} \oint_C dz \frac{e^{i (u-3 \pi)z}}{z^3} $$

where $C$ is a semicircular contour of radius $R$ with a small semicircular indentation of radius $\epsilon$ about the origin. Whether $C$ extends into the upper or lower half plane depends on the sign of $u+3 \pi$ or $u+\pi$. For example, consider the first integral, which may be rewritten as

$$\oint_C dz \frac{e^{i (u+3 \pi)z}}{z^3} = \int_{-R}^{-\epsilon} dx \frac{e^{i (u+3 \pi) x}}{x^3} + i \epsilon \int_{\pi}^0 d\phi\, e^{i \phi} \frac{e^{i (u+3 \pi) \epsilon e^{i \phi}}}{\epsilon^3 e^{i 3 \phi}} + \\ \int_{\epsilon}^R dx \frac{e^{i (u+3 \pi) x}}{x^3} + i R \int_0^{\pi} d\phi\, e^{i \phi} \frac{e^{i (u+3 \pi) R e^{i \phi}}}{R^3 e^{i 3 \phi}}$$

when $C$ is in the upper half plane. In this case, we consider the conditions under which the fourth integral vanishes as $R \to \infty$. In fact, the integral is bounded by

$$\frac{1}{R^2} \int_0^{\pi} d\phi \,e^{-(u+3 \pi) R \sin{\phi}} \le\frac{2}{R^2} \int_0^{\pi/2} d\phi \, e^{-2 (u+3 \pi) R \phi/\pi} \le \frac{\pi}{R^3 (u+3 \pi)}$$

Note that the integral vanishes as $R \to \infty$ only when $u+3 \pi \gt 0$.

To evaluate the second integral, we must expand the exponential in the numerator to $O\left(\epsilon^2\right)$ (this is analogous to computing a residue). It is only for that order of expansion at which we get a nonzero contribution to the integral as $\epsilon \to 0$:

$$ i \epsilon \int_{\pi}^0 d\phi\, e^{i \phi} \frac{e^{i (u+3 \pi) \epsilon e^{i \phi}}}{\epsilon^3 e^{i 3 \phi}} = i \pi \frac12 (u+3 \pi)^2 $$

Putting this all together and invoking Cauchy’s theorem (i.e., the integral over the closed contour is zero), we get

$$\frac{i}{16 \pi^3} \oint_C dz \frac{e^{i (u+3 \pi)z}}{z^3} = \frac{i}{16 \pi^3}PV\int_{-\infty}^{\infty} dx \frac{e^{i (u+3 \pi)x}}{x^3} – \frac{1}{32\pi^2} (u+3 \pi)^2 = 0$$

where $PV$ denotes the Cauchy principal value of the integral. When $u+3 \pi \lt 0$, the analysis is the same, except the sign is reversed because we must consider $C$ in the lower half-plane. Thus we have

$$\frac{i}{16 \pi^3}PV\int_{-\infty}^{\infty} dx \frac{e^{i (u+3 \pi)x}}{x^3} = \frac{1}{32\pi^2} (u+3 \pi)^2 \text{sgn}(u+3 \pi) $$

where sgn is the signum function (i.e., $\pm 1$ according to the sign of its argument).

It should be clear that the exact same analysis extends to the other three integrals so that we may now simply write down the result we seek:

$$\int_0^{\infty} dt \, \frac{\sin^3{\pi t}}{(\pi t)^3} \cos{u t} = \frac{1}{32 \pi^2} \left [(u+3 \pi)^2 \text{sgn}(u+3 \pi) – (u-3 \pi)^2 \text{sgn}(u-3 \pi) – 3 (u- \pi)^2 \text{sgn}(u- \pi) + 3 (u+ \pi)^2 \text{sgn}(u+ \pi) \right]$$

Note that we no longer need the $PV$ because the integral is defined at $t=0$. Through a little manipulation, we may rewrite this into a more transparent form:

$$\int_0^{\infty} dt \, \frac{\sin^3{\pi t}}{(\pi t)^3} \cos{u t} = \begin{cases} \\ 0 & u \lt -3 \pi \\ \frac{1}{16 \pi^2} (u+3 \pi)^2 & -3 \pi \lt u \lt -\pi \\ \frac{1}{8 \pi^2} (3 \pi^2-u^2) & -\pi \lt u \lt \pi \\ \frac{1}{16 \pi^2} (u-3 \pi)^2 &\pi \lt u \lt 3 \pi \\ 0 & u \gt 3 \pi\end{cases}$$

Here is a plot of the integral over $u$:

sinc3plot

(The gaps are an artifice of Mathematica and are not real.) It is interesting that the $1/t^3$ behavior of the integrand leads to a discontinuity in the second derivative – just as $1/t$ behavior leads to a discontinuity in the function value and $1/t^2$ behavior leads to discontinuity in the derivative.

CONVOLUTION THEOREM

The convolution theorem states that, for functions $f$ and $g$ defined on $L^2(\mathbb{R})$, we have

$$I(u) = \int_{-\infty}^{\infty} dt \, f(t) g(t) \, e^{i u t} = \frac{1}{2 \pi} \int_{-\infty}^{\infty} du’ \, \hat{f}(u’) \hat{g}(u-u’)$$

where $\hat{f}$ and $\hat{g}$ are the Fourier transforms of $f$ and $g$, respectively. The FT of a product of two functions is the convolution of the FTs of the functions.

Here, we have

$$f(t) = \frac{\sin{\pi t}}{\pi t}$$

$$g(t) = \frac{\sin^2{\pi t}}{(\pi t)^2}$$

I will assume that the FFTs of these functions are known; that is

$$\hat{f}(u) = \begin{cases} \\ 1 & |u| \lt \pi \\ 0 & |u| > \pi \end{cases}$$

$$\hat{g}(u) = \begin{cases} \\ 1-\frac{|u|}{2 \pi} & |u| \lt 2 \pi \\ 0 & |u| > 2 \pi \end{cases}$$

If the reader so desires, the reader may derive these in a manner similar to the contour integration above. Note that the ultimate answer will be $1/2$ of this convolution, as the original integral is over $[0,\infty)$.

The problem of evaluating the integral above reduces to evaluating the convolution of two very simple functions having finite support. The following picture illustrates what needs to be done:

conv

Here, the rectangle (i.e. $\hat{f}(u-u’)$) moves from left to right along the $u’$ axis, while $\hat{g}(u’)$ stays fixed. The convolution at a particular value of $u$ is the overlap area. The form of this overlap area will vary depending on the value of $u$. For example, it should be clear that the overlap area is zero when $|u| \gt 3 \pi$. (The near edges of the rectangle lie outside the triangle.)

Now, when $u \in [-3 \pi,-\pi]$, the overlap area is a right triangle extending from the left vertex of the triangle to the right side of the rectangle (see the figure above). That is,

$$\begin{align}I(u) &= \frac{1}{4 \pi} \int_{-2 \pi}^{u+\pi} du’ \left ( 1+ \frac{u’}{2 \pi}\right )\\ &= \frac{u+3 \pi}{4 \pi} + \frac{(u+\pi)^2-4 \pi^2}{16\pi^2}\\&= \frac{1}{16 \pi^2} (u+3 \pi)^2 \end{align} $$

When $u \in [\pi, 3 \pi]$, the situation is the mirror image:

$$\begin{align}I(u) &= \frac{1}{4 \pi} \int_{u- \pi}^{2\pi} du’ \left ( 1- \frac{u’}{2 \pi}\right )\\ &= \frac{u-3 \pi}{4 \pi} – \frac{4 \pi^2-(u-\pi)^2}{16\pi^2}\\&= \frac{1}{16 \pi^2} (u-3 \pi)^2 \end{align} $$

Finally, when $u \in [-\pi,\pi]$, the overlap area is bounded on both sides by the rectangle:

$$\begin{align} I(u) &=\frac{1}{4 \pi}\int_{u-\pi}^0 du’ \left ( 1+\frac{u’}{2 \pi}\right) +\frac{1}{4 \pi} \int_0^{u+\pi} du’ \left ( 1-\frac{u’}{2 \pi}\right) \\ &= \frac{\pi-u}{4 \pi} – \frac{(u-\pi)^2}{8 \pi^2} + \frac{u+\pi}{4 \pi} – \frac{(u+\pi)^2}{8 \pi^2} \\ &= \frac{3 \pi^2-u^2}{8 \pi^2}\end{align}$$

The reader should see that the final result here matches that from the direct evaluation above.

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