May 9, 2026
In my Pi Squared Over Six post, we proved the Basel problem using Fourier series and Parseval's theorem. We took a periodic square wave, computed its Fourier coefficients, and applied Parseval to show that \(\sum 1/n^2 = \pi^2/6\). The whole thing boiled down to counting energy across discrete harmonics.
That was the discrete world: periodic signals, integer-indexed frequencies, and a sum of squared coefficients. In this post, we move to the continuous world. We replace the Fourier series with the Fourier transform, replace sums with integrals, and replace discrete harmonics with a continuous spectrum. The reward? An entire family of improper integrals that look terrifying on paper:
\[ \int_0^\infty \frac{\sin x}{x}\,dx, \quad \int_0^\infty \left(\frac{\sin x}{x}\right)^2 dx, \quad \int_0^\infty \left(\frac{\sin x}{x}\right)^3 dx, \quad \dots \]
Each one of these turns out to be a simple geometric calculation involving rectangles and triangles. No residues, no contour integration, no differentiating under the integral sign. Let's see how.
We use the following Fourier transform convention throughout this post:
\[ X(f) = \int_{-\infty}^{\infty} x(t)\,e^{-j2\pi ft}\,dt, \qquad x(t) = \int_{-\infty}^{\infty} X(f)\,e^{j2\pi ft}\,df \]
Now define the unit rectangle function:
\[ \text{rect}(t) = \begin{cases} 1, & |t| < \tfrac{1}{2} \\ 0, & \text{otherwise} \end{cases} \]
Let's compute its Fourier transform directly:
\begin{align} X(f) &= \int_{-1/2}^{1/2} e^{-j2\pi ft}\,dt \\ \\ &= \left[\frac{e^{-j2\pi ft}}{-j2\pi f}\right]_{-1/2}^{1/2} \\ \\ &= \frac{e^{j\pi f} - e^{-j\pi f}}{j2\pi f} \\ \\ &= \frac{\sin(\pi f)}{\pi f} \end{align}
This is the normalized sinc function:
\[ \text{sinc}(f) = \frac{\sin(\pi f)}{\pi f} \]
We now have the Fourier pair that will serve as our main tool:
\[ \boxed{\text{rect}(t) \longleftrightarrow \text{sinc}(f)} \]
A rectangle in time becomes a sinc in frequency. Everything that follows is built on top of this one pair.
The inverse Fourier transform, evaluated at \(t = 0\), gives us:
\[ x(0) = \int_{-\infty}^{\infty} X(f)\,df \]
For our rectangle, \(x(0) = \text{rect}(0) = 1\). Therefore:
\[ 1 = \int_{-\infty}^{\infty} \text{sinc}(f)\,df = \int_{-\infty}^{\infty} \frac{\sin(\pi f)}{\pi f}\,df \]
Substitute \(x = \pi f\), so \(df = dx/\pi\):
\[ 1 = \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{\sin x}{x}\,dx \]
Therefore:
\[ \int_{-\infty}^{\infty} \frac{\sin x}{x}\,dx = \pi \]
Since \(\sin x / x\) is an even function:
\[ \boxed{\int_0^\infty \frac{\sin x}{x}\,dx = \frac{\pi}{2}} \]
Think about what just happened. That integral is notorious in calculus courses precisely because the antiderivative of \(\sin x / x\) cannot be expressed in closed form. Students are told to "just accept the result" or use contour integration from complex analysis. But from our perspective, the entire integral reduces to a single question: what is the height of the rectangle at zero? The answer is 1, and the \(\pi\) comes from the substitution. We never had to wrestle with the oscillation at all.
For the squared sinc, we bring in Parseval's theorem. Recall from the Basel post that Parseval equates time-domain energy with frequency-domain energy. In the Fourier transform setting, the statement is:
\[ \int_{-\infty}^{\infty} |x(t)|^2\,dt = \int_{-\infty}^{\infty} |X(f)|^2\,df \]
For the unit rectangle, the left side is immediate:
\[ \int_{-\infty}^{\infty} \text{rect}^2(t)\,dt = \int_{-1/2}^{1/2} 1\,dt = 1 \]
Since \(X(f) = \text{sinc}(f)\), the right side gives us:
\[ \int_{-\infty}^{\infty} \text{sinc}^2(f)\,df = 1 \]
Converting back to the standard calculus variable via \(x = \pi f\):
\[ \int_{-\infty}^{\infty} \left(\frac{\sin x}{x}\right)^2 dx = \pi \]
And for the half-line:
\[ \boxed{\int_0^\infty \left(\frac{\sin x}{x}\right)^2 dx = \frac{\pi}{2}} \]
This is the continuous-domain sibling of our Basel calculation. In the Basel post, we summed \(|c_n|^2\) over integers and got \(\pi^2/6\). Here, we integrate \(|\text{sinc}(f)|^2\) over the real line and get 1 (which becomes \(\pi\) after the substitution). Same philosophy: the energy of a simple shape in time tells us the integral of its complicated-looking spectrum.
Parseval gave us \(\text{sinc}^2\) for free. But how do we handle \(\text{sinc}^3\), \(\text{sinc}^4\), and beyond? For that, I want to introduce a different tool that will handle all higher powers in a uniform way. This tool is convolution.
Let's build up to it starting from discrete time. Given two sequences \(x[n]\) and \(h[n]\), their convolution is:
\[ y[n] = \sum_{k=0}^{\infty} x[k]\,h[n - k] \]
or equivalently:
\[ y[n] = \sum_{k=0}^{\infty} x[n - k]\,h[k] \]
What does this formula actually do? You take the sequence \(h\), flip it in time (replace \(k\) with \(-k\)), then slide the flipped version across \(x\) one position at a time. At each position \(n\), you multiply the overlapping samples together and sum them up. The result is \(y[n]\).
Where does this formula come from? It comes from two properties that most well-behaved systems satisfy: linearity and time-invariance.
A system is linear if scaling the input scales the output by the same factor, and if the sum of two inputs produces the sum of their individual outputs. A system is time-invariant if delaying the input by \(k\) samples simply delays the output by \(k\) samples, without changing its shape. A system satisfying both properties is called LTI (linear time-invariant).
Now here is the key insight. Any discrete input \(x[n]\) can be decomposed as a sum of shifted impulses:
\[ x[n] = \sum_{k=0}^{\infty} x[k]\,\delta[n - k] \]
where \(\delta[n]\) is the unit impulse (1 at \(n = 0\), zero elsewhere). If we feed a single impulse \(\delta[n]\) into our LTI system, we get some output. Call that output \(h[n]\), the impulse response. Now, by time-invariance, a delayed impulse \(\delta[n - k]\) produces a delayed copy of the impulse response: \(h[n - k]\). And by linearity, the response to the full input \(x[n] = \sum_k x[k]\,\delta[n-k]\) is just the weighted sum of those delayed copies:
\[ y[n] = \sum_{k=0}^{\infty} x[k]\,h[n - k] \]
This is convolution. It was not invented out of thin air; it is the inevitable consequence of linearity and time-invariance.
A finite impulse response (FIR) filter is simply an LTI system whose impulse response \(h[n]\) has finite length. If \(h[n]\) is nonzero only for \(n = 0, 1, \dots, M\), with values \(b_0, b_1, \dots, b_M\), then the output becomes:
\[ y[n] = \sum_{k=0}^{M} b_k\,x[n-k] \]
The output at each time step is a weighted sliding window over the input. The coefficients \(b_k\) are the taps of the filter, and they are the impulse response. Convolution is nothing more than "flip, slide, multiply, add."
The continuous version replaces the sum with an integral:
\[ (f * g)(t) = \int_{-\infty}^{\infty} f(\tau)\,g(t - \tau)\,d\tau \]
Same idea: flip one function, slide it by \(t\), multiply the overlapping portions, and integrate the product. Instead of counting discrete samples, we measure overlap area.
Now, why does this help us with sinc integrals? Because of the convolution theorem: multiplication in the frequency domain corresponds to convolution in the time domain. Let's prove it. If \(x(t) \leftrightarrow X(f)\) and \(y(t) \leftrightarrow Y(f)\), we want to show that the inverse Fourier transform of \(X(f) \cdot Y(f)\) is \((x * y)(t)\).
Start with the inverse transform of the product:
\begin{align} \mathcal{F}^{-1}\{X(f)Y(f)\} &= \int_{-\infty}^{\infty} X(f)\,Y(f)\,e^{j2\pi ft}\,df \\ \\ &= \int_{-\infty}^{\infty} \left[\int_{-\infty}^{\infty} x(\tau)\,e^{-j2\pi f\tau}\,d\tau\right] Y(f)\,e^{j2\pi ft}\,df \\ \\ &= \int_{-\infty}^{\infty} x(\tau) \left[\int_{-\infty}^{\infty} Y(f)\,e^{j2\pi f(t - \tau)}\,df\right] d\tau \\ \\ &= \int_{-\infty}^{\infty} x(\tau)\,y(t - \tau)\,d\tau \\ \\ &= (x * y)(t) \end{align}
The key step was swapping the order of integration in the middle (moving the \(d\tau\) integral outside and the \(df\) integral inside). This is the kind of move that makes a pure mathematician reach for their Fubini's theorem, but as engineers we just do it and keep moving. The inner integral in the fourth line is simply the inverse Fourier transform of \(Y(f)\) evaluated at \(t - \tau\), which is \(y(t - \tau)\). And there it is:
\[ \boxed{X(f) \cdot Y(f) \longleftrightarrow (x * y)(t)} \]
(If you noticed that this proof has the same structure as our Parseval proof from the Basel post, you are absolutely right. Both proofs amount to substituting the Fourier representation, swapping the order of summation/integration, and exploiting orthogonality or the inverse transform. Same playbook.)
This means that \(\text{sinc}^2(f) = \text{sinc}(f) \cdot \text{sinc}(f)\) corresponds to \(\text{rect} * \text{rect}\) in the time domain. And integrating a spectrum over all frequencies is equivalent to evaluating its inverse transform at \(t = 0\). So every sinc power integral becomes: convolve some shapes, then read off the result at zero. Let's see this in action.
Since \(\text{sinc}^2(f) \leftrightarrow \text{rect} * \text{rect}\), let's compute that convolution. We slide one copy of the rectangle across another and measure the overlap at each shift. When the two rectangles fully overlap (at zero shift), the overlap area is 1. As we slide further apart, the overlap decreases linearly until the rectangles no longer touch. The result is the triangle function:
\[ \Lambda(t) = \begin{cases} 1 - |t|, & |t| \le 1 \\ 0, & \text{otherwise} \end{cases} \]
A rectangle slides over a fixed rectangle. The purple overlap area traces out the triangle \(\Lambda(t)\). Peak overlap is 1 at zero shift.
Now, integrating \(\text{sinc}^2(f)\) over all \(f\) is equivalent to evaluating the inverse transform at zero:
\[ \int_{-\infty}^{\infty} \text{sinc}^2(f)\,df = (\text{rect} * \text{rect})(0) = \Lambda(0) = 1 \]
Same answer as before, arrived at differently. The integral of sinc squared equals the peak of the triangle, which is 1. After the \(\pi\) scaling, we again obtain \(\pi/2\) on the half-line.
Here is where convolution really shines, because Parseval alone cannot easily handle odd powers. We write:
\[ \text{sinc}^3(f) = \text{sinc}(f) \cdot \text{sinc}^2(f) \]
By the convolution theorem, the corresponding time-domain function is:
\[ \text{sinc}^3(f) \longleftrightarrow \text{rect} * \Lambda \]
So:
\[ \int_{-\infty}^{\infty} \text{sinc}^3(f)\,df = (\text{rect} * \Lambda)(0) \]
This is the overlap of the unit rectangle (which lives on \([-1/2, 1/2]\)) with the triangle function, evaluated at zero shift:
\begin{align} (\text{rect} * \Lambda)(0) &= \int_{-1/2}^{1/2} \Lambda(\tau)\,d\tau \\ \\ &= \int_{-1/2}^{1/2} (1 - |\tau|)\,d\tau \\ \\ &= 2\int_0^{1/2} (1 - \tau)\,d\tau \\ \\ &= 2\left[\tau - \frac{\tau^2}{2}\right]_0^{1/2} \\ \\ &= 2\left(\frac{1}{2} - \frac{1}{8}\right) = \frac{3}{4} \end{align}
A rectangle slides across the triangle. At zero shift, the overlap area is exactly 3/4.
Converting to the standard form (multiply by \(\pi\) from the substitution):
\[ \int_{-\infty}^{\infty} \left(\frac{\sin x}{x}\right)^3 dx = \frac{3\pi}{4} \]
And for the half-line (the integrand is even, so we simply halve):
\[ \boxed{\int_0^\infty \left(\frac{\sin x}{x}\right)^3 dx = \frac{3\pi}{8}} \]
The integral of sinc cubed is the area where a rectangle sits on top of a triangle. We just had to compute the area of the triangle between \(-1/2\) and \(1/2\). That's all the "calculus" that was needed.
For the fourth power, we have two natural routes that both lead to the same answer.
Since \(\Lambda(t) \leftrightarrow \text{sinc}^2(f)\), Parseval gives us:
\[ \int_{-\infty}^{\infty} \text{sinc}^4(f)\,df = \int_{-\infty}^{\infty} |\text{sinc}^2(f)|^2\,df = \int_{-\infty}^{\infty} \Lambda^2(t)\,dt \]
This is the energy of the triangle function:
\begin{align} \int_{-1}^{1} (1 - |t|)^2\,dt &= 2\int_0^1 (1 - t)^2\,dt \\ \\ &= 2\left[-\frac{(1-t)^3}{3}\right]_0^1 \\ \\ &= \frac{2}{3} \end{align}
Alternatively, \(\text{sinc}^4(f) = \text{sinc}^2(f) \cdot \text{sinc}^2(f)\), so:
\[ \text{sinc}^4(f) \longleftrightarrow \Lambda * \Lambda \]
Evaluating at zero:
\[ (\Lambda * \Lambda)(0) = \int_{-1}^{1} \Lambda(\tau)\,\Lambda(-\tau)\,d\tau = \int_{-1}^{1} (1 - |\tau|)^2\,d\tau = \frac{2}{3} \]
A word of caution: this is the integral of the pointwise product \(\Lambda(\tau) \cdot \Lambda(-\tau)\) over their common support. It is NOT the product of their individual areas (the triangle has area 1, and \(1 \times 1 = 1 \ne 2/3\)). Convolution at zero is an overlap integral, not a product of areas.
Two triangles overlap. At zero shift, the integral of their pointwise product is 2/3.
Both routes yield the same result. Converting:
\[ \int_{-\infty}^{\infty} \left(\frac{\sin x}{x}\right)^4 dx = \frac{2\pi}{3} \]
\[ \boxed{\int_0^\infty \left(\frac{\sin x}{x}\right)^4 dx = \frac{\pi}{3}} \]
| Integral (full line) | Geometry | Value | Half-line value |
|---|---|---|---|
| \(\displaystyle\int_{-\infty}^{\infty} \frac{\sin x}{x}\,dx\) | Rectangle value at zero | \(\pi\) | \(\pi/2\) |
| \(\displaystyle\int_{-\infty}^{\infty} \left(\frac{\sin x}{x}\right)^2 dx\) | Rectangle energy / Triangle peak | \(\pi\) | \(\pi/2\) |
| \(\displaystyle\int_{-\infty}^{\infty} \left(\frac{\sin x}{x}\right)^3 dx\) | Rectangle-triangle overlap at zero | \(3\pi/4\) | \(3\pi/8\) |
| \(\displaystyle\int_{-\infty}^{\infty} \left(\frac{\sin x}{x}\right)^4 dx\) | Triangle energy / Triangle-triangle overlap | \(2\pi/3\) | \(\pi/3\) |
Every single one of these integrals reduces to measuring something about rectangles and triangles. No contour integration, no residues, no tricks with differentiating under the integral sign. The Fourier transform converts the oscillatory nightmare into flat shapes with straight edges, and we just measure areas and heights.
In the Basel post, we discovered that \(\pi^2/6\) was hiding inside the energy of a square wave. Now we've seen that an entire family of sinc integrals was hiding inside the geometry of rectangles and triangles in the time domain. The pattern is the same: Fourier analysis lets us trade a difficult calculation in one domain for a trivial one in the other.
The lesson, I think, is that calculus itself is not the bottleneck. The bottleneck is perspective. When you stare at \(\int_0^\infty (\sin x / x)^3\,dx\), you see an oscillating integrand that decays slowly to zero. But once you adopt the Fourier viewpoint, you see a rectangle sitting on top of a triangle, and the answer is the overlap area at a single position. The right basis turns a hard problem into something you can sketch on the back of an envelope.
Basel came from orthogonality and energy in Fourier series. These sinc integrals come from the same idea, except now the harmonics form a continuum rather than a discrete set. One framework, two domains, and a surprising amount of mileage from the humble rectangle.