Convolution is fun!

Given two functions f and g, their convolution is

\displaystyle (f*g)(t) = \int_{-\infty}^{\infty} f(\tau) g(t - \tau) \, \mathrm{d} \tau.

Let us choose f(x) = x and g(x) = \frac{1}{1 + x^2}. Now we have that

\displaystyle (f * g)(t) = \int_{-\infty}^{\infty} \frac{\tau}{1 + (t - \tau)^2} \, \mathrm{d}\tau.

Let us set x = t - \tau, which implies that \tau = t - x, \frac{\mathrm{d}x}{\mathrm{d}\tau} = -1, and so \mathrm{d}\tau = -\mathrm{d}x. Also, as \tau \to \infty, x \to -\infty and vice versa. Finally, we can integrate:

\displaystyle \begin{aligned} (f*g)(t) &= \int_{\infty}^{-\infty} - \frac{t - x}{1 + x^2} \, \mathrm{d}x \\          &= \int_{-\infty}^{\infty} \frac{t - x}{1 + x^2} \, \mathrm{d}x \\          &= \int_{-\infty}^{\infty} \frac{t}{1 + x^2} - \frac{x}{1 + x^2} \, \mathrm{d}x \\          &= \Bigg[ t \arctan x \Bigg]_{x = -\infty}^{x = \infty} - \Bigg[  \frac{1}{2} \ln |x^2 + 1| \Bigg]_{x = -\infty}^{x = \infty} \\          &= t \Bigg[ \frac{\pi}{2} - \Bigg( - \frac{\pi}{2} \Bigg) \Bigg] - \lim_{x \to \infty} \Bigg[ \frac{1}{2} \ln |x^2 + 1| - \frac{1}{2} \ln |(-x)^2 + 1| \Bigg] \\          &= \pi t - \lim_{x \to \infty} [0] \\          &= \pi t. \end{aligned}

Now, what happens if we compute

\displaystyle (g * f)(t) = \int_{-\infty}^{\infty} f(t - \tau) g(\tau) \, \mathrm{d}\tau

instead? Let us take a look:

\displaystyle \begin{aligned} (g*f)(t) &= \int_{-\infty}^{\infty} \frac{t - \tau}{1 + \tau^2} \, \mathrm{d} \tau \\          &= \int_{-\infty}^{\infty} \frac{t}{1 + \tau^2} - \frac{\tau}{1 + \tau} \, \mathrm{d} \tau \\          &= \Bigg[ t \arctan \tau \Bigg]_{\tau = -\infty}^{\tau = \infty} -  \Bigg[ \frac{1}{2} \ln |1 + \tau^2| \Bigg]_{\tau = -\infty}^{\tau = \infty} \\          &= t \Bigg[ \frac{\pi}{2} - \Bigg( - \frac{\pi}{2} \Bigg) \Bigg] - \lim_{\tau \to \infty} \Bigg[ \frac{1}{2} \ln |1 + \tau^2| - \frac{1}{2} \ln |1 + (-\tau)^2|\Bigg] \\          &= \pi t - \lim_{\tau \to \infty} [0] \\          &= \pi t. \end{aligned}

This is not a coincidence, since convolution commutes. To see that, set x  = t - \tau, which implies \tau = t - x, \mathrm{d}\tau = -\mathrm{d}x. Note also that as \tau \to \infty, x \to -\infty. Now,

\displaystyle \begin{aligned} (f*g)(t) &= \int_{\infty}^{-\infty} -f(t-x) g(x) \, \mathrm{d}x \\          &= \int_{-\infty}^{\infty} f(t-x) g(x) \, \mathrm{d} x, \end{aligned}

which is (g*f)(t).

However, there exist pairs of functions for which convolution does not converge. Let us demonstrate an example. We will reuse g, yet redefine f to be f(x) = x^2. Here we go:

\displaystyle \begin{aligned} (f*g)(t) &= \int_{-\infty}^{\infty} f(\tau)g(t - \tau) \, \mathrm{d} \tau \\          &= \int_{-\infty}^{\infty} \frac{\tau^2}{1 + (t - \tau)^2} \, \mathrm{d} \tau. \end{aligned}

Setting t - \tau = x, we have that \tau = t - x, \frac{\mathrm{d}\tau}{\mathrm{d}x} = -1, and so \mathrm{d}\tau = -\mathrm{d}x. Once again, as \tau \to \infty, x \to -\infty, and vice versa. Finally

\displaystyle  \begin{aligned} \int_{\infty}^{-\infty} - \frac{(t - x)^2}{1 + x^2} \, \mathrm{d}x &=  \int_{-\infty}^{\infty} \frac{(t - x)^2}{1 + x^2} \, \mathrm{d}x \\ &= \int_{-\infty}^{\infty} \frac{t^2 - 2tx + x^2}{1 + x^2} \, \mathrm{d}x \\ &= \int_{-\infty}^{\infty} \frac{t^2}{ 1 + x^2 } \, \mathrm{d}x - \int_{-\infty}^{\infty} \frac{ 2tx }{ 1 + x^2 } \, \mathrm{d}x + \int_{-\infty}^{\infty} \frac{x^2 }{1 + x^2 } \, \mathrm{d}x  \\ &= \pi t^2 - \Bigg[ t \ln |1 + x^2| \Bigg]_{x = -\infty}^{x = \infty} + \int_{-\infty}^{\infty} \frac{x^2}{1 + x^2} \, \mathrm{d}x \\ &= \pi t^2 - \lim_{x \to \infty} \Bigg[ t \ln |1 + x^2| - t \ln |1 + (-x)^2| \Bigg] + \int_{-\infty}^{\infty} \frac{x^2}{1 + x^2} \, \mathrm{d}x \\ &= \pi t^2 = \lim_{x \to \infty} [0] + \int_{-\infty}^{\infty} \frac{x^2}{1 + x^2} \, \mathrm{d}x \\ &= \pi t^2 + \int_{-\infty}^{\infty} \frac{x^2}{1 + x^2} \, \mathrm{d}x, \end{aligned}

which does not converge. In order to prove divergence, first observe that \frac{x^2}{1+x^2} \in [0, 1). Next, let us choose \varepsilon \in (0, 1). Now,

\displaystyle \begin{aligned} \frac{x^2}{1 + x^2} &> \varepsilon   & \Longleftrightarrow \\ x^2 &> \varepsilon + \varepsilon x^2 & \Longleftrightarrow \\ (1 - \varepsilon) x^2 &> \varepsilon & \Longleftrightarrow \\ x^2 &> \frac{\varepsilon}{1 - \varepsilon} & \Longleftrightarrow \\ |x| &> \sqrt{\frac{\varepsilon}{1 - \varepsilon}} = \delta.& \end{aligned}

Since such \delta > 0 and \varepsilon exist, \int_{-\infty}^{\infty} \frac{x^2}{1 + x^2} \, \mathrm{d}x diverges by Lemma 1.

Lemma 1 Let f \colon \mathbb{R} \to \mathbb{R} be a continuous function. If there exist a \in \mathbb{R} and \varepsilon > 0 such that f(x) > \varepsilon when x < a and there exists b > a and \varepsilon_b \geq 0 such that f(x) \geq \varepsilon_b when x > b, then \int_{-\infty}^{\infty} f(x) \, \mathrm{d}x = \infty.
Proof

Let C = \int_a^b f(x) \, \mathrm{d}x. Since f is continuous, C is finite. Now

\displaystyle \int_{-\infty}^{\infty} f(x) \, \mathrm{d}x = \int_{-\infty}^a f(x) \, \mathrm{d}x + C + \int_b^{\infty} f(x) \, \mathrm{d}x.

Since f(x) \geq 0 when x > b, \int_b^{\infty} f(x) \, \mathrm{d}x \geq 0. Finally,

\displaystyle \int_{-\infty}^a f(x) \, \mathrm{d}x > \lim_{b \to -\infty} \varepsilon_a(a - b) = \infty .

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