Calculus Topic: Differentiability of the Integral of a Bounded Function

We saw that if \(\phi\) is any bounded function defined on an interval with only a finite number of discontinuities, then \(\phi\) has a continuous integral (antiderivative), that is, there is a continuous function \(\Phi\) defined on that interval such that \(\Phi'(x) = \phi(x)\) for all \(x\) in the interval except at the points of discontinuity.

In this case, the integral function \(\Phi\) is continuous, but may fail to be differentiable at some points of the interval. So we ask: When is the integral function \(\Phi\) differentiable? Given a point \(x = c\) in the interval, does \(\Phi'(c)\) exist?

Three possible cases arise:

1. \(\phi\) is continuous at \(x = c\). In this case, \(\Phi\) is differentiable at \(x = c\), that is, \(\Phi'(c)\) exists.
2. \(\phi\) has a jump discontinuity at \(x = c\), that is, \(\phi(c-) \neq \phi(c+)\). In this case, the right and left derivatives of \(\Phi\) exist separately but are unequal. We have \[ \begin{align*} \Phi'(c-) & = \lim_{h \to 0-} \frac{\Phi(c + h) - \Phi(c)}{h} = \phi(c-), \quad \text{and}\\ \Phi'(c+) & = \lim_{h \to 0+} \frac{\Phi(c + h) - \Phi(c)}{h} = \phi(c+), \end{align*} \] so \(\Phi\), while being continuous, has a corner point at \(x = c\), that is, the left-side and right-side slopes of the antiderivative \(\Phi\) at \(x = c\) are different. In other words, the left side of the graph of \(\Phi\) at \(x = c\) meets the right side at an angle, instead of lining up straight.
3. \(\phi\) has an essential discontinuity at \(x = c\), that is, either \(\phi(c-)\) or \(\phi(c+)\) does not exist. In this case the situation is quite subtle, and \(\Phi\) may or may not be differentiable at \(x = c\).

Case 3: Differentiability of the integral at an essential discontinuity of the integrand

We will illustrate the last case (case 3) with example functions.

Suppose that \(\phi\) is a bounded function with an essential discontinuity at \(x = c\). For definiteness, we will take \(c = 0\), and assume that the \(\phi\) has an essential discontinuity on the right side of \(x = 0\), that is, the right-hand limit \(\phi(0+)\) does not exist. (These assumptions do not cause any loss of generality.)

The figure below shows a typical example of such an essential discontinuity to the right of \(0\):

Note that as \(x\) approaches \(0\) from the right (\(x \to 0+\)), the function above oscillates infinitely many times with the amplitude steadily approaching \(1\), and with successive peaks occurring at \(x = x_1\), \(x = x_2\), \(x = x_3\), etc. The horizontal distance between successive peaks is thus \(x_n - x_{n+1}\) (\(n = 1, 2, \dots\)), and so the period here is variable and decreases as \(x \to 0+\).

We will give examples of two such functions \(f\) and \(g\) below, with subtly different behavior.

Example 1 : The function \(f\) and its antiderivative \(F\)

As a first example, consider the following function defined on \(0 \lt x \lt \infty\) (this is essentially our earlier example): \[ f(x) = 2x \sin \frac{1}{x} - \cos \frac{1}{x}, \quad 0 \lt x \lt \infty. \] The function \(f\) has an essential discontinuity to the right of \(0\) that is just like the one shown in Figure 1 above: As \(x\) approaches \(0\) from the right (\(x \to 0+\)), the function \(f\) oscillates infinitely many times with the amplitude steadily approaching \(1\), but with decreasing (variable) period.

A continuous integral function \(F\) for \(f\) (antiderivative of \(f\)) is obtained as: \[ F(x) = \begin{cases} 0, & x = 0 \\ x^2 \sin \frac{1}{x}, \quad & 0 \lt x \lt \infty \end{cases}. \] The integral function \(F\) here is right-differentiable at \(x = 0+\), since \[ \lim_{h \to 0+} \frac{F(0+h) - F(0)}{h} = \lim_{h \to 0+} h \sin \frac{1}{h} = 0, \] so \(F'(0+)\) exists and is \(= 0\).

As noted earlier, this can be visually understood from the following observation. The graph of the integral function \(F\) also oscillates infinitely many times as \(x\) approaches \(0\) from the right (\(x \to 0+\)), but the amplitude of these oscillations decay to zero at a fast quadratic rate. The fast damping rate not only makes the function \(F\) continuous at \(x = 0\), but also forces it to be differentiable at \(x = 0\).

Example 2 : The function \(g\) and its antiderivative \(G\)

Let's now take another similar example: \[ g(x) = \cos (\ln x - \pi/4), \quad 0 \lt x \lt \infty. \] The function \(g\) too has an essential discontinuity to the right of \(0\) which, like that of \(f\), is just like the essential discontinuity of the function shown in Figure 1 above: As \(x\) approaches \(0\) from the right (\(x \to 0+\)), the function \(g\) oscillates infinitely many times with a steady amplitude of \(1\), but with decreasing (variable) period.

A continuous integral function \(G\) for \(g\) (antiderivative of \(g\)) is obtained as: \[ G(x) = \begin{cases} 0, & x = 0 \\ \frac{1}{\sqrt{2}} x \sin \ln x, \quad & 0 \lt x \lt \infty \end{cases}. \] However, the integral function \(G\) here is no longer right-differentiable at \(x = 0+\), since \[ \lim_{h \to 0+} \frac{G(0+h) - G(0)}{h} = \frac{1}{\sqrt{2}} \lim_{h \to 0+} \sin \ln h \;\longleftarrow\; \text{Does Not Exist}, \] so \(G'(0+)\) does not exist!

This can be visually understood by examining the graph (not shown) of the integral function \(G\) near \(0+\): The integral function \(G\) is also oscillating infinitely many times as \(x\) approaches \(0\) from the right (\(x \to 0+\)), but with the amplitude of the oscillations decaying to zero at a linear rate. The linear damping rate of the oscillations ensures continuity at \(x = 0\), but breaks differentiability, since it is not a sufficiently fast decaying rate.

Explanation of the different behaviors

We thus have examples of two bounded functions \(f\) and \(g\), with apparently very similar essential discontinuities at the same point \(x = 0+\), but their integrals \(F\) and \(G\), although continuous, have different differentiablity properties at \(x = 0+\): \(F\) is differentiable at \(x = 0+\), but \(G\) is not!

It turns out that if a bounded function \(\phi\) has an essential discontinuity at \(x = c\), then whether its integral \(\Phi\) will be differentiable at \(x = c\) depends subtly on the nature of the essential discontinuity!

For the example function \(f\), we noted that as \(x\) approaches \(0\) from the right (\(x \to 0+\)), the function \(f\) oscillates infinitely many times with a steady amplitude of about \(1\) and with decreasing (variable) period. Looking more carefully at the oscillations of \(f\), we notice that the horizontal position of the peaks occurring at \(x = x_1, x_2, \dots,\) follow an approximately harmonic pattern as \(x\) approaches \(0+\), with the peaks occurring approximately at \[ x_n \approx \frac{1}{(2n + 1)\pi}, \quad n = 1, 2, \dots. \] Because the \(x\)-coordinates of the peaks form an approximate harmonic sequence, the successive periods of \(f\), that is the horizontal distances \(x_n - x_{n+1}\) between successive peaks, become equal asymptotically, in the sense that their ratio approaches \(1\) in the limit as \(n \to \infty\): \[ \lim_{n \to \infty} \frac{x_{n+1} - x_{n+2}}{x_n - x_{n+1}} = 1. \]

On the other hand, while the function \(g(x) = \cos (\pi/4 - \ln x)\) also oscillates infinitely many times with a steady amplitude of \(1\) and with decreasing (variable) period as \(x\) approaches \(0\) from the right (\(x \to 0+\)), its variable periods decrease in a way different from that of \(f\): The horizontal distance between two successive peaks of \(g\) decreases geometrically as \(x\) approaches \(0+\), with the peaks occurring at \[ x_n = \frac{e^{\pi/4}}{e^{2n\pi}}, \quad n = 1, 2, \dots. \] So for \(g\), the variable period is decreasing geometrically, with the ratio of two successive periods having the constant value \(1/e^{2\pi} \lt 1\): \[ \frac{x_{n+1} - x_{n+2}}{x_n - x_{n+1}} = \frac{1}{e^{2\pi}} \lt 1. \]

This subtle difference between the natures of the essential discontinuities of the two functions \(f\) and \(g\) at \(x = 0+\) is the reason that the integral of one of them (\(F(x) = \int f(x)\,dx\)) is differentiable at \(x = 0\), while the integral of the other (\(G(x) = \int g(x)\,dx\)) is not differentiable at \(x = 0\).

One final remark: The magnitude (absolute value) of the slope of the right-secant line through the origin for the function \(G\) was: \[ \left|\frac{G(0+h) - G(0)}{h}\right| = \frac{1}{\sqrt{2}} |\sin \ln h| \leq \frac{1}{\sqrt{2}}, \] so while the slope of this secant for \(G\) does not have a limit as \(h \to 0+\), the slope remains bounded. It turns out that this is true in general: The slope of a secant line for the integral of a bounded function cannot become unbounded. This follows from the following result: Suppose that a function \(\psi\) is continuous on the interval \(0 \leq x \lt a\) and differentiable on the interval \(0 \lt x \lt a\) with the magnitude of the derivative bounded by \(M\) (i.e., \(|\psi'(x)| \leq M\) whenever \(0 \lt x \lt a\)). Then \(\psi\) may not be differentiable from the right at \(x = 0\) (as we saw in the last example), but the magnitude of the slope of the right secant line at \(x = 0\) must remain bounded by \(M\) (i.e., \(|(\psi(h) - \psi(0))/h| \leq M\) whenever \(0 \lt h \lt a\)).