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Here is a statement of the second part of the Fundamental Theorem of Calculus (FTC2), from a well-known calculus text (James Stewart, Calculus, 4th ed):

If $f$ is continuous on $[a,b]$, then $\int_a^b f(x) \, dx = F(b)-F(a)$, where $F$ is any [emphasis mine] antiderivative of $f$, that is, a function such that $F'=f$.

The following, however, seems to give a counterexample.*
Can someone resolve this for me?:

Let $f(x) = \frac{1}{4 \sin (x)+5}$.

$f$ is continuous on $[0, 2 \pi]$:

enter image description here

Consider two antiderivatives of $f$, $F_1$ and $F_2$:

$$F_1(x) = \frac{x}{3}+\frac{2}{3} \tan^{-1}\left(\frac{\cos (x)}{\sin (x)+2}\right)$$

$$F_2(x)=\frac{1}{3} \left(\tan ^{-1}\left(2-\frac{3}{\tan \left(\frac{x}{2}\right)+2}\right)-\tan^{-1}\left(2-\frac{3}{\cot \left(\frac{x}{2}\right)+2}\right)\right).$$

Using Mathematica, I've confirmed that both $F_1'= f$ and $F_2'= f$. According to my reading of the above statement of FTC(2), $\int_0^{2\pi} f (x) \, dx = F_1(2\pi)-F_1(0)= F_2(2\pi)-F_2(0)$

However,

$F_1(2\pi)-F_1(0)=2\pi/3$

$F_2(2\pi)-F_2(0)=0$

Note from the plots below that $F_1$ is continuous on $[a,b]$, while $F_2$ is not. Given all of this, it seems the sufficient condition for $\int_a^b f (x) \, dx = F(b)-F(a)$ is that the antiderivative be continuous on $[a,b]$, not the integrand.

$F_1 =$

enter image description here

$F_2=$

enter image description here

*I've taken this example function from a wolfram.com blog.

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    $\begingroup$ It's obvious just from the graphs that $F_2$ is not an antiderivative. At the points of discontinuity, $F_2$ has a limit of its derivative, but that's not the same thing as being differentiable at that point. $\endgroup$
    – aschepler
    Jan 20, 2019 at 3:56
  • $\begingroup$ $F_1$ and $F_2$ can be made equal by adding constants, $c_i$, in each interval where $F_2$ is continuous. For instance $c_i\approx0.524$ when $x\in(-0.93, 4.07)$ and so on. These $c_i$ account for the discontinuities in $f(x)$ but as long as you're not integrating over any discontinuities, $F_2$ would be valid. This is all because the antiderivative of a discontinuous function may have multiple constants of integration (even in the case of $\int \frac1x\mathrm{d}x$). $\endgroup$
    – Jam
    Jan 20, 2019 at 20:10
  • $\begingroup$ Great Duck forgets about the Heavyside impulse function. $\endgroup$ Jan 24, 2019 at 5:33
  • $\begingroup$ @richard1941 The Heaviside step function is not the antiderivative of a function. The Dirac delta function is not a true function in the sense of real analysis (though it can be addressed by the theory of "generalized functions"). $\endgroup$ Feb 13, 2019 at 4:48

3 Answers 3

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A differentiable function is continuous. If your "antiderivative" is not continuous, it is not an antiderivative of a continuous function.

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  • $\begingroup$ Thanks Martin. That clears it up. I'll just wait the requisite additional day or so before accepting an answer. I'm curious: Is the proper phrasing to say that F2 is not differentiable, period, or can we say it is (piecewise?) differentiable on the domain x ∈ ℝ, where: x ≠ {- 2(arctan[2] + n𝜋) || - 2(cotan[2] + n𝜋)} & n ∈ ℤ? [I understand that, even if one could say the latter, it wouldn't change the fact that F2 is not an antiderivative of f.] $\endgroup$
    – theorist
    Jan 20, 2019 at 21:51
  • $\begingroup$ Glad I could help. And yes, you are right. It is piecewise differentiable. $\endgroup$ Jan 20, 2019 at 23:52
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@Martin Argerami posted a succinct and correct answer:

A differentiable function is continuous. If your "antiderivative" is not continuous, it is not an antiderivative of a continuous function.

To elaborate, $F_2$ is not continuous due to the multi-valued nature of $\arctan$. Since $\tan$ is a periodic function (with a period of $\pi$), this means that $\arctan$ has an infinite number of values. Its plot looks like this:

multi-valued arctan

To make it a proper function (one unique output value), we usually pick the primary branch (the one through the origin). However, notice that the primary branch is discontinuous at infinity - if we pass through positive infinity and come back out at negative infinity, we've jumped from $\pi/2$ to $-\pi/2$. What we really wanted was to move to the next branch up to preserve continuity, but to do that we'd no longer have a single valued function. We wouldn't have a function at all, really, since the common definition of function requires it to be single valued.

The discontinuities of $F_2$ are precisely where the argument of $\arctan$ becomes infinite - where $2+\cot\frac{x}{2}$ and $2+\tan\frac{x}{2}$ become zero - approximately 4.07 and 5.36.

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  • $\begingroup$ Since we are talking about functions on the real numbers, of which $\infty$ is a not a member, there no such thing as "discontinuous at infinity". $\endgroup$ Jan 20, 2019 at 1:35
  • $\begingroup$ @Paul Sinclair: True, I'm playing a bit loose with that term. 3/(2+tan(x/2)) approaches negative infinity (at x=4.07), becomes indeterminate, then reappears coming in from positive infinity. Applying arctan to two minus that value, it approaches -pi/2, becomes indeterminate, then reappears decreasing from pi/2. Strictly speaking, it's a non-removable discontinuity that can be understood by looking at the limiting behavior of arctan as it approaches $\pm\infty$ $\endgroup$ Jan 20, 2019 at 3:21
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The top paragraph apparently has an error. The integral of f(x) dx from a to b is F(b) - F(a), not F(a) - F(b), where F is any antiderivative of f.

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  • $\begingroup$ Thanks, fixed it! Though, for future reference, since this isn't an answer, according to site guidelines it should be posted as a comment on my post rather than an answer. [Stack Exchange likes to distinguish the two to maintain site readability.] $\endgroup$
    – theorist
    Jan 25, 2019 at 0:34

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