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Integrating by parts the following integral $$I=\int \frac{f'(x)}{f(x)}dx$$

gives us

$$\begin{align*} I&=\int \frac{f'(x)}{f(x)}\,dx\\ &=\int\frac1{f(x)}f'(x)\,dx\\ &=\frac1{f(x)}f(x)-\int\left(\frac1{f(x)}\right)'f(x)\,dx\\ &=1+\int \frac{f'(x)}{f(x)}\,dx, \end{align*} $$ so $$I=1+I\implies0=1,$$

which seems like a contradiction but is in reality a mistake as we can see by being somewhat more rigorous:

$$ \begin{align*} I&=\int_{a}^x \frac{f'(t)}{f(t)}\,dt\\ &=\int_{a}^x\frac1{f(t)}f'(t)\,dt\\ &=\left[\frac1{f(t)}f(t)\right]_a^x-\int_{a}^x\left(\frac1{f(t)}\right)'f(t)\,dt, \end{align*}$$ so $I=I.$


How do I explain this to a student-of economics for what it's worth-who has not still learned about definite Integrals?

(I suspect I could insert some constant on the upper part of the "failed" relations but I am not sure where or how to properly explain it. I also understand that in a couple of lessons we will talk about definitive integrals but what can I say now-That indefinite integrals are in reality a form of definite ones? )

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    $\begingroup$ For any function $f$, $\int f=1+\int f$ because antiderivatives are unique only up to a constant. The error is that you've forgotten $\int f$ denotes a family of functions (any two of which differ by a constant), not a unique function. (No need to talk about definite integrals; you only need to understand the meaning of an indefinite integral, which is an antiderivative.) $\endgroup$ – symplectomorphic Mar 28 '17 at 18:55
  • $\begingroup$ @symplectomorphic I do know that $\int f$ denotes a family of functions but I had some difficulty explaining it in the form presented above-perhaps I should note that this was given "as is" by my student. In any case thank you for the comment. Coupled with tilper's answer it is more than enough. $\endgroup$ – MathematicianByMistake Mar 28 '17 at 19:06
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Integrating by parts the following integral $$I=\int \frac{f'(x)}{f(x)}dx$$

gives us

$$I=\int \frac{f'(x)}{f(x)}dx=\int\frac1{f(x)}f'(x)dx=\\\frac1{f(x)}f(x)-\int\Big(\frac1{f(x)}\Big)'f(x)dx=\\1+\int \frac{f'(x)}{f(x)}dx\Rightarrow\\ I=1+I$$

Be careful here. The $I$ on the left and the $I$ on the right are not exactly the same quantity.

How can that make sense? It makes sense because indefinite integrals have an arbitrary constant of integration, and for any given indefinite integral, its arbitrary constant is not guaranteed or required to have the same value everywhere the integral is written.

In fact, the process you followed shows that the arbitrary constant for the $I$ on the left is necessarily $1 + $ the arbitrary constant for the $I$ on the right.

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It isn't a paradox at all. Evaluating an indefinite integral amounts to finding an antiderivative of the integrand; i.e., $$\int f(x) \, dx = F(x) \iff F'(x) = f(x).$$ Therefore, the antiderivative is unique only up to addition by a constant.* Different methods of integration may yield different antiderivatives.

If we think of differentiation as an operator on a set of differentiable functions, then integration is merely an inverse operator, much like taking a square root is the inverse operation of squaring. But there is no guarantee that such operations are one-to-one: indeed, if I say $x^2 = 4$, there are two real numbers for which this is true, namely $x = 2$ and $x = -2$. That this is the case does not imply that we must have $2 = -2$.


*Moreover, one can create antiderivatives that differ by an arbitrary constant on different intervals of the integrand's domain; e.g., the function $$F(x) = \begin{cases} \log x + C, & x > 0 \\ \log(-x) + K, & x < 0, \end{cases}$$ satisfies the relationship $F'(x) = \frac{1}{x}$ for all $x \ne 0$, for any constant choices of $C, K$. Thus the so-called "constant of integration" does not itself encompass all antiderivatives of a given function; one could have different constants of integration for each interval of an integrand on which it is defined.

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  • $\begingroup$ Thank you for this. The second part has been quite enlightening in pedagogical purposes. $\endgroup$ – MathematicianByMistake Mar 28 '17 at 19:23
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It is actually should be $$I=\int \frac { f'(x) }{ f(x) } dx=I=\int \frac { d\left( f\left( x \right) \right) }{ f\left( x \right) } =\ln { \left| f\left( x \right) \right| +C } $$

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  • $\begingroup$ I know that. The problem is how to explain the "paradox" above. $\endgroup$ – MathematicianByMistake Mar 28 '17 at 19:01
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    $\begingroup$ It is not paradox, you can call it "mathematical sophism" $\endgroup$ – haqnatural Mar 28 '17 at 19:28
  • $\begingroup$ Yeah, that sounds better! :-) Kinda of like a "false proof".. $\endgroup$ – MathematicianByMistake Mar 28 '17 at 19:34

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