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I can't understand where the mistake in the following steps is:

$$\int \frac{1}{x}dx = \int x \frac{1}{x^2}dx = -\int x (-\frac{1}{x^2})dx = -\int x(\frac{1}{x})'dx = -x\frac{1}{x} + \int \frac{1}{x}dx = -1 + \int \frac{1}{x}dx$$

So: $$\int \frac{1}{x}dx = \int \frac{1}{x}dx -1$$

Also: Why is it that the integral of this function is $\ln x$? How did people conclude that?

Blue
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Antonis
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    Arbitrary constant? – Angina Seng Dec 19 '19 at 19:04
  • Oh well... silly me. I should probably only ask the second question. – Antonis Dec 19 '19 at 19:07
  • Indefinite integrals aren't functions, per se, they're sets of functions that all have the same derivative. The constant is eliminated after differentiation so it doesn't really matter. Regarding how we know that $\int\frac1x\ \mathrm{d}x$ is $\ln x$, that's effectively just the name we give that integral. It's commonly used as a definition of $\ln x$, when you include the limits. – Jam Dec 19 '19 at 19:08
  • Many take that as the definition of $\ln x$, i.e. $\ln x = \int _1^x 1/t dt$ – David P Dec 19 '19 at 19:10
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    @Antonis: You should ask the second question as a separate question. – Blue Dec 19 '19 at 19:23
  • @AnginaSeng It seems that your suggestion does not solves the problem: adding $+C$ (with arbitrary $C$), wouldn't the conclusion be $C=1$ for all $C\in\mathbb R$? – Pedro Apr 21 '24 at 14:58

3 Answers3

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$\int\frac{1}{x}dx$ denotes the set of all antiderivatives of $\frac{1}{x}$. You've shown a function $f$ is such antiderivative iff $f-1$ is too. You need another method to show it's a logarithm. You can note for example that$$\frac{dy}{dx}=\frac1x\implies\frac{dx}{dy}=x\implies x\propto\exp y\implies y=\ln x+C.$$(Due to the discontinuity at $x=0$, $C$ in this context is locally constant. For example, if $x$ is a real variable, we can choose $y-\ln|x|$ to have a $\operatorname{sign}x$-dependent value that's otherwise constant.)

J.G.
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  • Thanks. I just wanted to see a method for integrating this. Also in your second step what is the symbol after x, and in the definite case what is the method? – Antonis Dec 19 '19 at 19:16
  • @Antonis $u\propto v$ means $u$ is proportional to $v$. (If you want to use this symbol, write \propto.) You would compute the definite integral over an interval not containing $0$ using an arbitrary antiderivative. – J.G. Dec 19 '19 at 19:25
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The reason for "contradictory looking" result is because you are trying to estimate an indefinite integral. An indefinite integral such as $$ \int \frac1x dx $$ gives a set of answers. So $$ \int \frac1x dx = \{f: f(x) = \log x + a \text{ where } a \in \mathbb{R} \} $$ So the equality you got is actually a set equality. That is $$ \{f: f(x) = \log x + a \text{ where } a \in \mathbb{R} \} - 1 = \{f: f(x) = \log x + a \text{ where } a \in \mathbb{R} \} $$ and there is nothing wrong mathematically.

Your steps aren't wrong, they just do not account for the set equality that you are showing.

The reason why $$ \int \frac1x dx = \log x + a $$ is mainly due to the definition of exponential function $\exp$ and its relation to the logarithm $\log$.

Semzem
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  • Your notation for the set of antiderivatives is slightly incorrect. Due to the discontinuity at $x=0$, there can be two independent constants of integration. So the set would actually be $\left{f: f(x) = \log |x|+\begin{cases}a^+,&x>0\a^-,&x<0\end{cases},\ \text{where}\ a^+,a^-\in\mathbb{R}\right}$. See Tom Leinster: Reader Survey: log|x| + C. – Jam Dec 19 '19 at 20:11
  • Ah, thanks for the fix. – Semzem Dec 20 '19 at 14:59
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Elementary answer (for students who have not yet formalized the indefinite integral as a set of function or as a definite integral with variable upper limit):

(1) The basic rules for indefinite integrals that the students are used to are:

  • $\displaystyle\int cf(x)\,dx=c\int f(x)\,dx$
  • $\displaystyle\int [f(x)+g(x)]\,dx=\int f(x)\,dx+\int g(x)\,dx$
  • $\displaystyle\int k\,dx=kx+C$ ($k$ constant)

Therefore, $$\int f(x)\,dx-\int f(x)\,dx=\int f(x)\,dx+\int -f(x)\,dx=\int [f(x)-f(x)]\,dx=\int 0\,dx=C.$$

In other words:

In the context of operations with indefinite integrals, $\int f(x)\,dx-\int f(x)\,dx$ is not zero, but $C$ (constant).

This is generally not discussed because this expression rarely appears.

(2) If we look at the derivation of the integration by parts formula, we see that the correct formula is

$$\int u\,dv=uv-\int v\,du+C.$$

Usually, the $C$ is omitted because it is written after the calculation of $\int v\,du$ (formally, there would be two constants that are written as a single one). But since in your case $\int v\,du$ was not calculated, $C$ cannot be omitted.

(3) Putting these things together, the correct reasoning would be

$$\int \frac{1}{x}dx = \int \frac{1}{x}dx -1+C_1$$ $$\int \frac{1}{x}dx - \int \frac{1}{x}dx= -1+C_1$$

$$C_2= -1+C_1$$

This just gives us a relationship between the arbitrary constants coming from (1) and (2) for the particular case considered.

Pedro
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