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I read the following in a book on differential equations

$$\int \frac{1}{y-1} dy = \log |y - 1|$$

If I put $\int \frac{1}{y-1} dy$ into Wolfram Alpha it gives $\log (y - 1)$, i.e. the argument of the function is not an absolute value unlike the first equation. So which is correct and why am I seeing conflicting results for the same integral?

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  • $\begingroup$ the first one is correct $\endgroup$
    – Aaron Maroja
    Sep 8, 2014 at 21:17
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    $\begingroup$ The first is correct but I think that Wolfram Alpha assumes that $y>1$. $\endgroup$
    – user63181
    Sep 8, 2014 at 21:18
  • $\begingroup$ Did you try to check each of them yourself? $\endgroup$
    – Did
    Sep 8, 2014 at 21:20
  • $\begingroup$ 1)it should be $\log|y-1| +C$, 2) real log function is defined only for positive arguments $\endgroup$
    – Alex
    Sep 8, 2014 at 21:21
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    $\begingroup$ Neither is really completely right. If we want all the functions with largest (real domain) whose derivative is $y-1$, it should be $F(y)=\ln(y-1)+C$ when $y\gt 1$; $F(y)=\ln(1-y)+D$ when $y\lt 1$. Here $C$ and $D$ are arbitrary constants, not necessarily equal. $\endgroup$
    – André Nicolas
    Sep 8, 2014 at 21:34

1 Answer 1

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That happens because $\log (y - 1)$ only exists in $(1, +\infty)$, hence its derivative would be $1/(y - 1)$, but with considering only values of $y$ that are greater than $1$.

This is of course different from what you are trying to integrate, because that $1/(y - 1)$ inside the integral is a function that is defined everywhere in $\mathbb{R}$ (with the sole exception of $1$). Then its indefinite integral is $\log |y - 1|$ because not only its derivative is $1/(y - 1)$, but its domain is also exactly $\mathbb{R} - \{1\}$, that is the same of the function you're trying to integrate.

This happens because functions with different domains are different functions even if their written expression is the same, so $1/(y - 1)$ may yield different results when integrating, according to what its domain is.

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