# The deep reason why \int \frac{1}{x}\operatorname{d}x\int \frac{1}{x}\operatorname{d}x is a transcendental function (\log\log) [duplicate]

In general, the indefinite integral of $x^n$ has power $n+1$. This is the standard power rule. Why does it “break” for $n=-1$? In other words, the derivative rule fails to hold for $n=0$.
Is there some deep reason for this discontinuity?

I’ll try to give a soft answer to what I see as the spirit of the question, which is not why you get exactly log, but why the behaviour is different when integrating $x^{k}$ for $k=-1$.
The way I see it is that the logarithm would actually be there for other powers of $x$ too, but it’s “hidden” by the fact that in a geometric series one term “gobbles up” all the others together, except in the very special case when all terms are equal. Put in other words, the intuition is very close to the reason why $\sum_{i=a}^b c^i$ can be reasonably approximated by just the first term of the sum ($c^a$) if $c<1$, and just by the last ($c^b$) if $c>1$, regardless of how large $b-a$ is, i.e. of how many terms you have in the sum. But if $c=1$ no single term dominates all the others, and that's when you have to count them all, and you end up seeing the $b-a$ term "emerge" in $\sum_{i=a}^b 1^i=b-a+1$.
Informally, you can see how this applies to the case at hand writing $\int_{x_0}^{x_f} x^k dx$ as $\int_{x_0}^{2 x_0} x^k dx + \int_{2x_0}^{4x_0} x^k dx + ...$. Each of the $\approx \log_2 \frac{x_f}{x_0}$ terms has the same weight as the others if, and only if, $k$ has a very specific value (which?), in which case $\int_{x_0}^{x_f} x^k dx$ equals $\approx \log_2 \frac{x_f}{x_0}$ times $\int_{x_0}^{2 x_0} x^k dx$. If it's just an $\epsilon$ smaller, or larger, you get the sum within a constant factor of either $\int_{x_0}^{2 x_0} x^k dx$ or of $\int_{\frac{1}{2}x_f}^{x_f} x^k dx$, respectively, regardless of $\frac{x_f}{x_0}$. Note that instead of using $2$ as a base, we could have used $3$, or $e$, or $7.24$, or $\pi$, and the critical value of $k$ would have remained the same: it's the value that ensures that if you integrate $x^k$ over an interval $7.24$ longer, but with a starting point $7.24$ times larger, the integral does not change, $k=-1$.