Any deeper insights for why $\int \frac{1}{x}dx = \ln|x|+C$?

Question

I have already learned that $$\int x^\alpha \ dx = \cases{\frac{1}{\alpha+1}x^{\alpha+1}+C & if $\alpha\neq-1$ \\ \ln|x|+C & if $\alpha=-1$} \ . $$

I am just curious if there is any deeper insight to explain the abrupt jump from powers of $x$ to the natural logarithm $\ln|x|$ at $\alpha=-1$, rather than just saying because the coefficient $1/(\alpha+1)$ is undefined when $\alpha=-1$.

I have tried plotting things out using Desmos, and the 'jump' at $k=-1$ still doesn't go away. Here is what I did: I plotted a function $$f_k(x) := x^k$$ with parameter $k$. As I adjust $k$, I can see all the $x^k$, but none of them are close to the plot of $\ln{x}$.

My guess: I suspect we need function of two variables to visualize the 'jump'. I tried to construct $g(x, y) := x^y$ and observed what happen at $y=0$. It seems like there is some 'saddle line' parametrized by $(t,0,1)$ for all $t\in\mathbb{R}$. But I'm still not sure how $\ln{x}$ can come into play.

Answer 1

If we replace the indefinite integral with a definite one so as to keep one point (say, $(1, 0)$) fixed as we vary the exponent, then there's no jump at all. For all $s \neq 0$, define $$f_s(x) := \int_1^x\, t^{s-1}\, dt = \frac{x^s - 1}{s}.$$

Then by l'Hopital,

$$\lim_{s \to 0} f_s(x) = \left . \frac{d}{ds} x^s \right|_{s = 0} = \left . \frac{d}{ds} e^{s \ln x} \right|_{s = 0} = \ln x.$$

Answer 2

The constant is in fact a function of parameter alpha. Try defined integral, say, from 1 to x. You’ll see a continuous transition from power function to log. x^a - 1 = e^(a ln x) - 1 -> a ln x, for a -> 0

Answer 3

Another way to see this is to note that, by a change of variable,

$\begin{array}\\ \int_1^x\dfrac{dt}{t}+\int_1^y\dfrac{dt}{t} &=\int_1^x\dfrac{dt}{t}+\int_x^{xy}\dfrac{dt}{t}\\ &=\int_1^{xy}\dfrac{dt}{t}\\ \end{array} $

Therefore, if $f(x) = \int_1^x\dfrac{dt}{t}$, then $f(x)+f(y) =f(xy)$, which is the functional equation for log.