In general,
$$ \int x^{n} dx = \frac{x^{n+1}}{n+1} + C $$
However, this rule does not work when $n=-1$ because it leads to division by zero. Instead, $\int x^{-1} dx = \ln(|x|) + C$. This standard result made me wonder if the natural logarithm could be approximated by using $n$-values that are close to, but not equal to, $-1$. For example, setting n as equal to $-0.99$:
$$ \int x^{-0.99} dx = \frac{x^{0.01}}{0.01} +C = 100x^{0.01}+C $$
The $C$ value that translated the graph so that it became a good approximation was $-100$. Here is the general case. My suspicion is that as $k$ tends to infinity, the approximation becomes better and better:
$$ \int x^{-1+1/k} dx = \frac{x^{1/k}}{1/k}+C=kx^{1/k}+C $$
Using $-k$ as the constant of integration, I produced the following result:
As you can see, it is difficult to see any difference between the two graphs. I have two questions:
- Why does $C$ have to equal $-k$ in order for the approximation to work?
- Can this approach be formalised by using limits? E.g. as $k \to \infty$, $kx^{1/k}-k$ becomes arbitrarily close to $\ln(x)$?
You can define the function $$f(a)=\int_1^xt^a\,dt= \begin{cases} \dfrac{x^{a+1}-1}{a+1} & a\ne -1 \\ \log(x) & a=-1 \end{cases} $$ (where $x>0$) and show that $f$ is continuous at $a=-1$. Using $$\lim_{y\to 0}\frac{e^y-1}{y}=1 $$ we have $$\lim_{a\to -1}\frac{x^{a+1}-1}{a+1}=\lim_{a\to -1}\left(\frac{e^{\log(x)(a+1)}-1}{\log(x)(a+1)}\cdot\log(x)\right)=\log(x) $$ So you can approximate the graph of $\log(x)$ by taking $a$ to be close to $-1$, as you've done in your plot.