5
$\begingroup$

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.

$\endgroup$
1
5
$\begingroup$

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,

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.