Trying to visualise the integral of $1/x$ given division by zero is undefined.

Question

I know the integral of $1/x$ is $\ln(x)$. Now, I also know what these graphs look like visually.

I tried to "add power, divide by new power" and of course it's undefined.

My logic was to consider $x^{-1.1}$ and $x^{-0.9}$ whos integrals are $\frac{-x^{-0.1}}{0.1}$ and $\frac{x^{0.1}}{0.1}$. And then visualise that $\ln(x)$ is inbetween and therefore the integral of $1/x$. When I plot $\frac{-x^{-0.1}}{0.1}$ and $\frac{x^{0.1}}{0.1}$ I do not get much resembalence to $\ln(x)$, why is this? And can anyone follow my thoughts and help me show, visually, that the curve of the integral of $1/x$ is $\ln(x)$? Thanks.

Answer 1

As I pointed out in a comment, you considered the wrong secong graph. Here are how the three graphs look in the interval $(0,5]$.

As you can see, they compare quite closely.

Answer 2

Your logic isn't so crazy: you think $x^{-1.1}$, $x^{-1}$ and $x^{-0.9}$ behave very similarly so you expect their anti-derivatives to do the same.

Take a look at the plots of these three functions: they are indeed very similar.

The problem with taking 'an' anti-derivative (based on the standard table: the power rule for $x^a$ with $a \ne -1$ and $\ln x$ for the other one) is that you make an implicit choice in the contant of integration and this may prevent you from getting the graphs you expect.

You can take a specific anti-derivative, e.g. by forcing all of them to go through $(1,0)$ for easier comparison. For $\ln x$, without an extra constant of integration, this is the default but for the other two it requires a different constant of integration. Note that you can also find them via: $$\int_1^x t^{-1.1} \,\mbox{d}t = 10-10x^{-0.1} \quad\mbox{and}\quad \int_1^x t^{-0.9} \,\mbox{d}t = -10+10x^{0.1} $$ So we have:

• $\left( \color{blue}{10-10x^{-0.1}} \right)' = x^{-1.1}$
• $\left( \color{blue}{\ln x} \right)' = x^{-1}$
• $\left( \color{blue}{-10+10x^{0.1}} \right)' = x^{-0.9}$

Now compare the graphs of the three functions in blue above; I guess this is close(r) to what you were expecting?