1 Integrating Logs

There's a sneaky technique for integrating the natural logarithm function which involves writing $\ln (x)$ as $1\times \ln (x)$ and invoking the product rule:

There are a variety of ways of explaining this, but the ones that I know don't tend to be all that enlightening on how to come up with something like this in the first place. For example, I usually say that we try it because we know how to differentiate $\ln (x)$ and its derivative is considerably nicer than $\ln (x)$. But that's a bit unsatisfactory, since the reason the second integral is so nice is because of the interaction between the derivative of $\ln (x)$ and the integral of $1$ rather than just the niceness of the derivative of $\ln (x)$ by itself.

Also, it still feels like a justification after the fact. Without seeing it work for $\ln (x)$ then I wouldn't think to try it for anything else. To put that another way, knowing it works for $\ln (x)$ means I might try it for, say, ${\sin }^{-1}(x)$, but I still can't envision coming up with it out of a vacuum.

By the way, did you know that it also works for ${\sin }^{-1}(x)$? Just in case not, here it is:

2 Inverses?

Both $\ln (x)$ and ${\sin }^{-1}(x)$ can be thought of as inverses of more well-known functions1. So maybe this trick is about inverses?

This doesn't seem all that enlightening, but let's do a cunning substitution2 and see what happens.

Let $y=f^{-1}(x)$, so $x=f(y)$ and "$dx=f\text{'}(y)dy$". Then:

Integrating this relies on being able to integrate $f(y)$, but with inverse functions this might well be a more possible prospect.

This substitution, though, has more to it than is apparent at first. Let's do the substitution throughout but also being a little choosy about when to substitute.

We have $y=f^{-1}(x)$ so we can write the initial integral as just:

The integrated term is $x{f}^{-1}(x)=xy$. Then the last term is:

So the whole equation reads:

Or, a bit more symmetrically,

3 Wait!? What was integration again?

Integration calculates (signed) area. Given a function $y=g(x)$ then ${\int }_a^{b}ydx$ calculates the area between $g(x)$ and the $x$–axis over the interval $[a,b]$.

Also given a function $x=h(y)$ then ${\int }_c^{d}xdy$ calculates the area between $h(y)$ and the $y$–axis "over" the interval $[c,d]$. These are illustrated in the two graphs in Figure 1.

Now, what would happen if those two curves happened to be the same, as in Figure 2?

Well, then we would have:

This looks nice! When does this happen?

It happens precisely when the two functions are inverse to each other. That is, $x=f(y)$ and $y=f^{-1}(x)$.

Which is where we came in.

The product rule is definitely lurking there in the background, but I find that this does neatly explain the connection between the two integrations. There is a clear connection between the integral of an inverse function and the integral of the original.

The final piece is to relate the limits, but that is straightforward. Since our focus is on $f^{-1}$, it is most natural to write them as:

4 Back to Logs

Let's work it through with $\ln (x)$. We'll do the definite integral, from which we can recover the indefinite by appropriate substitution of variables, but the definite helps keep the various letters in their correct roles.

Our relationship states that:

Substituting in $y=\ln (x)$ with limits $d=\ln (b)$ and $c=\ln (a)$, we get:

Switching to indefinite integrals, we get:

as expected.

5 Sign of the Times

One should always be a bit sceptical of a "proof by pictures". Here, the underlying assumptions in my pictures were that everything was positive: my limits $a$, $b$, $c$, $d$ were all positive and my function was increasing3. What if it weren't so?

One thing we must have is that the function be monotonic since it has an inverse. But all bets are off on everything else.

I shan't cover all possible cases, but here's an extreme to show that this still works. Consider the diagram in Figure 3.

With $A_i$ being the positive area of its region, we have:

Note that since $d<c$ then the second integral is the negative of what one might expect.

Adding these gives:

Since, in this example, $a<0$ and $c>0$, the positive area $A_1+A_2$ is equal to $-ac$. Similarly, the positive area $A_4+A_5$ is equal to $-bd$. Putting those together yields:

exactly as before.

6 Afterword

This came about from a conversation with a student about integration and ${\sin }^{-1}$. In fact, it was a misunderstanding on my part in that they meant the integration that has answer ${\sin }^{-1}$ whereas I thought they were asking about integrating ${\sin }^{-1}$ and it got me thinking as to how to actually do it. I'd just been teaching about inverse functions (to a different set of students) so the relationship between the graphs was at the forefront of my mind and the connection became very clear.

I'll need to think it through carefully, but this does feel like something quite teachable and might make that oNe wEiRd tRicK perhaps a little less mystifying.

Finally, everything I write on this website comes with the disclaimer that I make no claims to originality – all I require is that it catch my eye for some reason and that there be something for me to work through to make it worth recording.

I guess I'm keeping a record of the times when I was one of today's ten thousand and I publish them in case there's someone else out there in the same position.