5
$\begingroup$

In one of my textbooks, the expansion of a logarithm product is proved using integrals.

$$\ln xy = \ln x + \ln y\iff \int_1^\left(xy\right)dt/t$$

$$\ = \int_1^xdt/t + \int_x^\left(xy\right)dt/t$$

Then, let $ u = t/x $ and substitute in the second integral:

$$= \int_1^x dt/t + \int_1^y du/u = ln x + ln y $$

While the expansion is rather quaint, I find the u-substitution difficult to follow:

If $u = t/x$ then $du/dt = 1/x$ and $du = dt/x $ ... I am unsure how $ du/u $ is obtained for substitution.

Any help or hints are appreciated!

$\endgroup$
  • $\begingroup$ That's all very interesting, but why would they choose to prove the log properties using integration? Surely the relationship to the power rules is much clearer? If you just make both sides powers of e, the relationship is proved almost instantly. $\endgroup$ – daviewales May 28 '13 at 16:15
  • $\begingroup$ @daviewales Conventionally, $\ln(x)$ is defined as the inverse of $e^x$. If on is to use this definition, their are some problems that must be overcome. The first of which is "what is the number $e$?" One this is done, we can define $e^x$ for any rational number $x$. But we now need a definition for $e^x$ when $x$ is irrational, which is not impossible, but which is also a little bit of work. So no the your proof is not instant as some steps are swept under the rug, although I think that this way is more conceptual. One way to subvert this possess is to define $\ln(x)=\int_1^x \frac{dt}{t}$. $\endgroup$ – Baby Dragon May 28 '13 at 16:42
  • $\begingroup$ OK. That's very interesting. $\endgroup$ – daviewales May 29 '13 at 1:13
6
$\begingroup$

Because $ u = \frac{t}{x} $, we have that $ t = ux \implies dt = x \cdot du $. Hence, $ \frac{1}{t} \ dt = \frac{1}{ux} \cdot x \ du = \frac{1}{u} \ du $. The bounds follow from the transformation and calculated independently of the integrand in this case.

$\endgroup$
  • $\begingroup$ Of course! Thanks! $\endgroup$ – d0rmLife May 28 '13 at 16:09

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.