2
$\begingroup$

First, this is not homework; I just decided to try a classic integral in a non-standard way and came out with a strange result.

The integral $I:=\int\frac{dx}{x\ln x}$ is well-known to equal $\ln\ln x + C$, and the result is easily obtained with the substitution $u = \ln x$. I tried the alternate method of integration by parts:

$$I=\int\frac{dx}{x\ln x} = \int\frac{1}{\ln x}\frac{dx}{x}$$

Integrating by parts with $u = \frac{1}{\ln x} \implies du = -\frac{dx}{x\ln^2 x}$ and $dv = \frac{dx}{x} \implies v = \ln x$ gives:

$$\int\frac{1}{\ln x}\frac{dx}{x} = \frac{\ln x}{\ln x} - \int \ln x(-\frac{dx}{x\ln^2 x}) = 1 + \int\frac{dx}{x\ln x} = 1 + I$$

This leads to $I=1+I$. Since this is clearly wrong, I'm going to assume I've made a stupid arithmetic mistake somewhere down the line; but after trying it twice I came to this same barrier. What have I done wrong?

$\endgroup$
1
  • 2
    $\begingroup$ Hint: Constant of integration! $\endgroup$ Jun 29 '14 at 21:59
1
$\begingroup$

Nothing; remember that that antiderivatives always have that arbitrary constant $+C$ at the end.

$\endgroup$
2
  • $\begingroup$ ...oh. I didn't actually gain any information about the primitive from that, did I? I see. $\endgroup$
    – theage
    Jun 29 '14 at 22:01
  • $\begingroup$ Nope, none whatsoever. Like so many other things in math, what you've done is absolute Gospel truth, but utterly useless :) $\endgroup$
    – user61527
    Jun 29 '14 at 22:02

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.