# Thread: Integral of ln (x)

1. ## Integral of ln (x)

First, I must disclose that this is a homework problem.

Here is the integral exactly as stated:

Evaluate

$\displaystyle \int\limits_0^1 {\ln x} dx$

So this is what I did:

$\displaystyle \int\limits_0^1 {\ln x} dx = x\ln x - x|_0^1$

However, ln 0 is undefined. So how do I proceed? Or, is it possible to proceed?

Any help is greatly appreciated.

2. Originally Posted by kid funky fried
First, I must disclose that this is a homework problem.

Here is the integral exactly as stated:

Evaluate

$\displaystyle \int\limits_0^1 {\ln x} dx$

So this is what I did:

$\displaystyle \int\limits_0^1 {\ln x} dx = x\ln x - x|_0^1$

However, ln 0 is undefined. So how do I proceed? Or, is it possible to proceed?

Any help is greatly appreciated.
No, This is called an "improper integral", It should be solved in terms of limits.
$\displaystyle \int_0^1 ln(x) dx=\lim_{t \to 0^+} \int_t^1 ln(x) dx$.

3. Originally Posted by kid funky fried
First, I must disclose that this is a homework problem.

Here is the integral exactly as stated:

Evaluate

$\displaystyle \int\limits_0^1 {\ln x} dx$

So this is what I did:

$\displaystyle \int\limits_0^1 {\ln x} dx = x\ln x - x|_0^1$

However, ln 0 is undefined. So how do I proceed? Or, is it possible to proceed?

Any help is greatly appreciated.
This is an improper integral.

$\displaystyle \int_0^1ln(x)dx=\lim_{t->0^+}\int_t^1ln(x)dx$

Replace the 0 by t in your equation you obtained by IBP and then calculate this as a limit.