# Integral of ln (x)

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• Jan 26th 2010, 02:51 PM
kid funky fried
Integral of ln (x)
First, I must disclose that this is a homework problem.

Here is the integral exactly as stated:

Evaluate

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

So this is what I did:

$
\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.
• Jan 26th 2010, 02:55 PM
General
Quote:

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

Here is the integral exactly as stated:

Evaluate

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

So this is what I did:

$
\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.
$\int_0^1 ln(x) dx=\lim_{t \to 0^+} \int_t^1 ln(x) dx$.
• Jan 26th 2010, 02:56 PM
adkinsjr
Quote:

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

Here is the integral exactly as stated:

Evaluate

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

So this is what I did:

$
\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.

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

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