Question

integral(ln x dx) from [0,1].
Normally. you'd get negative infinity for that, right? ( Because 0 - infinity is - infinity ), or |-inf| = inf, since some say that area should be absolute.
Anyway, is there any way to play with the form as to get -1 instead?

Answer 1

Sorry, I misread it the first time. The value of that integral is -1, what's your question?

Answer 2

\[ \int_0^1 \ln(x) dx = \left[ x\ln(x)-x\right]^1_0 = -1 - 0 = -1 \]

Answer 3

\[\int\limits \ln (x)*1 dx=\ln x \int\limits 1 dx-\int\limits \frac{ d }{ dx }\left( \ln x \right)\int\limits 1dx=(\ln x)x-\int\limits \frac{ 1 }{ x }*x dx\]
\[=x \ln x-x\]

Answer 4

I guess my question is why 0*(ln 0) - 0 = 0
doesn't ln 0 approach negative infinity?
0 * - inf = what? Isn't that undetermined and dependent on which is faster anyhow? Or does it really just simply reduce to zero?

Answer 5

@LanHikari22 the issue occurs when you reach the point `0*infinity`

that's an indeterminate form (one of many)

https://en.wikipedia.org/wiki/Indeterminate_form

Answer 6

Ah, yeah. It turns out that \(x\) approaches zero "faster" than \(\ln(x)\) approaches negative infinity, and the product goes to zero. This can be demonstrated in a number of ways, for example via L'Hospital's Rule.

Answer 7

you'll have to rewrite
\[\Large x\ln(x)\]
into
\[\Large \frac{\ln(x)}{1/x}\]
then use L'Hospital's rule

Answer 8

That just takes you into an infinite deriving loop though due to it all being indefinite

Answer 9

No it doesn't. Applying L'Hospital's rule gives
\[\frac{1/x}{-1/x^2} =-x \]
which approaches zero as \(x\rightarrow 0\).

Answer 10

Oh crap. For whatever reason, I never thought of cancelling, I just took myself into an infinite loop. This all makes sense now, thanks!