# Logarithm Integral

• November 15th 2009, 06:34 AM
bram kierkels
Logarithm Integral
For every $x \in (0,\infty)$ we have $ln'(x) = \frac{1}{x}$
So by the fundemental theorem of calculus we have $\int_{[a,b]}\frac{1}{x}dx=ln(b)-ln(a)$ for any interval $[a,b]$ in $(0,\infty)$
How can I proof this using the inverse function theorem?
Thanks
• November 15th 2009, 07:04 AM
tonio
Quote:

Originally Posted by bram kierkels
For every $x \in (0,\infty)$ we have $ln'(x) = \frac{1}{x}$
So by the fundemental theorem of calculus we have $\int_{[a,b]}\frac{1}{x}dx=ln(b)-ln(a)$ for any interval $[a,b]$ in $(0,\infty)$
How can I proof this using the inverse function theorem?
Thanks

"Inverse function theorem"? What is that and what does inverse functions have to do with anything related to this problem? The value of that integral is that because of the Fundamental Theorem of Integral Calculus and because $\ln'x=\frac{1}{x}$ in every closed interval $[a,b]\subset(0,\infty)$, as you wrote...there's nothing else to it

Tonio
• November 15th 2009, 07:49 AM
HallsofIvy
The "inverse function theorm" asserts that if $\frac{df}{dx}(a)\ne 0$, then $f^{-1}$ exists on some neighborhood of a.

But I also do not see what that has to do with proving the fundamental theorem of calculus as applied to 1/x.
• November 15th 2009, 07:50 AM
bram kierkels
Quote:

Originally Posted by tonio
"Inverse function theorem"? What is that and what does inverse functions have to do with anything related to this problem? The value of that integral is that because of the Fundamental Theorem of Integral Calculus and because $\ln'x=\frac{1}{x}$ in every closed interval $[a,b]\subset(0,\infty)$, as you wrote...there's nothing else to it

Tonio

Ah, I was a bit confused myself. The second phrase was just a remark. The question is only to proof $ln'(x)=\frac{1}{x}$ i guess.
The inverse function theorem says that if f is continuously differentiable en has derrivative at a which is not equal to 0 then f is invertible in a neighborhood of a. The inverse is then also continuously differentiable.

And $(f^{-1})(f(a))=\frac{1}{f'(a)}$

So the answer is like this i think;
if $y=ln(x)$
$\Rightarrow x=e^y$
$\Rightarrow e^y\frac{dy}{dx}=1$
$\Rightarrow \frac{dy}{dx}=\frac{1}{e^y}$
$\Rightarrow \frac{d}{dx}ln(x)=\frac{1}{x}$
• November 15th 2009, 08:03 AM
tonio
Quote:

Originally Posted by bram kierkels
Ah, I was a bit confused myself. The second phrase was just a remark. The question is only to proof $ln'(x)=\frac{1}{x}$ i guess.
The inverse function theorem says that if f is continuously differentiable en has derrivative at a which is not equal to 0 then f is invertible in a neighborhood of a. The inverse is then also continuously differentiable.

And $(f^{-1})(f(a))=\frac{1}{f'(a)}$

So the answer is like this i think;
if $y=ln(x)$
$\Rightarrow x=e^y$
$\Rightarrow e^y\frac{dy}{dx}=1$
$\Rightarrow \frac{dy}{dx}=\frac{1}{e^y}$
$\Rightarrow \frac{d}{dx}ln(x)=\frac{1}{x}$

Exactly...good!

Tonio