For every we have

So by the fundemental theorem of calculus we have for any interval in

How can I proof this using the inverse function theorem?

Thanks

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- November 15th 2009, 07:34 AMbram kierkelsLogarithm Integral
For every we have

So by the fundemental theorem of calculus we have for any interval in

How can I proof this using the inverse function theorem?

Thanks - November 15th 2009, 08:04 AMtonio

"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 in every closed interval , as you wrote...there's nothing else to it

Tonio - November 15th 2009, 08:49 AMHallsofIvy
The "inverse function theorm" asserts that if , then 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, 08:50 AMbram kierkels
Ah, I was a bit confused myself. The second phrase was just a remark. The question is only to proof 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

So the answer is like this i think;

if

- November 15th 2009, 09:03 AMtonio