# Thread: Higher Level Integration Marathon & Questions

1. ## Re: Higher Level Integration Marathon & Questions

Show that

$\int_0^{\tfrac{\sqrt{6}-\sqrt{2}-1}{\sqrt{6}-\sqrt{2}+1}} \dfrac{\ln x}{(x-1)\sqrt{x^2-2(15+8\sqrt{3})x+1}}dx=\dfrac{2}{3}(2-\sqrt{3})G$

where G is Catalan's constant.

2. ## Re: Higher Level Integration Marathon & Questions

@Paradoxica I think I finally figured what it should've been.

$\int_1^e x \sqrt[6]{x^{-1} \sqrt[20]{x \sqrt[42]{x^{-1} \dots}}}\,dx$

Well, hopefully.

3. ## Re: Higher Level Integration Marathon & Questions

$\int_0^{10} x^{\lim_{n\to \infty} f^n(x)}\,dx \text{ for }f(x) = \frac12 \left( x + \frac2x\right)$

$\text{where }f^n(x) = \underbrace{f \cdots f}_{n\text{-times}}(x)$

Shouldn't be hard assuming I didn't mess up typing the question.

4. ## Re: Higher Level Integration Marathon & Questions

Originally Posted by leehuan
$\int_0^{10} x^{\lim_{n\to \infty} f^n(x)}\,dx \text{ for }f(x) = \frac12 \left( x + \frac2x\right)$

$\text{where }f^n(x) = \underbrace{f \cdots f}_{n\text{-times}}(x)$

Shouldn't be hard assuming I didn't mess up typing the question.
For all positive real numbers, the iteration derived from Newton's Method applied to the equation q(x)=x²-2 results in the f(x) described above, and converges to the positive real root of the equation.

Hence, fn(x) converges to √ ̅2 for x>0

The point x=0 can be discarded, as it is a set of zero measure.

Hence, the integral is equivalent to

$\int_0^{10} x^{\sqrt{2}} \text{d}x \\\\ = \frac{10^{\sqrt{2}+1}}{\sqrt{2}+1} = 10^{\sqrt{2}+1}\left(\sqrt{2}-1\right)$

My only problem with this question is that students don't know how to handle problematic points on the domain of an integral.

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