Higher Level Integration Marathon & Questions (1 Viewer)

calamebe

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Really hope I'm right coz this took me forever. Also sorry for bad LaTeX.







Let



where

This is easily evaluated depending on the different cases:

If or , we have , if then and , and if then and .

Case 1:

hence for some constant .

Evaluating the integral for , we obtain that:

when .

Case or .

(which follows from Case 1).

when or .

Case 3:



Let

By symmetry,

Let



By symmetry,

Therefore , so .

Subbing the values back in, we obtain .

Case 4:



This can be done by the same method as case 3, and so .

Hence, when , , and otherwise.
 
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Paradoxica

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Bonus Challenge: Prove the last two integrals are symmetric in (a,b) <=> (b,a) without evaluating it to the final result.
 
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leehuan

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...but unfortunately I don't know how to prove that final integral equals to negative gamma.
 

leehuan

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



Well, hopefully.
 

leehuan

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

Paradoxica

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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



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

blyatman

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Would've been so much nicer if it was a cos instead of a sin lol.

Been a while since I've done one of these, but I'm guessing: take the CCW quarter-circle contour C around the first quadrant with radius R approaching infinity, so that:

where C1 is the quarter circle arc (from z = R to z = iR), C2 goes from iR to 0, and C3 goes from 0 to R?

The contour over C can then be evaluated using Residue theorem, with with z = +i being the only singularity within C.
The contours over C1 and C2 can be evaluted by using an appropriate parameterisation (alternatively, I'm thinking that the integral over C1 just be 0 since the denominator goes as R^2?).

The original integral can then be extracted (or at least I think it can...) by taking imaginary parts of both sides, so that


Is this correct? I'm a bit hesitant on that last step, not entirely convinced myself if that last expression holds.

I don't normally spend time thinking about integration questions, so that's enough integration for me for the next year haha.
 
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leehuan

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Lol my bad. Idk why I decided to switch the cos out for a sin when posting
 

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