# Higher Level Integration Marathon & Questions (1 Viewer)

##### -insert title here-
I think you guys should enrol in Extreme Integration (MATH1052), or if you're feeling ambitious try Insane Integration (MATH3025)!

PM me for details.
this feels like a meme

• He-Mann

#### leehuan

##### Well-Known Member
At least quick partial fractions is possible this time round lol

• Last edited:

#### calamebe

##### Active Member
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.

Last edited:
• leehuan and fluffchuck

##### -insert title here-
Really hope I'm right coz this took me forever. Also sorry for bad LaTeX.
Fourier Expansion • leehuan

#### calamebe

##### Active Member
Fourier Expansion No idea what that is, I'm not all that far beyond HSC maths ##### -insert title here-

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

Last edited:

#### leehuan

##### Well-Known Member

______________________________________________

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

#### Kingom

##### Member
Show that

where G is Catalan's constant.

#### leehuan

##### Well-Known Member
@Paradoxica I think I finally figured what it should've been.

Well, hopefully.

• #### leehuan

##### Well-Known Member

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

##### -insert title here-

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.

• leehuan

#### blyatman

##### Well-Known Member
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.

Last edited:
• • leehuan and Drdusk

#### leehuan

##### Well-Known Member
Lol my bad. Idk why I decided to switch the cos out for a sin when posting

#### blyatman

##### Well-Known Member
Lol my bad. Idk why I decided to switch the cos out for a sin when posting
Lol yeh would've taken 2 lines with the cos.