• Best of luck to the class of 2019 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here

Higher Level Integration Marathon & Questions (1 Viewer)

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,671
Gender
Male
HSC
2007
^ That was good practice!

I need to go out for dinner v soon, so will supply gory details later. In short, the answer can be expressed solely in terms of Apery's constant zeta(3).

1. IBP to essentially reduce the integral to log^2(1+x)/x. (The bulk of the difficulty of the originally posted integral lies in this integral as I see it.)

2. Expand this out to obtain an expression in terms of polylogs of order 2 and 3, but we basically only need to know the value of these polylogs at 1/2 and 1.

3. Special functions funtimes proving some reflection formulae etc to compute these guys.
 

Paradoxica

-insert title here-
Joined
Jun 19, 2014
Messages
2,477
Location
Outside reality
Gender
Male
HSC
2016
^ That was good practice!

I need to go out for dinner v soon, so will supply gory details later. In short, the answer can be expressed solely in terms of Apery's constant zeta(3).

1. IBP to essentially reduce the integral to log^2(1+x)/x. (The bulk of the difficulty of the originally posted integral lies in this integral as I see it.)

2. Expand this out to obtain an expression in terms of polylogs of order 2 and 3, but we basically only need to know the value of these polylogs at 1/2 and 1.

3. Special functions funtimes proving some reflection formulae etc to compute these guys.
I was looking for a polylog-free solution, but I guess this puts the nail on the coffin :(
 

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,671
Gender
Male
HSC
2007
I was looking for a polylog-free solution, but I guess this puts the nail on the coffin :(
Eh I wouldn't say that necessarily, I did the integral in a not particularly imaginative way...there could be a clever shortcut.

To me it does seem like computing this integral is roughly the same mathematical "depth" as computing some particular values of the di/trilogarithm though.

Anyway, is there a reason why you were seeking to avoid them?
 

Paradoxica

-insert title here-
Joined
Jun 19, 2014
Messages
2,477
Location
Outside reality
Gender
Male
HSC
2016
Eh I wouldn't say that necessarily, I did the integral in a not particularly imaginative way...there could be a clever shortcut.

To me it does seem like computing this integral is roughly the same mathematical "depth" as computing some particular values of the di/trilogarithm though.

Anyway, is there a reason why you were seeking to avoid them?
because I like a nice balance of low tech and efficient solutions to integrals
 

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,671
Gender
Male
HSC
2007
because I like a nice balance of low tech and efficient solutions to integrals
Fair enough, seems a rather subjective matter...to me polylog and polygamma manipulations are okay, but I also have no qualms about using contour integration etc.
 

He-Mann

Vexed?
Joined
Sep 18, 2016
Messages
279
Location
Antartica
Gender
Male
HSC
N/A
I think you guys should enrol in Extreme Integration (MATH1052), or if you're feeling ambitious try Insane Integration (MATH3025)!

PM me for details.
 

Paradoxica

-insert title here-
Joined
Jun 19, 2014
Messages
2,477
Location
Outside reality
Gender
Male
HSC
2016




Prove the integral diverges if the argument of the exponential does not describe an ellipse.

Challenge: Formulate a generalisation to the above integral in higher dimensions. (Experience with Linear Algebra will be very handy)
 
Last edited:

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,097
Gender
Male
HSC
N/A




Prove the integral diverges if the argument of the exponential does not describe an ellipse.

Challenge: Formulate a generalisation to the above integral in higher dimensions. (Experience with Linear Algebra will be very handy)


 

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,671
Gender
Male
HSC
2007
Oh didn't see you edited it, was the divergence thing always there and I just blind before?

In any case, we can still represent the quadratic form by a symmetric, hence diagonalisable matrix Q. If the quadratic form does not represent an ellipse, then at least one eigenvalue is non-positive, which implies that at least one of the integrals in the above product of single-variable integrals is infinite and we are done.
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,097
Gender
Male
HSC
N/A
Oh didn't see you edited it, was the divergence thing always there and I just blind before?

In any case, we can still represent the quadratic form by a symmetric, hence diagonalisable matrix Q. If the quadratic form does not represent an ellipse, then at least one eigenvalue is non-positive, which implies that at least one of the integrals in the above product of single-variable integrals is infinite and we are done.
I think it was there before. I remember answering it in my head but then forgetting to comment on it in my post. Haha
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top