Calculating Singular Integrals using Hadamard Finite Part Method

[zetafunction]

Could someone provide a reference to calculate this kind of integrals ? for example

\int_{0}^{2}dx \frac{cos(x)}{x-1}

or in 3-D \iiint_{D}dx \frac{x-y+z^{2})}{x+y+z}

Where 'D' is the cube [-1,1]x[-1,1]x[-1,1]=D

as you can see there is a singularity at x=1 or whenever x+y+z=0 , perhaps the other integral is easier to define if we use polar coordinates , so the singularities appear when r=0

 

[Cyosis]

According to mathematica the integral does not converge.

 

[tiny-tim]

Yes, ∫cosx/(x-1) dx near x = 1 is (cos1)∫dx/(x-1) = (cos1)[log(x-1)], which obviously is infinite.

 

[zetafunction]

they exists in an especial sense known as hadamard integral

for the case 1-D you integrate and simply ignore the terms giving you an infinite Answer this is why is called 'finite part'

some more info http://books.google.es/books?id=E5a...X&oi=book_result&ct=result&resnum=6#PPA511,M1

for a brief description of what this integral is.

 

[Gib Z]

Are you sure you didn't mean something else? Another variable is in the integral of the usual definition: http://en.wikipedia.org/wiki/Hadamard_finite_part_integral .

 

[zetafunction]

yes that is the definition , but in general you drop the divergent term dvided by epsilon and take only the finite value , that is for 1-D for 3-D or similar i do not know what can be done, or if the integral is divergent at infinity for example

\int_{0}^{\infty}dxx^{3}cos(x)