Hadamard finite part integral

**zetafunction** · Jun7-09, 05:00 AM

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

$LaTeX Code: \\int_{0}^{2}dx \\frac{cos(x)}{x-1}$

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

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** · Jun7-09, 05:16 AM

According to mathematica the integral does not converge.

**tiny-tim** · Jun7-09, 05:41 AM

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

**zetafunction** · Jun7-09, 05:47 AM

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=E5a1...um=6#PPA511,M1

for a brief description of what this integral is.

**Gib Z** · Jun7-09, 07:53 AM

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

**zetafunction** · Jun7-09, 09:49 AM

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

$LaTeX Code: \\int_{0}^{\\infty}dxx^{3}cos(x)$