# Hadamard finite part integral

1. Jun 7, 2009

### 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

2. Jun 7, 2009

### Cyosis

According to mathematica the integral does not converge.

3. Jun 7, 2009

### tiny-tim

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

4. Jun 7, 2009

5. Jun 7, 2009

6. Jun 7, 2009

### 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

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

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