Cauchy principal value integral



Edwin Woollett wrote:

> 
> On 4-8-09 Michel Talon wrote

> 
> 2. The symbolic (operator) expression you have written down above actually
>     should be written:
> 
>     limit ( 1/(x - %i*eps), eps, 0, plus ) = P( 1/x ) +  %i*%pi*delta(x)
> 
>     which only has meaning when both sides are multiplied by f(x)*dx and
> integrated over
>     an interval which includes x = 0, in which case it means
> 
>     limit( integrate ( f(x)/(x - %i*eps), x, a, b), eps, 0, plus ) = pri (
> f(x), x, 0, a, b) + %i*%pi*f(0)
> 
>     provided a < 0 < b and f(x) is a smooth function near x = 0.
> 
>     (Refs: a. Paul Roman, Advanced Quantum Theory, Addison-Wesley, 1965,
>     p.
> 718
>               b. R. Shankar, Principles of Quantum Mechanics, 2nd ed,
>               Plenum
> Press, 1994,
>                             page 662.
> 

This is perfectly true and well documented. It happens that in the physics
context you mention (mostly quantum field theory) the small %i*eps
is always here implicitely or explicitely, and is specified by causality.
Mathematically it allows to close the Cauchy contour on the appropriate half
plane to pick the pole. People don't care one instant if the integral at
infinity vanishes, or other similar hair splitting, in fact the formula is
used in a context where everything is highly singular...
 

> 3. Since integrate ( 1/x,  x,  -1, 2 ) involves the integration of a real
> function
>    along the real axis,  the result  must be a real number, and there is
>    no physics magic which can turn a real number into an imaginary number.

This integral doesn't exist, in the mathematical sense, period. You can find
any real answer you want if you approach 0 in an unsymmetric way. The
principal part prescription (approach 0 from right and left in a symmetric
way) is a totally ad-hoc prescription which has no justification. At least
the physicist formula you explained above has justification, by the residue
theorem. I think i have read here that Maxima does some definite integrals
using the residue theorem, this may explain the answer it finds. However i
am puzzled by the factor 2 in 2*%i*%pi.

> 
> Ted Woollett

-- 
Michel Talon