zetafunction
Jan22-11, 03:46 AM
let be the integral \int_{0}^{2}dx \frac{f(x)}{(x-1)} is singular at x=1 so i use the Shokothsky formula to set the 'regularized' value
\int_{0}^{2}dx \frac{f(x)}{(x+i\epsilon -1)} +i\pi f(1)
is this correct? i mean since the 'epsilon' is positive let us say approximately 0.001 the first integral now exists in Riemann sense so this can be taken as the regularized value
\int_{0}^{2}dx \frac{f(x)}{(x+i\epsilon -1)} +i\pi f(1)
is this correct? i mean since the 'epsilon' is positive let us say approximately 0.001 the first integral now exists in Riemann sense so this can be taken as the regularized value