Cauchy integral also for b< e^(1/e)?
#14
Ansus Wrote:\( \Delta[f]=\exp f - f \)

Since \( \Delta[f] = -\frac{1}{2\pi i}
\int_{-1-i\infty}^{-1+i\infty} \frac{f(z+x)}{z(z-1)}\, dz \), we derive f(x).

But now I doubt the formula is true.
Setting for example \( f=\exp \).
Then
\( \exp(\exp(0)) - \exp(0) = -\frac{1}{2\pi i}
\int_{-1-i\infty}^{-1+i\infty} \frac{\exp(z)}{z(z-1)}\, dz \)
But if I compute this numerially I get on the right side something close to 0.
While the left side is \( e - 1 \).

Where did you get this formula? Is it applicable only to certain functions?
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