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If we have an entire function $f(z)$, can we apply the Cauchy integration formula for derivative for $f(z)$ and integrate over $\Bbb R$ instead of simple closed curve, i.e.

is this formula is true:

f'(a_{0}) = \frac{1}{2\pi i}\int_{\Bbb R}\frac{f(t)}{(t-a_{0})^{2}}dt

where $f(t)$ is just $f(z)$ with $z=t\in \Bbb R$.

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    You will need some growth behaviour at infinity.2011-04-01

3 Answers 3

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It's not true in general. It would be true for $Im(a_0) > 0$ if you could ensure that the integral over a suitable return path in the upper half plane goes to 0. For example, it's true for $f(z) = e^{ikz}$ if $k \ge 0$.

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I know that the following analog to the Cauchy integral formula holds.

For $z$ in the upper half-plane, and $ h_z(\zeta) = \frac{1}{2 \pi i} \left( \frac{1}{\zeta - z} - \frac{1}{\zeta - \overline{z}}\right) $ we have $ f(z) = \int_{-\infty}^{\infty} f(t)h_z(t) dt $

To prove this, evaluate the integral via residue calculus. I think another way to prove this is to start with the Cauchy integral formula for a circle and use a conformal mapping between the unit disc and the upper half-plane to get the above formula.

I would think a similar analog of the Cauchy integral formula for derivatives holds.

Note the upper half-plane excludes the real line.

I know this does not address your exact question, but I hope it is helpful in some way.

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    Again, this is not true in general. The integral might not even converge.2011-05-01
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That's not True Sivaram; $f$ is entire so f' is also entire function, but it doesn't say that $f$ (or $f'$) is real for real $z$!

(I don't see the "add comment icon on my screen")

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    I think you need around 50 reputation points before you can leave a comment2011-03-02