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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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Note: $a_{0}\in \Bbb R$. –  user7531 Mar 2 '11 at 2:15
    
In other words, prove or disprove that: If $f(z)$ is an entire function, then $$f'(a_{0}) = \frac{1}{2\pi i}\int_{-\infty}^{\infty} \frac{f(t)}{(t-a_{0})^{2}}dt$$ where $f(t)$ is just $f(z)$ with $z=t\in \Bbb R$, and $a_{0}\in \Bbb R$. –  user7531 Mar 2 '11 at 2:40
    
You can edit the original question rather than make comments to update it. I'm not sure if you knew that. –  admchrch Mar 2 '11 at 2:59
    
Take f(t) = exp(t). Isn't that a counterexample? –  Johan Mar 2 '11 at 12:39
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You will need some growth behaviour at infinity. –  Thomas Rot Apr 1 '11 at 11:02
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3 Answers

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. –  Robert Israel May 1 '11 at 10:15
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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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Oops... My mistake I thought it was just a real valued function. Didnt see the function was entire. Deleted the post. –  user17762 Mar 2 '11 at 4:20
    
I think you need around 50 reputation points before you can leave a comment –  user17762 Mar 2 '11 at 4:21
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