Difference between revisions of "Hadamard gamma"

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(Properties)
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$$H(x)=\dfrac{\psi(1-\frac{x}{2})-\psi(\frac{1}{2}-\frac{x}{2})}{2\Gamma(1-x)},$$
 
$$H(x)=\dfrac{\psi(1-\frac{x}{2})-\psi(\frac{1}{2}-\frac{x}{2})}{2\Gamma(1-x)},$$
 
where $\psi$ is the [[digamma function]].
 
where $\psi$ is the [[digamma function]].
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<strong>Proof:</strong> proof goes here █
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<strong>Theorem:</strong> The function $H$ is an [[entire function]].
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<div class="mw-collapsible-content">
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<strong>Proof:</strong> proof goes here █
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</div>
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<strong>Theorem:</strong> The function $H$ satisfies the formula
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$$H(x+1)=xH(x)+\dfrac{1}{\Gamma(1-x)}.$$
 
<div class="mw-collapsible-content">
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong> proof goes here █  
 
<strong>Proof:</strong> proof goes here █  

Revision as of 22:59, 13 January 2015

The Hadamard gamma function is defined by the formula $$H(x)=\dfrac{1}{\Gamma(1-x)} \dfrac{d}{dx} \log \left( \dfrac{\Gamma(\frac{1}{2}-\frac{x}{2})}{\Gamma(1-\frac{x}{2})} \right),$$ where $\Gamma$ denotes the gamma function.

Properties

Theorem: We can write $$H(x)=\dfrac{\psi(1-\frac{x}{2})-\psi(\frac{1}{2}-\frac{x}{2})}{2\Gamma(1-x)},$$ where $\psi$ is the digamma function.

Proof: proof goes here █

Theorem: The function $H$ is an entire function.

Proof: proof goes here █

Theorem: The function $H$ satisfies the formula $$H(x+1)=xH(x)+\dfrac{1}{\Gamma(1-x)}.$$

Proof: proof goes here █

References

Is the Gamma function misdefined?