Difference between revisions of "Hadamard gamma"
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[http://www.jstor.org/discover/10.2307/2309786?sid=21105065140641&uid=3739256&uid=2129&uid=70&uid=3739744&uid=4&uid=2 Leonhard Euler's Integral: A Historical Profile of the Gamma Function]<br /> | [http://www.jstor.org/discover/10.2307/2309786?sid=21105065140641&uid=3739256&uid=2129&uid=70&uid=3739744&uid=4&uid=2 Leonhard Euler's Integral: A Historical Profile of the Gamma Function]<br /> | ||
[http://link.springer.com/article/10.1007%2Fs12188-008-0009-5 A superadditive property of Hadamard's gamma function]<br /> | [http://link.springer.com/article/10.1007%2Fs12188-008-0009-5 A superadditive property of Hadamard's gamma function]<br /> | ||
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Latest revision as of 18:28, 24 May 2016
The Hadamard gamma function is defined by the formula $$H(x)=\dfrac{1}{\Gamma(1-x)} \dfrac{\mathrm{d}}{\mathrm{d}x} \log \left( \dfrac{\Gamma(\frac{1}{2}-\frac{x}{2})}{\Gamma(1-\frac{x}{2})} \right),$$ 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 █
Theorem: Define $\alpha_0 \approx 1.5031\ldots$ to be the only solution of the equation $H(2t)=2H(t)$. Then the inequality $H(x)+H(y) \leq H(x+y)$ holds for all real numbers $x,y$ with $x,y \geq \alpha$ if and only if $\alpha \geq \alpha_0$.
Proof: proof goes here █
See Also
References
Is the Gamma function misdefined?
Leonhard Euler's Integral: A Historical Profile of the Gamma Function
A superadditive property of Hadamard's gamma function
