Let the function (of $z$) be defined by the formula $$ \int \limits_1^\infty\frac{dt}{t^{\large z}\sqrt{t^2-1}}. $$ Is it known as a special function?
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If it converges, it is known as particular case of a constant function. – Git Gud Aug 23 '13 at 17:54
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3Please observe t^z in the denominator. – limanac Aug 23 '13 at 17:55
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It looked like a $2$. Let me help you with that. – Git Gud Aug 23 '13 at 17:55
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1It is a ratio of two gamma functions. – Start wearing purple Aug 23 '13 at 17:55
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Ok, I will write the details of the derivation. – Start wearing purple Aug 23 '13 at 18:01
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And if $Re, z>1$, it seems the same formula holds? – limanac Aug 23 '13 at 18:04
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This integral can be written in terms of beta/gamma functions. Namely, making change of variables $u=\frac{t^2-1}{t^2}$ transforms it into \begin{align} \int_0^{1}\underbrace{(1-u)^{z/2}}_{=t^{-z}}\times \underbrace{\sqrt{\frac{1-u}{u}}}_{=(t^2-1)^{-1/2}}\times \underbrace{\frac{1}{2}(1-u)^{-3/2}du}_{=dt}=\\ =\frac12\int_0^1u^{-1/2}(1-u)^{z/2-1}du=\\ =\frac12 B\left(\frac12,\frac{z}{2}\right) =\frac{1}{2}\frac{ \Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{z}{2}\right)}{\Gamma\left(\frac{z+1}{2}\right)} =\frac{\sqrt{\pi}}{2}\frac{ \Gamma\left(\frac{z}{2}\right)}{\Gamma\left(\frac{z+1}{2}\right)}. \end{align}
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