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Martin Sleziak
Martin Sleziak
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(I'm adding this as an answer since my original post won't let me add any more info, and the formula is too big for a comment.)

There is a variant on the formula as follows \begin{equation} \frac{\Gamma(a ) \Gamma(1-a-b)}{ \Gamma(1-b)} + \frac{\Gamma(b) \Gamma(1-a-b)}{ \Gamma(1-a)} - \frac{\Gamma(a) \Gamma(b)}{ \Gamma(a+b)} = \pi^{\frac12} \frac{\Gamma\left(\frac{ 1-a-b}{2}\right) }{\Gamma\left(\frac{a+b}{2}\right)} \frac{\Gamma\left(\frac{1+a}{2}\right)}{\Gamma\left(\frac{2-a}{2}\right)} \frac{\Gamma\left(\frac{1+b}{2}\right)}{\Gamma\left(\frac{2-b}{2}\right)}. \end{equation}

By adding/subtracting it with the original formula one gets an identity with the sum of two terms on each side. I didn't find it helpful but perhaps it sparks an idea with someone.

In the context of my original problem, the identity with + corresponds to a sum over even Dirichlet characters, while the identity in this answer corresponds to odd characters. The gamma factors on the right hand side are relevant for the functional equation of the Dirichlet L-functionsfunctional equation of the Dirichlet L-functions which depends on the parity of the character.

(I'm adding this as an answer since my original post won't let me add any more info, and the formula is too big for a comment.)

There is a variant on the formula as follows \begin{equation} \frac{\Gamma(a ) \Gamma(1-a-b)}{ \Gamma(1-b)} + \frac{\Gamma(b) \Gamma(1-a-b)}{ \Gamma(1-a)} - \frac{\Gamma(a) \Gamma(b)}{ \Gamma(a+b)} = \pi^{\frac12} \frac{\Gamma\left(\frac{ 1-a-b}{2}\right) }{\Gamma\left(\frac{a+b}{2}\right)} \frac{\Gamma\left(\frac{1+a}{2}\right)}{\Gamma\left(\frac{2-a}{2}\right)} \frac{\Gamma\left(\frac{1+b}{2}\right)}{\Gamma\left(\frac{2-b}{2}\right)}. \end{equation}

By adding/subtracting it with the original formula one gets an identity with the sum of two terms on each side. I didn't find it helpful but perhaps it sparks an idea with someone.

In the context of my original problem, the identity with + corresponds to a sum over even Dirichlet characters, while the identity in this answer corresponds to odd characters. The gamma factors on the right hand side are relevant for the functional equation of the Dirichlet L-functions which depends on the parity of the character.

(I'm adding this as an answer since my original post won't let me add any more info, and the formula is too big for a comment.)

There is a variant on the formula as follows \begin{equation} \frac{\Gamma(a ) \Gamma(1-a-b)}{ \Gamma(1-b)} + \frac{\Gamma(b) \Gamma(1-a-b)}{ \Gamma(1-a)} - \frac{\Gamma(a) \Gamma(b)}{ \Gamma(a+b)} = \pi^{\frac12} \frac{\Gamma\left(\frac{ 1-a-b}{2}\right) }{\Gamma\left(\frac{a+b}{2}\right)} \frac{\Gamma\left(\frac{1+a}{2}\right)}{\Gamma\left(\frac{2-a}{2}\right)} \frac{\Gamma\left(\frac{1+b}{2}\right)}{\Gamma\left(\frac{2-b}{2}\right)}. \end{equation}

By adding/subtracting it with the original formula one gets an identity with the sum of two terms on each side. I didn't find it helpful but perhaps it sparks an idea with someone.

In the context of my original problem, the identity with + corresponds to a sum over even Dirichlet characters, while the identity in this answer corresponds to odd characters. The gamma factors on the right hand side are relevant for the functional equation of the Dirichlet L-functions which depends on the parity of the character.

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Matt Young
Matt Young
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(I'm adding this as an answer since my original post won't let me add any more info, and the formula is too big for a comment.)

There is a variant on the formula as follows \begin{equation} \frac{\Gamma(a ) \Gamma(1-a-b)}{ \Gamma(1-b)} + \frac{\Gamma(b) \Gamma(1-a-b)}{ \Gamma(1-a)} - \frac{\Gamma(a) \Gamma(b)}{ \Gamma(a+b)} = \pi^{\frac12} \frac{\Gamma\left(\frac{ 1-a-b}{2}\right) }{\Gamma\left(\frac{a+b}{2}\right)} \frac{\Gamma\left(\frac{1+a}{2}\right)}{\Gamma\left(\frac{2-a}{2}\right)} \frac{\Gamma\left(\frac{1+b}{2}\right)}{\Gamma\left(\frac{2-b}{2}\right)}. \end{equation}

By adding/subtracting it with the original formula one gets an identity with the sum of two terms on each side. I didn't find it helpful but perhaps it sparks an idea with someone.

In the context of my original problem, the identity with + corresponds to a sum over even Dirichlet characters, while the identity in this answer corresponds to odd characters. The gamma factors on the right hand side are relevant for the functional equation of the Dirichlet L-functions which depends on the parity of the character.