Recently I came across this identity: $$ \prod _{n=1}^{{\infty }}\:\tanh \:\left(\frac{\pi n}{2}\right)=\frac{\Gamma \left(\frac{1}{4}\right)}{\sqrt[4]{8\pi ^3}} $$ The question is: how to prove this identity? Is there any theorem that can help prove this relation?
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2A strange connection with the formula there – Jean Marie Mar 21 '20 at 13:24
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An almost identical question here. See as well this – Jean Marie Mar 21 '20 at 13:26
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Nice one. Let us start as follows. $$ \prod _{n=1}^{{\infty }}\:\tanh \:\left(\frac{\pi n}{2}\right)=\left.\prod _{n=1}^{{\infty }}\frac{1-q^n}{1+q^n}\right|_{q=e^{-\pi}}. $$ You may now observe (term by term) that the latter product actually reads as follows $$ \prod _{n=1}^{{\infty }}\frac{1-q^n}{1+q^n} = 1-2q+2q^4 + ... = 1+2\sum_{k=1}^{\infty}(-1)^k q^{k^2}=\vartheta_4(0,q) $$ where the last equality follows directly from the definition of Jacobi theta function $\vartheta_4 (z,q)$. Now, your identity follows from known results for $\vartheta_4(0,q)$. First, let us use connection to Dedekind eta: $$\vartheta_4(0,q=e^{\pi i\tau})=\frac{\eta^2(\tau/2)}{\eta(\tau)},$$ where we have $\tau=i$. The final step is to use special values of Dedekind eta to obtain your identity.
stokes-line
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Thank you for your solution! Do you know how special values of theta function are derived,is there a proof? – mathlover123 Mar 21 '20 at 17:30
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Jean Marie posted several links above, it turns out that this identity is popular. – stokes-line Mar 21 '20 at 17:47