advanced-calculus-prove-that-n-0-n-1-2-n-1-2-2-n-n-2pi-ln-2-

Question Number 133228 by mnjuly1970 last updated on 20/Feb/21

\dots . a d v a n c e d c a l c u l u s \dots .

p r o v e t h a t ::

\underset{n = 0}{\sum^{\infty}} \frac{Γ (n + \frac{1}{2}) ψ (n + \frac{1}{2})}{2^{n} . n!} = - \sqrt{2 π} (γ + l n (2)) \dots .

Answered by Dwaipayan Shikari last updated on 20/Feb/21

\underset{n = 0}{\sum^{\infty}} \frac{Γ^{'} (n + \frac{1}{2})}{2^{n} n!} = \underset{n = 0}{\sum^{\infty}} \frac{1}{2^{n} n!} \int_{0}^{\infty} x^{n - \frac{1}{2}} e^{- x} l o g (x)

= \int_{0}^{\infty} \underset{n = 0}{\sum^{\infty}} \frac{{(\frac{x}{2})}^{n}}{n!} x^{- \frac{1}{2}} e^{- x} l o g (x)

= \int_{0}^{\infty} e^{\frac{x}{2}} x^{- \frac{1}{2}} e^{- x} l o g (x) d x = \int_{0}^{\infty} x^{- \frac{1}{2}} e^{- \frac{x}{2}} l o g (x) d x

= \sqrt{2} \int_{0}^{\infty} u^{- \frac{1}{2}} e^{- u} l o g (2 u) = \sqrt{2 π} l o g (2) + \sqrt{2} Γ^{'} (\frac{1}{2})

= \sqrt{2 π} l o g (2) + \sqrt{2 π} (- γ - 2 l o g (2))

= - \sqrt{2 π} (γ + l o g (2))

Commented by mnjuly1970 last updated on 20/Feb/21

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