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

$$....advancedcalculus....$$$$provethat::$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{\Gamma }(n+\frac{1}{2})\psi (n+\frac{1}{2})}{2^n.n!}=-\sqrt{2\pi }(\gamma +ln\left(2\right))....$$$$$$

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

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{\Gamma }}^{\prime }(n+\frac{1}{2})}{2^{n}n!}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{2^{n}n!}{\int }_0^{\mathrm{\infty }}{x}^{n-\frac{1}{2}}{e}^{-x}log\left(x\right)$$$$={\int }_0^{\mathrm{\infty }}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{x}{2}\right)}^n}{n!}{x}^{-\frac{1}{2}}{e}^{-x}log\left(x\right)$$$$={\int }_0^{\mathrm{\infty }}{e}^{\frac{x}{2}}{x}^{-\frac{1}{2}}{e}^{-x}log\left(x\right)dx={\int }_0^{\mathrm{\infty }}{x}^{-\frac{1}{2}}{e}^{-\frac{x}{2}}log\left(x\right)dx$$$$=\sqrt{2}{\int }_0^{\mathrm{\infty }}{u}^{-\frac{1}{2}}{e}^{-u}log\left(2u\right)=\sqrt{2\pi }log\left(2\right)+\sqrt{2}{\mathrm{\Gamma }}^{\prime }\left(\frac{1}{2}\right)$$$$=\sqrt{2\pi }log\left(2\right)+\sqrt{2\pi }(-\gamma -2log\left(2\right))$$$$=-\sqrt{2\pi }(\gamma +log\left(2\right))$$

Commented by mnjuly1970 last updated on 20/Feb/21

$$verynice$$$$thankyoumrpayan...$$

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