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Question Number 133228 by mnjuly1970 last updated on 20/Feb/21
....advanced calculus.... prove that :: Σ_(n=0) ^∞ ((Γ(n+(1/2))ψ(n+(1/2)))/(2^n .n!))=−(√(2π)) (γ+ln(2))....
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….{advanced}\:\:\:\:{calculus}…. \\ $$$$\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\psi\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}^{{n}} .{n}!}=−\sqrt{\mathrm{2}\pi}\:\left(\gamma+{ln}\left(\mathrm{2}\right)\right)…. \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 20/Feb/21
Σ_(n=0) ^∞ ((Γ′(n+(1/2)))/(2^n n!))=Σ_(n=0) ^∞ (1/(2^n n!))∫_0 ^∞ x^(n−(1/2)) e^(−x) log(x) =∫_0 ^∞ Σ_(n=0) ^∞ ((((x/2))^n )/(n!))x^(−(1/2)) e^(−x) log(x) =∫_0 ^∞ e^(x/2) x^(−(1/2)) e^(−x) log(x)dx =∫_0 ^∞ x^(−(1/2)) e^(−(x/2)) log(x)dx =(√2)∫_0 ^∞ u^(−(1/2)) e^(−u) log(2u)=(√(2π)) log(2)+(√2)Γ′((1/2)) =(√(2π)) log(2)+(√(2π)) (−γ−2log(2)) =−(√(2π)) (γ+log(2))
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\Gamma'\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}^{{n}} {n}!}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{n}} {n}!}\int_{\mathrm{0}} ^{\infty} {x}^{{n}−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{x}} {log}\left({x}\right) \\ $$$$=\int_{\mathrm{0}} ^{\infty} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{{x}}{\mathrm{2}}\right)^{{n}} }{{n}!}{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{x}} {log}\left({x}\right) \\ $$$$=\int_{\mathrm{0}} ^{\infty} {e}^{\frac{{x}}{\mathrm{2}}} {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{x}} {log}\left({x}\right){dx}\:=\int_{\mathrm{0}} ^{\infty} {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−\frac{{x}}{\mathrm{2}}} {log}\left({x}\right){dx}\:\:\:\:\:\: \\ $$$$=\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {u}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{u}} {log}\left(\mathrm{2}{u}\right)=\sqrt{\mathrm{2}\pi}\:{log}\left(\mathrm{2}\right)+\sqrt{\mathrm{2}}\Gamma'\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\sqrt{\mathrm{2}\pi}\:{log}\left(\mathrm{2}\right)+\sqrt{\mathrm{2}\pi}\:\left(−\gamma−\mathrm{2}{log}\left(\mathrm{2}\right)\right) \\ $$$$=−\sqrt{\mathrm{2}\pi}\:\left(\gamma+{log}\left(\mathrm{2}\right)\right) \\ $$
Commented by mnjuly1970 last updated on 20/Feb/21
very nice thank you mr payan...
$${very}\:{nice}\: \\ $$$${thank}\:{you}\:{mr}\:{payan}… \\ $$

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