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Question-196757




Question Number 196757 by MrGHK last updated on 31/Aug/23
Answered by qaz last updated on 31/Aug/23
(√π)Σ(((((Γ(n+1))/(Γ(n+(3/2)))))′)/(n+1))=Σ(1/(n+1))∫_0 ^1 x^n (1−x)^(−1/2) lnxdx  =−∫_0 ^1 ((lnxln(1−x))/(x(√(1−x))))dx=−(∂^2 /(∂a∂b))∣_(a=(1/2) ,b=0) ∫_0 ^1 x^(a−1) (1−x)^(b−1) dx  =6ζ(2)ln2−7ζ(3)
$$\sqrt{\pi}\Sigma\frac{\left(\frac{\Gamma\left({n}+\mathrm{1}\right)}{\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\right)'}{{n}+\mathrm{1}}=\Sigma\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} \left(\mathrm{1}−{x}\right)^{−\mathrm{1}/\mathrm{2}} {lnxdx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnxln}\left(\mathrm{1}−{x}\right)}{{x}\sqrt{\mathrm{1}−{x}}}{dx}=−\frac{\partial^{\mathrm{2}} }{\partial{a}\partial{b}}\mid_{{a}=\frac{\mathrm{1}}{\mathrm{2}}\:,{b}=\mathrm{0}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{b}−\mathrm{1}} {dx} \\ $$$$=\mathrm{6}\zeta\left(\mathrm{2}\right){ln}\mathrm{2}−\mathrm{7}\zeta\left(\mathrm{3}\right) \\ $$
Commented by MrGHK last updated on 31/Aug/23
wow nice solution
$${wow}\:{nice}\:{solution} \\ $$
Commented by MathedUp last updated on 01/Sep/23
How did you know ((d  )/ds) ((𝚪(s+1))/(𝚪(s+(3/2))))=(1/( (√π)))∫_0 ^1   ((u^s ∙ln(u))/( (√(1−u))))du ??
$$\mathrm{How}\:\mathrm{did}\:\mathrm{you}\:\mathrm{know}\:\frac{\mathrm{d}\:\:}{\mathrm{d}{s}}\:\frac{\boldsymbol{\Gamma}\left({s}+\mathrm{1}\right)}{\boldsymbol{\Gamma}\left({s}+\frac{\mathrm{3}}{\mathrm{2}}\right)}=\frac{\mathrm{1}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{u}^{{s}} \centerdot\mathrm{ln}\left({u}\right)}{\:\sqrt{\mathrm{1}−{u}}}\mathrm{d}{u}\:?? \\ $$

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