1-0-x-1-x-1-lnx-dx-

Question Number 179844 by mathlove last updated on 03/Nov/22

$$\underset{\mathrm{0}} {\int}^{\mathrm{1}} \:\frac{{x}−\mathrm{1}}{\left({x}+\mathrm{1}\right){lnx}}{dx}=? \\ $$

Answered by Peace last updated on 03/Nov/22

∫_0 ^1 ((x−x^s )/((x+1)ln(x)))=f(s) f(0)=Δ=∫_0 ^1 ((x−1)/(ln(x)(x+1))) f(1)=0 f′(s)=∫_0 ^1 (x^s /(1+x))dx=∫_0 ^1 ((x^s −x^(s+1) )/(1−x^2 ))dx =∫_0 ^1 ((t^((s−1)/2) −t^(s/2) )/(2(1−t)))dt=(1/2)(Ψ((s/2)+1)−Ψ(((s+1)/2))) Ψ(s)=−γ+∫_0 ^1 ((1−x^(s−1) )/(1−x))dx Ψ(s)=((Γ′(s))/(Γ(s))) f(s)=ln(((Γ(((s+2)/2)))/(Γ(((s+1)/2)))))+c f(1)=ln(((Γ((3/2)))/(Γ(1))))+c=0⇒c=ln((1/(Γ((3/2)))))=ln((2/( (√π)))) Δ=f(0)=ln(((Γ(1))/(Γ((1/2)))))+ln((2/( (√π))))=ln((1/( (√π))))+ln((2/( (√π))))=ln((2/π))

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}−{x}^{{s}} }{\left({x}+\mathrm{1}\right){ln}\left({x}\right)}={f}\left({s}\right) \\ $$$${f}\left(\mathrm{0}\right)=\Delta=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}−\mathrm{1}}{{ln}\left({x}\right)\left({x}+\mathrm{1}\right)} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${f}'\left({s}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{s}} }{\mathrm{1}+{x}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{s}} −{x}^{{s}+\mathrm{1}} }{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\frac{{s}−\mathrm{1}}{\mathrm{2}}} −{t}^{\frac{{s}}{\mathrm{2}}} }{\mathrm{2}\left(\mathrm{1}−{t}\right)}{dt}=\frac{\mathrm{1}}{\mathrm{2}}\left(\Psi\left(\frac{{s}}{\mathrm{2}}+\mathrm{1}\right)−\Psi\left(\frac{{s}+\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$\Psi\left({s}\right)=−\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}^{{s}−\mathrm{1}} }{\mathrm{1}−{x}}{dx} \\ $$$$\Psi\left({s}\right)=\frac{\Gamma'\left({s}\right)}{\Gamma\left({s}\right)} \\ $$$${f}\left({s}\right)={ln}\left(\frac{\Gamma\left(\frac{{s}+\mathrm{2}}{\mathrm{2}}\right)}{\Gamma\left(\frac{{s}+\mathrm{1}}{\mathrm{2}}\right)}\right)+{c} \\ $$$${f}\left(\mathrm{1}\right)={ln}\left(\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{1}\right)}\right)+{c}=\mathrm{0}\Rightarrow{c}={ln}\left(\frac{\mathrm{1}}{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}\right)={ln}\left(\frac{\mathrm{2}}{\:\sqrt{\pi}}\right) \\ $$$$\Delta={f}\left(\mathrm{0}\right)={ln}\left(\frac{\Gamma\left(\mathrm{1}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}\right)+{ln}\left(\frac{\mathrm{2}}{\:\sqrt{\pi}}\right)={ln}\left(\frac{\mathrm{1}}{\:\sqrt{\pi}}\right)+{ln}\left(\frac{\mathrm{2}}{\:\sqrt{\pi}}\right)={ln}\left(\frac{\mathrm{2}}{\pi}\right) \\ $$$$ \\ $$

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