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Question Number 130135 by mathmax by abdo last updated on 22/Jan/21

calculate ∫_0 ^∞ ((lnx)/((x+1)^6 ))dx

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnx}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{6}} }\mathrm{dx} \\ $$

Answered by Dwaipayan Shikari last updated on 22/Jan/21

I(a)=∫_0 ^∞ ((x^(a+1−1) )/((1+x)^(6+a+1−a−1) ))dx ⇒I(a)=((Γ(a+1)Γ(5−a))/(Γ(6))) I′(a)=(1/(120))(Γ′(a+1)Γ(5−a)−Γ′(5−a)Γ(a+1)) I′(0)=−(γ/5)−(1/(120))(−24γ+24(1+(1/2)+(1/3)+(1/4))) =−(5/(12)) Means in General ∫_0 ^∞ ((logx)/((1+x)^n ))dx=−(1/n)Σ_(n=1) ^(n−2) (1/n)

$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{a}+\mathrm{1}−\mathrm{1}} \:}{\left(\mathrm{1}+{x}\right)^{\mathrm{6}+{a}+\mathrm{1}−{a}−\mathrm{1}} }{dx}\:\Rightarrow{I}\left({a}\right)=\frac{\Gamma\left({a}+\mathrm{1}\right)\Gamma\left(\mathrm{5}−{a}\right)}{\Gamma\left(\mathrm{6}\right)} \\ $$$${I}'\left({a}\right)=\frac{\mathrm{1}}{\mathrm{120}}\left(\Gamma'\left({a}+\mathrm{1}\right)\Gamma\left(\mathrm{5}−{a}\right)−\Gamma'\left(\mathrm{5}−{a}\right)\Gamma\left({a}+\mathrm{1}\right)\right) \\ $$$${I}'\left(\mathrm{0}\right)=−\frac{\gamma}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{120}}\left(−\mathrm{24}\gamma+\mathrm{24}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\right)\right) \\ $$$$=−\frac{\mathrm{5}}{\mathrm{12}} \\ $$$${Means}\:{in}\:\:{General}\:\int_{\mathrm{0}} ^{\infty} \frac{{logx}}{\left(\mathrm{1}+{x}\right)^{{n}} }{dx}=−\frac{\mathrm{1}}{{n}}\underset{{n}=\mathrm{1}} {\overset{{n}−\mathrm{2}} {\sum}}\frac{\mathrm{1}}{{n}}\: \\ $$

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