Question Number 111025 by mathmax by abdo last updated on 01/Sep/20
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{x}^{\mathrm{2}} \mathrm{ln}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{4}} }\mathrm{dx} \\ $$
Answered by mathdave last updated on 01/Sep/20
![solution let I=∫_0 ^∞ ((x^2 ln(x))/((1+x)^4 ))dx I(a)=(∂/∂a)∫_0 ^∞ (x^(2+a) /((1+x)^4 ))dx (∂/∂a)∣_(a=0) I(a)=(∂/∂a)β(3+a,1−a)=(∂/∂a)[((Γ(3+a)Γ(1−a))/(Γ(4)))] I^′ (a)=((Γ(3+a)Γ(1−a))/(Γ(4)))[ψ(3+a)−ψ(1−a)] I^′ (0)=((Γ(3)Γ(1))/(Γ(4)))[ψ(3)−ψ(1)] I^′ (0)=(1/3)[(3/2)−γ+γ]=(1/2) ∵∫_0 ^∞ ((x^2 lnx)/((1+x)^4 ))dx=(1/2) mathdave](https://www.tinkutara.com/question/Q111031.png)
$${solution} \\ $$$${let}\:{I}=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} \mathrm{ln}\left({x}\right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{4}} }{dx} \\ $$$${I}\left({a}\right)=\frac{\partial}{\partial{a}}\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}+{a}} }{\left(\mathrm{1}+{x}\right)^{\mathrm{4}} }{dx} \\ $$$$\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{0}} {I}\left({a}\right)=\frac{\partial}{\partial{a}}\beta\left(\mathrm{3}+{a},\mathrm{1}−{a}\right)=\frac{\partial}{\partial{a}}\left[\frac{\Gamma\left(\mathrm{3}+{a}\right)\Gamma\left(\mathrm{1}−{a}\right)}{\Gamma\left(\mathrm{4}\right)}\right] \\ $$$${I}^{'} \left({a}\right)=\frac{\Gamma\left(\mathrm{3}+{a}\right)\Gamma\left(\mathrm{1}−{a}\right)}{\Gamma\left(\mathrm{4}\right)}\left[\psi\left(\mathrm{3}+{a}\right)−\psi\left(\mathrm{1}−{a}\right)\right] \\ $$$${I}^{'} \left(\mathrm{0}\right)=\frac{\Gamma\left(\mathrm{3}\right)\Gamma\left(\mathrm{1}\right)}{\Gamma\left(\mathrm{4}\right)}\left[\psi\left(\mathrm{3}\right)−\psi\left(\mathrm{1}\right)\right] \\ $$$${I}^{'} \left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{3}}\left[\frac{\mathrm{3}}{\mathrm{2}}−\gamma+\gamma\right]=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\because\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} \mathrm{ln}{x}}{\left(\mathrm{1}+{x}\right)^{\mathrm{4}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${mathdave} \\ $$