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Question Number 79096 by mathmax by abdo last updated on 22/Jan/20
calculate ∫_0 ^∞ ((ln(x))/((1+x)^3 ))dx
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left({x}\right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} }{dx} \\ $$
Commented by mind is power last updated on 22/Jan/20
sam idee will worck for ∫_0 ^(+∞) ((ln(x)^p )/((x+1)^r ))dx withe r>1,p∈N ln(x)^p =(∂^p /∂t^p )x^t ∣_(t=0) and sam idea like previous one but we will have mor complicated function Ψ_p (x)....
$${sam}\:{idee}\:{will}\:{worck} \\ $$$${for} \\ $$$$\int_{\mathrm{0}} ^{+\infty} \frac{{ln}\left({x}\right)^{{p}} }{\left({x}+\mathrm{1}\right)^{{r}} }{dx} \\ $$$${withe}\:{r}>\mathrm{1},{p}\in\mathbb{N} \\ $$$${ln}\left({x}\right)^{{p}} =\frac{\partial^{{p}} }{\partial{t}^{{p}} }{x}^{{t}} \mid_{{t}=\mathrm{0}} \:\:\:{and}\:{sam}\:{idea}\:{like}\:{previous}\:{one} \\ $$$${but}\:{we}\:{will}\:{have}\:{mor}\:{complicated}\:{function}\:\Psi_{{p}} \left({x}\right)…. \\ $$
Answered by mind is power last updated on 22/Jan/20
ln(x)=(d/dx)x^t ∣_(t=0) f(t)=∫_0 ^(+∞) (x^t /((1+x)^3 ))dx,we want f′(0) x=tg^2 (y)⇒ f(t)=∫_0 ^(π/2) ((tg^(2t) (y))/((1+tg^2 (y))^3 )).2(1+tg^2 (y))tg(y)dy =2∫_0 ^(π/2) ((tg^(2t+1) (y))/((1+tg^2 (y))^2 ))dy =2∫_0 ^(π/2) sin^(2t+1) (y)cos^(−2t+3) (y)dy =β(t+1,−t+2)=((Γ(t+1)Γ(2−t))/(Γ(3)))=((Γ(t+1)Γ(2−t))/2) f(t)=((Γ(t+1)Γ(2−t))/2) f′(t)=(1/2){Γ′(t+1)Γ(2−t)−Γ′(2−t)Γ(t+1)} =(1/2){Ψ(t+1)Γ(t+1)Γ(2−t)−Ψ(2−t)Γ(2−t)Γ(t+1)} f′(0)=(1/2){Ψ(1)−Ψ(2)}=(1/2){−γ−(1−γ)}=−(1/2)=−0.5
$${ln}\left({x}\right)=\frac{{d}}{{dx}}{x}^{{t}} \mid_{{t}=\mathrm{0}} \\ $$$${f}\left({t}\right)=\int_{\mathrm{0}} ^{+\infty} \frac{{x}^{{t}} }{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} }{dx},{we}\:{want}\:{f}'\left(\mathrm{0}\right) \\ $$$${x}={tg}^{\mathrm{2}} \left({y}\right)\Rightarrow \\ $$$${f}\left({t}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{tg}^{\mathrm{2}{t}} \left({y}\right)}{\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({y}\right)\right)^{\mathrm{3}} }.\mathrm{2}\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({y}\right)\right){tg}\left({y}\right){dy} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{tg}^{\mathrm{2}{t}+\mathrm{1}} \left({y}\right)}{\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({y}\right)\right)^{\mathrm{2}} }{dy} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{t}+\mathrm{1}} \left({y}\right){cos}^{−\mathrm{2}{t}+\mathrm{3}} \left({y}\right){dy} \\ $$$$=\beta\left({t}+\mathrm{1},−{t}+\mathrm{2}\right)=\frac{\Gamma\left({t}+\mathrm{1}\right)\Gamma\left(\mathrm{2}−{t}\right)}{\Gamma\left(\mathrm{3}\right)}=\frac{\Gamma\left({t}+\mathrm{1}\right)\Gamma\left(\mathrm{2}−{t}\right)}{\mathrm{2}} \\ $$$${f}\left({t}\right)=\frac{\Gamma\left({t}+\mathrm{1}\right)\Gamma\left(\mathrm{2}−{t}\right)}{\mathrm{2}} \\ $$$${f}'\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left\{\Gamma'\left({t}+\mathrm{1}\right)\Gamma\left(\mathrm{2}−{t}\right)−\Gamma'\left(\mathrm{2}−{t}\right)\Gamma\left({t}+\mathrm{1}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\Psi\left({t}+\mathrm{1}\right)\Gamma\left({t}+\mathrm{1}\right)\Gamma\left(\mathrm{2}−{t}\right)−\Psi\left(\mathrm{2}−{t}\right)\Gamma\left(\mathrm{2}−{t}\right)\Gamma\left({t}+\mathrm{1}\right)\right\} \\ $$$${f}'\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left\{\Psi\left(\mathrm{1}\right)−\Psi\left(\mathrm{2}\right)\right\}=\frac{\mathrm{1}}{\mathrm{2}}\left\{−\gamma−\left(\mathrm{1}−\gamma\right)\right\}=−\frac{\mathrm{1}}{\mathrm{2}}=−\mathrm{0}.\mathrm{5} \\ $$$$ \\ $$
Commented by msup trace by abdo last updated on 23/Jan/20
thank you sir now try to solve it complex analysis...
$${thank}\:{you}\:{sir}\:{now}\:{try}\:{to}\:{solve}\:{it} \\ $$$${complex}\:{analysis}… \\ $$

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