Question Number 196209 by mnjuly1970 last updated on 19/Aug/23
$$ \\ $$$$\:\:\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\left({x}−\mathrm{1}\right)^{\:\mathrm{2}} }{\mathrm{ln}^{\mathrm{2}} \left({x}\right)}\:{dx}=\:? \\ $$$$\:\:\:\:\:−−−− \\ $$
Answered by Mathspace last updated on 20/Aug/23
![lnx=−t ⇒Φ= ∫_0 ^∞ (((e^(−t) −1)^2 )/t^2 )e^(−t) dt =∫_0 ^∞ ((e^(−t) (e^(−2t) −2e^(−t) +1))/t^2 )dt =∫_0 ^∞ ((e^(−3t) −2e^(−2t) +e^(−t) )/t^2 )dt =[−(1/t)(e^(−3t) −2e^(−2t) +e^(−t) )]_0 ^∞ +∫_0 ^∞ (1/t)(−3e^(−3t) +4e^(−2t) −e^(−t) )dt =λ_0 −∫_0 ^∞ ((3e^(−3t) −4e^(−2t) +e^(−t) )/t)dt with=λ_0 =lim_(t→0) ((e^(−3t) −2e^(−2t) +e^(−t) )/t) =lim_(t→0) ((−3e^(−3t) +4e^(−2t) −e^(−t) )/1)(hospital) =0 ⇒ I=−∫_0 ^∞ ((3e^(−3t) −4e^(−2t) +e^(−t) )/t)dt we consider ϕ(λ)=∫_0 ^∞ (((3e^(−3t) −4e^(−2t) +e^(−t) )e^(−λt) )/t) (λ>0) ϕ′(λ)=−∫_0 ^∞ (3e^(−3t) −4e^(−2t) +e^(−t) )e^(−λt) dt =−3∫_0 ^∞ e^(−(3+λ)t) dt+4∫_0 ^∞ e^(−(2+λ)t) dt −∫_0 ^∞ e^(−(1+λ)t) dt =[((−1)/(3+λ))e^(−(3+λ)t) ]_0 ^∞ +4[−(1/(2+λ))e^(−(2+λ)t) ]_0 ^∞ −[−(1/(1+λ)) e^(−(1+λ)t) ]_0 ^∞ =(1/(3+λ))+(4/(2+λ))−(1/(1+λ)) ⇒ ϕ(λ)=ln(λ+3)+4ln(λ+2)−ln(λ+1) +c ⇒I=−lim_(λ→0) ϕ(λ) =−ln3−4ln2−c −ϕ(λ)=−ln(λ+3)−4ln(λ+2)+ln(λ+1)−c =ln(((λ+1)/(λ+3)))+ln((1/((λ+2)^4 ))) ⇒ lim−ϕ(λ)=−c=0 ⇒ I=−ln3−4ln2...! perhaps there is arror of sign?](https://www.tinkutara.com/question/Q196215.png)
$${lnx}=−{t}\:\Rightarrow\Phi= \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\left({e}^{−{t}} −\mathrm{1}\right)^{\mathrm{2}} }{{t}^{\mathrm{2}} }{e}^{−{t}} {dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{t}} \left({e}^{−\mathrm{2}{t}} −\mathrm{2}{e}^{−{t}} +\mathrm{1}\right)}{{t}^{\mathrm{2}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−\mathrm{3}{t}} −\mathrm{2}{e}^{−\mathrm{2}{t}} +{e}^{−{t}} }{{t}^{\mathrm{2}} }{dt} \\ $$$$=\left[−\frac{\mathrm{1}}{{t}}\left({e}^{−\mathrm{3}{t}} −\mathrm{2}{e}^{−\mathrm{2}{t}} +{e}^{−{t}} \right)\right]_{\mathrm{0}} ^{\infty} \\ $$$$+\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{t}}\left(−\mathrm{3}{e}^{−\mathrm{3}{t}} +\mathrm{4}{e}^{−\mathrm{2}{t}} −{e}^{−{t}} \right){dt} \\ $$$$=\lambda_{\mathrm{0}} −\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{3}{e}^{−\mathrm{3}{t}} −\mathrm{4}{e}^{−\mathrm{2}{t}} +{e}^{−{t}} }{{t}}{dt} \\ $$$${with}=\lambda_{\mathrm{0}} ={lim}_{{t}\rightarrow\mathrm{0}} \frac{{e}^{−\mathrm{3}{t}} −\mathrm{2}{e}^{−\mathrm{2}{t}} +{e}^{−{t}} }{{t}} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \frac{−\mathrm{3}{e}^{−\mathrm{3}{t}} +\mathrm{4}{e}^{−\mathrm{2}{t}} −{e}^{−{t}} }{\mathrm{1}}\left({hospital}\right) \\ $$$$=\mathrm{0}\:\Rightarrow \\ $$$${I}=−\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{3}{e}^{−\mathrm{3}{t}} −\mathrm{4}{e}^{−\mathrm{2}{t}} +{e}^{−{t}} }{{t}}{dt} \\ $$$${we}\:{consider}\:\varphi\left(\lambda\right)=\int_{\mathrm{0}} ^{\infty} \frac{\left(\mathrm{3}{e}^{−\mathrm{3}{t}} −\mathrm{4}{e}^{−\mathrm{2}{t}} +{e}^{−{t}} \right){e}^{−\lambda{t}} }{{t}}\:\left(\lambda>\mathrm{0}\right) \\ $$$$\varphi'\left(\lambda\right)=−\int_{\mathrm{0}} ^{\infty} \left(\mathrm{3}{e}^{−\mathrm{3}{t}} −\mathrm{4}{e}^{−\mathrm{2}{t}} +{e}^{−{t}} \right){e}^{−\lambda{t}} {dt} \\ $$$$=−\mathrm{3}\int_{\mathrm{0}} ^{\infty} {e}^{−\left(\mathrm{3}+\lambda\right){t}} {dt}+\mathrm{4}\int_{\mathrm{0}} ^{\infty} {e}^{−\left(\mathrm{2}+\lambda\right){t}} {dt} \\ $$$$−\int_{\mathrm{0}} ^{\infty} {e}^{−\left(\mathrm{1}+\lambda\right){t}} {dt} \\ $$$$=\left[\frac{−\mathrm{1}}{\mathrm{3}+\lambda}{e}^{−\left(\mathrm{3}+\lambda\right){t}} \right]_{\mathrm{0}} ^{\infty} +\mathrm{4}\left[−\frac{\mathrm{1}}{\mathrm{2}+\lambda}{e}^{−\left(\mathrm{2}+\lambda\right){t}} \right]_{\mathrm{0}} ^{\infty} \\ $$$$−\left[−\frac{\mathrm{1}}{\mathrm{1}+\lambda}\:{e}^{−\left(\mathrm{1}+\lambda\right){t}} \right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}+\lambda}+\frac{\mathrm{4}}{\mathrm{2}+\lambda}−\frac{\mathrm{1}}{\mathrm{1}+\lambda}\:\Rightarrow \\ $$$$\varphi\left(\lambda\right)={ln}\left(\lambda+\mathrm{3}\right)+\mathrm{4}{ln}\left(\lambda+\mathrm{2}\right)−{ln}\left(\lambda+\mathrm{1}\right)\:+{c} \\ $$$$\Rightarrow{I}=−{lim}_{\lambda\rightarrow\mathrm{0}} \varphi\left(\lambda\right) \\ $$$$=−{ln}\mathrm{3}−\mathrm{4}{ln}\mathrm{2}−{c} \\ $$$$−\varphi\left(\lambda\right)=−{ln}\left(\lambda+\mathrm{3}\right)−\mathrm{4}{ln}\left(\lambda+\mathrm{2}\right)+{ln}\left(\lambda+\mathrm{1}\right)−{c} \\ $$$$={ln}\left(\frac{\lambda+\mathrm{1}}{\lambda+\mathrm{3}}\right)+{ln}\left(\frac{\mathrm{1}}{\left(\lambda+\mathrm{2}\right)^{\mathrm{4}} }\right)\:\Rightarrow \\ $$$${lim}−\varphi\left(\lambda\right)=−{c}=\mathrm{0}\:\Rightarrow \\ $$$${I}=−{ln}\mathrm{3}−\mathrm{4}{ln}\mathrm{2}…! \\ $$$${perhaps}\:{there}\:{is}\:{arror}\:{of}\:{sign}? \\ $$
Commented by mnjuly1970 last updated on 20/Aug/23
$$\: \\ $$
Answered by witcher3 last updated on 20/Aug/23
![Ω=∫_0 ^1 ((x(x−1)^2 )/(xln^2 (x)))dx u=x(x−1)^2 ,u′=(x−1)^2 +2x(x−1),v′=(1/(xln^2 (x))) v=−(1/(ln(x))) . Ω=lim_((a,b)→(0,1)) [−(((x−1)^2 )/(ln(x)))]_a ^b +∫_0 ^1 (((x−1)(3x−1))/(ln(x))) =∫_0 ^1 ((x−1)/(ln(x))).(3x−1)dx ∫_0 ^1 (3x−1)∫_0 ^1 x^t dtdx =∫_0 ^1 ∫_0 ^1 (3x−1)x^t dxdt ∫_0 ^1 ∫_0 ^1 3x^(t+1) −x^t dxdt =∫_0 ^1 (3/(t+2))−(1/(t+1))=3ln(3)−3ln(2)−ln(2) =3ln(3)−4ln(2)](https://www.tinkutara.com/question/Q196254.png)
$$\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{xln}^{\mathrm{2}} \left(\mathrm{x}\right)}\mathrm{dx} \\ $$$$\mathrm{u}=\mathrm{x}\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} ,\mathrm{u}'=\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2x}\left(\mathrm{x}−\mathrm{1}\right),\mathrm{v}'=\frac{\mathrm{1}}{\mathrm{xln}^{\mathrm{2}} \left(\mathrm{x}\right)} \\ $$$$\mathrm{v}=−\frac{\mathrm{1}}{\mathrm{ln}\left(\mathrm{x}\right)} \\ $$$$. \\ $$$$\Omega=\underset{\left(\mathrm{a},\mathrm{b}\right)\rightarrow\left(\mathrm{0},\mathrm{1}\right)} {\mathrm{lim}}\left[−\frac{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{ln}\left(\mathrm{x}\right)}\right]_{\mathrm{a}} ^{\mathrm{b}} +\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{3x}−\mathrm{1}\right)}{\mathrm{ln}\left(\mathrm{x}\right)} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}−\mathrm{1}}{\mathrm{ln}\left(\mathrm{x}\right)}.\left(\mathrm{3x}−\mathrm{1}\right)\mathrm{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{3x}−\mathrm{1}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{t}} \mathrm{dtdx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{3x}−\mathrm{1}\right)\mathrm{x}^{\mathrm{t}} \mathrm{dxdt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{3x}^{\mathrm{t}+\mathrm{1}} −\mathrm{x}^{\mathrm{t}} \mathrm{dxdt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{3}}{\mathrm{t}+\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{t}+\mathrm{1}}=\mathrm{3ln}\left(\mathrm{3}\right)−\mathrm{3ln}\left(\mathrm{2}\right)−\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$=\mathrm{3ln}\left(\mathrm{3}\right)−\mathrm{4ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$$$ \\ $$