Question Number 118292 by Lordose last updated on 16/Oct/20
$$\boldsymbol{\mathrm{P}}\mathrm{rove}\:\mathrm{that}: \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{x}^{\mathrm{n}} −\mathrm{1}}{\mathrm{lnx}}\:=\:\boldsymbol{\mathrm{ln}}\mid\boldsymbol{\mathrm{n}}+\mathrm{1}\mid \\ $$
Answered by TANMAY PANACEA last updated on 16/Oct/20
![I(n)=∫_0 ^1 ((x^n −1)/(lnx))dx (dI/dn)=∫_0 ^1 (∂/∂n)(((x^n −1)/(lnx)) )dx =∫_0 ^1 ((x^n lnx)/(lnx))dx=∣(x^(n+1) /(n+1))∣_0 ^1 =(1/(n+1)) ∫dI(n)=∫(dn/(n+1)) I(n)=ln(n+1)+C when n=0 the value of I(n)=0 so C=0 I(n)=ln(n+1) proved [I(n)=∫_0 ^1 ((x^n −1)/(lnx))dx I(0)=∫_0 ^1 (0/(lnx))dx=0 ]](https://www.tinkutara.com/question/Q118294.png)
$$ \\ $$$${I}\left({n}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}} −\mathrm{1}}{{lnx}}{dx} \\ $$$$\frac{{dI}}{{dn}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\partial}{\partial{n}}\left(\frac{{x}^{{n}} −\mathrm{1}}{{lnx}}\:\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}} {lnx}}{{lnx}}{dx}=\mid\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\mid_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$$\int{dI}\left({n}\right)=\int\frac{{dn}}{{n}+\mathrm{1}} \\ $$$${I}\left({n}\right)={ln}\left({n}+\mathrm{1}\right)+{C} \\ $$$${when}\:{n}=\mathrm{0}\:\:{the}\:{value}\:{of}\:{I}\left({n}\right)=\mathrm{0} \\ $$$${so}\:{C}=\mathrm{0} \\ $$$${I}\left({n}\right)={ln}\left({n}+\mathrm{1}\right)\:\:{proved} \\ $$$$\left[{I}\left({n}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}} −\mathrm{1}}{{lnx}}{dx}\right. \\ $$$$\left.{I}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{0}}{{lnx}}{dx}=\mathrm{0}\:\:\:\:\right] \\ $$
Commented by mnjuly1970 last updated on 16/Oct/20
$${very}\:{nice}\:{solution}.{thank} \\ $$$${you}\:{sir}.. \\ $$
Commented by TANMAY PANACEA last updated on 16/Oct/20
$${most}\:{welcome}\:{sir} \\ $$
Answered by Bird last updated on 17/Oct/20
![A_n =∫_0 ^1 ((x^n −1)/(lnx))dx we do the changement lnx =−t ⇒A_n =−∫_0 ^∞ ((e^(−nt) −1)/(−t))(−e^(−t) )dt =−∫_0 ^∞ ((e^(−(n+1)t) −e^(−t) )/t) dt =∫_0 ^∞ ((e^(−t) −e^(−(n+1)t) )/t)dt=lim_(ξ→0^+ ) ∫_ξ ^∞ )...)dt let I(ξ)=∫_ξ ^∞ ((e^(−t) −e^(−(n+1)t) )/t)dt we hsve I(ξ)=∫_ξ ^(∞ ) (e^(−t) /t)dt −∫_ξ ^∞ (e^(−(n+1)t) /t)dt(→(n+1)t=u) =∫_ξ ^(∞ ) (e^(−t) /t)dt−∫_((n+1)ξ) ^∞ (e^(−u) /(u/(n+1)))(du/(n+1)) =∫_ξ ^((n+1)ξ) (e^(−t) /t)dt ∃c ∈]ξ,(n+1)ξ[ /I(ξ) =e^(−ξ) ∫_ξ ^((n+1)ξ) (dt/t) =e^(−ξ) ln∣n+1∣⇒lim_(ξ→0^+ ) I(ξ)=ln∣n+1∣ finally A_n =ln∣n+1∣](https://www.tinkutara.com/question/Q118342.png)
$${A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{{n}} −\mathrm{1}}{{lnx}}{dx}\:{we}\:{do}\:{the}\:{changement} \\ $$$${lnx}\:=−{t}\:\Rightarrow{A}_{{n}} =−\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{nt}} −\mathrm{1}}{−{t}}\left(−{e}^{−{t}} \right){dt} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−\left({n}+\mathrm{1}\right){t}} −{e}^{−{t}} }{{t}}\:{dt} \\ $$$$\left.=\left.\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{t}} −{e}^{−\left({n}+\mathrm{1}\right){t}} }{{t}}{dt}={lim}_{\xi\rightarrow\mathrm{0}^{+} } \:\int_{\xi} ^{\infty} \:\right)…\right){dt} \\ $$$${let}\:{I}\left(\xi\right)=\int_{\xi} ^{\infty} \:\frac{{e}^{−{t}} −{e}^{−\left({n}+\mathrm{1}\right){t}} }{{t}}{dt} \\ $$$${we}\:{hsve}\:{I}\left(\xi\right)=\int_{\xi} ^{\infty\:} \frac{{e}^{−{t}} }{{t}}{dt} \\ $$$$−\int_{\xi} ^{\infty} \:\frac{{e}^{−\left({n}+\mathrm{1}\right){t}} }{{t}}{dt}\left(\rightarrow\left({n}+\mathrm{1}\right){t}={u}\right) \\ $$$$=\int_{\xi} ^{\infty\:} \frac{{e}^{−{t}} }{{t}}{dt}−\int_{\left({n}+\mathrm{1}\right)\xi} ^{\infty} \:\frac{{e}^{−{u}} }{\frac{{u}}{{n}+\mathrm{1}}}\frac{{du}}{{n}+\mathrm{1}} \\ $$$$\left.=\int_{\xi} ^{\left({n}+\mathrm{1}\right)\xi} \:\frac{{e}^{−{t}} }{{t}}{dt}\:\:\exists{c}\:\in\right]\xi,\left({n}+\mathrm{1}\right)\xi\left[\right. \\ $$$$/{I}\left(\xi\right)\:={e}^{−\xi} \:\:\int_{\xi} ^{\left({n}+\mathrm{1}\right)\xi} \:\frac{{dt}}{{t}} \\ $$$$={e}^{−\xi} \:{ln}\mid{n}+\mathrm{1}\mid\Rightarrow{lim}_{\xi\rightarrow\mathrm{0}^{+} } \:{I}\left(\xi\right)={ln}\mid{n}+\mathrm{1}\mid \\ $$$${finally}\:\:{A}_{{n}} ={ln}\mid{n}+\mathrm{1}\mid \\ $$