Question Number 45117 by rahul 19 last updated on 09/Oct/18
$${Prove}\:{that}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{{a}} −\mathrm{1}}{\mathrm{log}\:{x}}\:{dx}\:=\:\mathrm{log}\:\left({a}+\mathrm{1}\right). \\ $$
Commented by maxmathsup by imad last updated on 09/Oct/18
![for a>−1 let f(a) =∫_0 ^1 ((x^a −1)/(ln(x)))dx changement ln(x) =−t give f(a) =−∫_0 ^(+∞) ((e^(−at) −1)/(−t)) (−1)e^(−t) dt =−∫_0 ^∞ ((e^(−(a+1)t) −e^(−t) )/t) dt =∫_0 ^∞ ((e^(−t) −e^(−(a+1)t) )/t)dt ⇒f^′ (a) =∫_0 ^∞ e^(−(a+1)t) dt=[−(1/(a+1))e^(−(a+1)t) ]_(t=0) ^(+∞) =(1/(a+1)) ⇒f(a) =ln∣a+1∣ +c =ln(a+1) +c but f(0) =0 =ln(1)+c ⇒c=0 ⇒f(a)=ln(a+1) .](https://www.tinkutara.com/question/Q45142.png)
$${for}\:{a}>−\mathrm{1}\:{let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{{a}} −\mathrm{1}}{{ln}\left({x}\right)}{dx}\:{changement}\:{ln}\left({x}\right)\:=−{t}\:{give} \\ $$$${f}\left({a}\right)\:=−\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{e}^{−{at}} \:−\mathrm{1}}{−{t}}\:\left(−\mathrm{1}\right){e}^{−{t}} {dt}\:=−\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−\left({a}+\mathrm{1}\right){t}} −{e}^{−{t}} }{{t}}\:{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{t}} \:−{e}^{−\left({a}+\mathrm{1}\right){t}} }{{t}}{dt}\:\Rightarrow{f}^{'} \left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−\left({a}+\mathrm{1}\right){t}} {dt}=\left[−\frac{\mathrm{1}}{{a}+\mathrm{1}}{e}^{−\left({a}+\mathrm{1}\right){t}} \right]_{{t}=\mathrm{0}} ^{+\infty} \\ $$$$=\frac{\mathrm{1}}{{a}+\mathrm{1}}\:\Rightarrow{f}\left({a}\right)\:={ln}\mid{a}+\mathrm{1}\mid\:+{c}\:={ln}\left({a}+\mathrm{1}\right)\:+{c}\:\:{but}\: \\ $$$${f}\left(\mathrm{0}\right)\:=\mathrm{0}\:={ln}\left(\mathrm{1}\right)+{c}\:\Rightarrow{c}=\mathrm{0}\:\Rightarrow{f}\left({a}\right)={ln}\left({a}+\mathrm{1}\right)\:. \\ $$$$ \\ $$$$ \\ $$
Commented by rahul 19 last updated on 09/Oct/18
thanks prof Abdo ☺️
Commented by turbo msup by abdo last updated on 09/Oct/18
$${you}\:{are}\:{welcome}\:{sir}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Oct/18
$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{a}} −\mathrm{1}}{{lnx}}{dx} \\ $$$$\frac{{dI}}{{da}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{lnx}}\frac{\partial}{\partial{a}}\left({x}^{{a}} −\mathrm{1}\right){dx} \\ $$$$\:\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{lnx}}×{x}^{{a}} ×{lnxdx} \\ $$$$\:\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}} {dx} \\ $$$$\:\:\:\:\:\:=\mid\frac{{x}^{{a}+\mathrm{1}} }{{a}+\mathrm{1}}\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{{a}+\mathrm{1}} \\ $$$$\int{dI}=\int\frac{{da}}{{a}+\mathrm{1}}\:\:\:{I}={ln}\left({a}+\mathrm{1}\right) \\ $$
Commented by rahul 19 last updated on 09/Oct/18
Sir , how will this come to my mind first we have to differentiate!! ����
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Oct/18
$${i}\:{shall}\:{post}\:{some}\:{example}…{this}\:{type}\:{intregal} \\ $$$${are}\:{advanced}\:{intregal}…{from}\:{book}\:{i}\:{shall}\:{post}.. \\ $$
Commented by rahul 19 last updated on 12/Oct/18
$${Sir},{plss}\:{post}\:\mathrm{2}−\mathrm{3}\:{examples}….. \\ $$