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Question Number 45117 by rahul 19 last updated on 09/Oct/18

Prove that ∫_0 ^1  ((x^a −1)/(log x)) dx = log (a+1).

$${Prove}\:{that}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{{a}} −\mathrm{1}}{\mathrm{log}\:{x}}\:{dx}\:=\:\mathrm{log}\:\left({a}+\mathrm{1}\right). \\ $$

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) .

$${for}\:{a}>−\mathrm{1}\:{let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{{a}} −\mathrm{1}}{{ln}\left({x}\right)}{dx}\:{changement}\:{ln}\left({x}\right)\:=−{t}\:{give} \\ $$$${f}\left({a}\right)\:=−\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{e}^{−{at}} \:−\mathrm{1}}{−{t}}\:\left(−\mathrm{1}\right){e}^{−{t}} {dt}\:=−\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−\left({a}+\mathrm{1}\right){t}} −{e}^{−{t}} }{{t}}\:{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{t}} \:−{e}^{−\left({a}+\mathrm{1}\right){t}} }{{t}}{dt}\:\Rightarrow{f}^{'} \left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−\left({a}+\mathrm{1}\right){t}} {dt}=\left[−\frac{\mathrm{1}}{{a}+\mathrm{1}}{e}^{−\left({a}+\mathrm{1}\right){t}} \right]_{{t}=\mathrm{0}} ^{+\infty} \\ $$$$=\frac{\mathrm{1}}{{a}+\mathrm{1}}\:\Rightarrow{f}\left({a}\right)\:={ln}\mid{a}+\mathrm{1}\mid\:+{c}\:={ln}\left({a}+\mathrm{1}\right)\:+{c}\:\:{but}\: \\ $$$${f}\left(\mathrm{0}\right)\:=\mathrm{0}\:={ln}\left(\mathrm{1}\right)+{c}\:\Rightarrow{c}=\mathrm{0}\:\Rightarrow{f}\left({a}\right)={ln}\left({a}+\mathrm{1}\right)\:. \\ $$$$ \\ $$$$ \\ $$

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.

$${you}\:{are}\:{welcome}\:{sir}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 09/Oct/18

I(a)=∫_0 ^1 ((x^a −1)/(lnx))dx  (dI/da)=∫_0 ^1 (1/(lnx))(∂/∂a)(x^a −1)dx       =∫_0 ^1 (1/(lnx))×x^a ×lnxdx       =∫_0 ^1 x^a dx        =∣(x^(a+1) /(a+1))∣_0 ^1         =(1/(a+1))  ∫dI=∫(da/(a+1))   I=ln(a+1)

$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{a}} −\mathrm{1}}{{lnx}}{dx} \\ $$$$\frac{{dI}}{{da}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{lnx}}\frac{\partial}{\partial{a}}\left({x}^{{a}} −\mathrm{1}\right){dx} \\ $$$$\:\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{lnx}}×{x}^{{a}} ×{lnxdx} \\ $$$$\:\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}} {dx} \\ $$$$\:\:\:\:\:\:=\mid\frac{{x}^{{a}+\mathrm{1}} }{{a}+\mathrm{1}}\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{{a}+\mathrm{1}} \\ $$$$\int{dI}=\int\frac{{da}}{{a}+\mathrm{1}}\:\:\:{I}={ln}\left({a}+\mathrm{1}\right) \\ $$

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..

$${i}\:{shall}\:{post}\:{some}\:{example}...{this}\:{type}\:{intregal} \\ $$$${are}\:{advanced}\:{intregal}...{from}\:{book}\:{i}\:{shall}\:{post}.. \\ $$

Commented by rahul 19 last updated on 12/Oct/18

Sir,plss post 2−3 examples.....

$${Sir},{plss}\:{post}\:\mathrm{2}−\mathrm{3}\:{examples}..... \\ $$

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