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Question Number 118292 by Lordose last updated on 16/Oct/20
Prove that: ∫_0 ^( 1) ((x^n −1)/(lnx)) = ln∣n+1∣
Provethat:∫01xn−1lnx=ln∣n+1∣
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 ]
I(n)=∫01xn−1lnxdxdIdn=∫01∂∂n(xn−1lnx)dx=∫01xnlnxlnxdx=∣xn+1n+1∣01=1n+1∫dI(n)=∫dnn+1I(n)=ln(n+1)+Cwhenn=0thevalueofI(n)=0soC=0I(n)=ln(n+1)proved[I(n)=∫01xn−1lnxdxI(0)=∫010lnxdx=0]
Commented by mnjuly1970 last updated on 16/Oct/20
very nice solution.thank you sir..
verynicesolution.thankyousir..
Commented by TANMAY PANACEA last updated on 16/Oct/20
most welcome sir
mostwelcomesir
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∣
An=∫01xn−1lnxdxwedothechangementlnx=−t⇒An=−∫0∞e−nt−1−t(−e−t)dt=−∫0∞e−(n+1)t−e−ttdt=∫0∞e−t−e−(n+1)ttdt=limξ→0+∫ξ∞)…)dtletI(ξ)=∫ξ∞e−t−e−(n+1)ttdtwehsveI(ξ)=∫ξ∞e−ttdt−∫ξ∞e−(n+1)ttdt(→(n+1)t=u)=∫ξ∞e−ttdt−∫(n+1)ξ∞e−uun+1dun+1=∫ξ(n+1)ξe−ttdt∃c∈]ξ,(n+1)ξ[/I(ξ)=e−ξ∫ξ(n+1)ξdtt=e−ξln∣n+1∣⇒limξ→0+I(ξ)=ln∣n+1∣finallyAn=ln∣n+1∣

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