Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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∣

Terms of Service

Privacy Policy

Contact: info@tinkutara.com