Menu Close

nice-calculus-prove-that-0-1-1-1-x-1-2-x-2-ln-1-x-dx-2-1-log-2-

Tinku Tara 0 Comments
Pin




Question Number 124064 by mnjuly1970 last updated on 30/Nov/20
... nice calculus... prove that::: ∫_0 ^( 1) {((1+(1−x)^(1/2) )/x) +(2/(ln(1−x)))}dx =^(???) 2(γ−1+log(2))
…nicecalculus…provethat:::∫01{1+(1−x)12x+2ln(1−x)}dx=???2(γ−1+log(2))
Answered by mindispower last updated on 01/Dec/20
let ln(1−x)=−t ⇔∫_0 ^∞ {((1+e^(−(t/2)) )/(1−e^(−t) ))−(2/t)}e^(−t) dt =∫_0 ^∞ −((2e^(−t) )/t)+((1+e^(−(t/2)) )/(1−e^(−t) ))e^(−t) dt =−2∫{(e^(−t) /t)−(1/2)((1+e^(−(t/2)) )/(1−e^(−t) ))e^(−t) }dt =−2∫(e^(−t) /t)−(e^(−3(t/(2 ))) /(1−e^(−t) ))+(1/2).((e^(−(t/2)) −1)/(1−e^(−t) )).e^(−t) dt =−2∫_0 ^∞ {(e^(−t) /t)−(e^(−3(t/2)) /(1−e^(−t) ))}dt+∫_0 ^∞ ((1−e^(−(t/2)) )/(1−e^(−t) ))e^(−t) dt Ψ(z)=∫_0 ^∞ ((e^(−t) /t)−(e^(−zt) /(1−e^(−t) )))dt we get −2Ψ((3/2))+∫_0 ^∞ ((e^(−t) dt)/(1+e^(−(t/2)) ))=−2Ψ((3/2))+A A=∫_0 ^∞ {e^(−(t/2)) −1+(1/(1+e^(−(t/2)) ))}dt A=lim_(x→∞) ∫_0 ^x {e^(−(t/2)) −1+(1/(1+e^(−(t/2)) ))}dt =lim_(x→∞) [−2e^(−(x/2)) +2−x+2ln(e^(x/2) +1)−2ln(2)] =lim_(x→∞) [2−x+2.(x/2)+2ln(1+e^(−(x/2)) )−2ln(2)] =2−2ln(2) Ψ((3/2))=Ψ((1/2))+2=−2ln(2)−γ+2 we get −2(−2ln(2)−γ)+2−2ln(2) =2ln(2)+2γ−2=2(−1+γ+ln(2)) ∫((1+(√(1−x)))/x)+(2/(ln(1−x)))dx=2(γ−1+ln(2))
letln(1−x)=−t⇔∫0∞{1+e−t21−e−t−2t}e−tdt=∫0∞−2e−tt+1+e−t21−e−te−tdt=−2∫{e−tt−121+e−t21−e−te−t}dt=−2∫e−tt−e−3t21−e−t+12.e−t2−11−e−t.e−tdt=−2∫0∞{e−tt−e−3t21−e−t}dt+∫0∞1−e−t21−e−te−tdtΨ(z)=∫0∞(e−tt−e−zt1−e−t)dtweget−2Ψ(32)+∫0∞e−tdt1+e−t2=−2Ψ(32)+AA=∫0∞{e−t2−1+11+e−t2}dtA=limx→∞∫0x{e−t2−1+11+e−t2}dt=limx→∞[−2e−x2+2−x+2ln(ex2+1)−2ln(2)]=limx→∞[2−x+2.x2+2ln(1+e−x2)−2ln(2)]=2−2ln(2)Ψ(32)=Ψ(12)+2=−2ln(2)−γ+2weget−2(−2ln(2)−γ)+2−2ln(2)=2ln(2)+2γ−2=2(−1+γ+ln(2))∫1+1−xx+2ln(1−x)dx=2(γ−1+ln(2))
Commented by mnjuly1970 last updated on 01/Dec/20
grateful sir mindspower...
gratefulsirmindspower…
Commented by mindispower last updated on 02/Dec/20
withe pleasur sir
withepleasursir
Answered by mnjuly1970 last updated on 01/Dec/20
solution we know that .. i: ∫_0 ^( 1) ((1/(ln(x)))+(1/(1−x)))dx=_([IMAGE⇓⇓]) ^(why?) γ ✓ ii:∫_0 ^( 1) ((t^n −1)/(ln(t)))dt=^(why?) ln(n+1) ✓ Φ=∫_0 ^( 1) (((1+(√(1−x)))/x)+(2/(ln(1−x))))dx =^(1−x=t^2 ) 2∫_0 ^( 1) {((1+t)/(1−t^2 ))+(2/(2ln(t)))}tdt =2∫_0 ^( 1) ((1/(1−t))+(1/(ln(t))))tdt =2∫_0 ^( 1) (((t−1+1)/(1−t))+((t+1−1)/(ln(t))))dt =2∫_0 ^( 1) ((1/(1−t))+(1/(ln(t))))dt−2+∫_0 ^( 1) ((t^1 −1)/(ln(t)))dt =2γ−2+ln(2)=2(γ−1+ln(2))✓
solutionweknowthat..i:∫01(1ln(x)+11−x)dx=why?[IMAGE⇓⇓]γ✓ii:∫01tn−1ln(t)dt=why?ln(n+1)✓Φ=∫01(1+1−xx+2ln(1−x))dx=1−x=t22∫01{1+t1−t2+22ln(t)}tdt=2∫01(11−t+1ln(t))tdt=2∫01(t−1+11−t+t+1−1ln(t))dt=2∫01(11−t+1ln(t))dt−2+∫01t1−1ln(t)dt=2γ−2+ln(2)=2(γ−1+ln(2))✓
Commented by mnjuly1970 last updated on 01/Dec/20
Commented by mnjuly1970 last updated on 01/Dec/20

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *