Menu Close

find-0-1-e-2x-ln-1-t-e-x-dx-with-0-lt-t-lt-1-

Tinku Tara 0 Comments
Pin




Question Number 28073 by abdo imad last updated on 20/Jan/18
find ∫_0 ^1 e^(−2x) ln(1+t e^(−x) )dx with 0<t<1 .
find∫01e−2xln(1+te−x)dxwith0<t<1.
Commented by abdo imad last updated on 26/Jan/18
let put f(t) = ∫_0 ^1 e^(−2x) ln(1+t e^(−x) )dx f^′ (t)= ∫_0 ^1 e^(−2x) (e^(−x) /(1+t e^(−x) ))dx = ∫_0 ^1 e^(−3x) (Σ_(n=0) ^∝ (−1)^n t^n e^(−nx) )dx = Σ_(n=0) ^∝ (−1)^n t^n ∫_0 ^1 e^(−(n+3)x) dx = Σ_(n=0) ^∞ (((−1)^n t^n )/(−(n+3))) [ e^(−(n+3)x) ]_0 ^1 =−Σ_(n=0) ^∞ (((−1)^n t^n )/(n+3)) (e^(−(n+3)) −1) = Σ_(n=0) ^(+∞) (((−1)^n )/(n+3)) t^n − Σ_(n=0) ^(+∞) (((−1)^n t^n e^(−(n+3)) )/(n+3)) ⇒ f(t) = ∫_0 ^t (....)du +λ = Σ_(n=0) ^∝ (((−1)^n )/((n+1)(n+3))) t^(n+1) −Σ_(n.0) ^∝ (((−1)^n t^(n+1) e^(−(n+3)) )/((n+1)(n+3))) +λ λ=f(0)=0 and f(x)= (1/2) Σ_(n=0) ^∝ (−1)^n ( (1/(n+1)) −(1/(n+3)))t^(n+1) − (e^(−2) /2)Σ_(n=0) ^∝ (−1)^n ( (1/(n+1)) − (1/(n+3)))(te^(−1) )^(n+1) = (1/(2 ))Σ_(n=0) ^∝ (((−1)^n )/(n+1))t^(n+1) −(1/2) Σ_(n=0) ^∝ (((−1)^n t^(n+1) )/(n+3)) −(e^(−2) /2) Σ_(n=0) ^∝ (((−1)^n (te^(−1) )^(n+1) )/(n+1)) +(e^(−2) /2) Σ_(n=0) ^∝ (((−1)^n (t e^(−1) )^(n+1) )/(n+3)) and all those sum are calculable ...be contiued.
letputf(t)=∫01e−2xln(1+te−x)dxf′(t)=∫01e−2xe−x1+te−xdx=∫01e−3x(∑n=0∝(−1)ntne−nx)dx=∑n=0∝(−1)ntn∫01e−(n+3)xdx=∑n=0∞(−1)ntn−(n+3)[e−(n+3)x]01=−∑n=0∞(−1)ntnn+3(e−(n+3)−1)=∑n=0+∞(−1)nn+3tn−∑n=0+∞(−1)ntne−(n+3)n+3⇒f(t)=∫0t(….)du+λ=∑n=0∝(−1)n(n+1)(n+3)tn+1−∑n.0∝(−1)ntn+1e−(n+3)(n+1)(n+3)+λλ=f(0)=0andf(x)=12∑n=0∝(−1)n(1n+1−1n+3)tn+1−e−22∑n=0∝(−1)n(1n+1−1n+3)(te−1)n+1=12∑n=0∝(−1)nn+1tn+1−12∑n=0∝(−1)ntn+1n+3−e−22∑n=0∝(−1)n(te−1)n+1n+1+e−22∑n=0∝(−1)n(te−1)n+1n+3andallthosesumarecalculable…becontiued.

Pin

Leave a Reply

Your email address will not be published.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