Question Number 28073 by abdo imad last updated on 20/Jan/18
$${find}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−\mathrm{2}{x}} {ln}\left(\mathrm{1}+{t}\:{e}^{−{x}} \right){dx}\:\:\:{with}\:\:\mathrm{0}<{t}<\mathrm{1}\:\:. \\ $$
Commented byabdo 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.](Q28476.png)
$${let}\:{put}\:\:{f}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−\mathrm{2}{x}} {ln}\left(\mathrm{1}+{t}\:{e}^{−{x}} \right){dx} \\ $$ $${f}^{'} \left({t}\right)=\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−\mathrm{2}{x}} \:\frac{{e}^{−{x}} }{\mathrm{1}+{t}\:{e}^{−{x}} }{dx}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−\mathrm{3}{x}} \:\left(\sum_{{n}=\mathrm{0}} ^{\propto} \:\left(−\mathrm{1}\right)^{{n}} \:{t}^{{n}} \:{e}^{−{nx}} \right){dx} \\ $$ $$=\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\:\left(−\mathrm{1}\right)^{{n}} \:{t}^{{n}} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−\left({n}+\mathrm{3}\right){x}} {dx} \\ $$ $$=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {t}^{{n}} }{−\left({n}+\mathrm{3}\right)}\:\:\left[\:\:{e}^{−\left({n}+\mathrm{3}\right){x}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$ $$=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{t}^{{n}} }{{n}+\mathrm{3}}\:\left({e}^{−\left({n}+\mathrm{3}\right)} \:−\mathrm{1}\right) \\ $$ $$=\:\sum_{{n}=\mathrm{0}} ^{+\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{3}}\:{t}^{{n}} \:\:−\:\sum_{{n}=\mathrm{0}} ^{+\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} {t}^{{n}} \:{e}^{−\left({n}+\mathrm{3}\right)} }{{n}+\mathrm{3}} \\ $$ $$\Rightarrow\:{f}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{{t}} \left(....\right){du}\:+\lambda \\ $$ $$=\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{3}\right)}\:{t}^{{n}+\mathrm{1}} \:\:−\sum_{{n}.\mathrm{0}} ^{\propto} \:\frac{\left(−\mathrm{1}\right)^{{n}} {t}^{{n}+\mathrm{1}} \:{e}^{−\left({n}+\mathrm{3}\right)} }{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{3}\right)}\:+\lambda \\ $$ $$\lambda={f}\left(\mathrm{0}\right)=\mathrm{0}\:\:{and} \\ $$ $${f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\left(−\mathrm{1}\right)^{{n}} \left(\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\:−\frac{\mathrm{1}}{{n}+\mathrm{3}}\right){t}^{{n}+\mathrm{1}} \:\: \\ $$ $$\:\:−\:\frac{{e}^{−\mathrm{2}} }{\mathrm{2}}\sum_{{n}=\mathrm{0}} ^{\propto} \:\:\left(−\mathrm{1}\right)^{{n}} \left(\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\:−\:\frac{\mathrm{1}}{{n}+\mathrm{3}}\right)\left({te}^{−\mathrm{1}} \right)^{{n}+\mathrm{1}} \\ $$ $$=\:\frac{\mathrm{1}}{\mathrm{2}\:}\sum_{{n}=\mathrm{0}} ^{\propto} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}{t}^{{n}+\mathrm{1}} \:\:−\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} {t}^{{n}+\mathrm{1}} }{{n}+\mathrm{3}} \\ $$ $$−\frac{{e}^{−\mathrm{2}} }{\mathrm{2}}\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\frac{\left(−\mathrm{1}\right)^{{n}} \left({te}^{−\mathrm{1}} \right)^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:\:+\frac{{e}^{−\mathrm{2}} }{\mathrm{2}}\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\frac{\left(−\mathrm{1}\right)^{{n}} \left({t}\:{e}^{−\mathrm{1}} \right)^{{n}+\mathrm{1}} }{{n}+\mathrm{3}} \\ $$ $${and}\:{all}\:{those}\:{sum}\:{are}\:{calculable}\:...{be}\:{contiued}. \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$