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Question Number 26756 by abdo imad last updated on 28/Dec/17

prove that ∫_0 ^1 (dx/(x+ e^x )) = Σ_(n=0) ^∝ (((−1)^n )/((n+1)^(n+1) )) A_n with A_n = ∫_0 ^(n+1) t^n e^(−t) dt .

$${prove}\:{that}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{{x}+\:{e}^{{x}} }\:=\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:{A}_{{n}} \\ $$$${with}\:\:{A}_{{n}} \:=\:\int_{\mathrm{0}} ^{{n}+\mathrm{1}} \:{t}^{{n}} \:{e}^{−{t}} {dt}\:. \\ $$

Commented by abdo imad last updated on 01/Jan/18

let put I= ∫_0 ^1 (dx/(x+e^x )) I= ∫_0 ^1 (e^(−x) /(1+x e^(−x) ))dx but /xe^(−x) /≤1 I= ∫_0 ^1 ( Σ_(n=0) ^∝ (−1)^n x^n e^(−nx) )e^(−x) dx = Σ_(n=0) ^∝ (−1)^n ∫_0 ^1 x^n e^(−(n+1)x) dx and by the chsngement (n+1)x=t ∫_0 ^1 x^n e^(−(n+1)x) dx = ∫_0 ^(n+1) ((t/(n+1)))^n e^(−t) (dt/(n+1)) = (1/((n+1)^(n+1) )) ∫_0 ^1 t^n e^(−t) dt= (A_n /((n+1)^(n+1) )) and finally ? I= Σ_(n=0) ^(n=∝) (((−1)^n )/((n+1)^(n+1) )) A_n with A_n = ∫_0 ^1 t^n e^(−t) dt .

$${let}\:{put}\:{I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{{x}+{e}^{{x}} } \\ $$$${I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{e}^{−{x}} }{\mathrm{1}+{x}\:{e}^{−{x}} }{dx}\:\:{but}\:/{xe}^{−{x}} /\leqslant\mathrm{1} \\ $$$${I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\:\sum_{{n}=\mathrm{0}} ^{\propto} \left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \:{e}^{−{nx}} \right){e}^{−{x}} {dx} \\ $$$$=\:\sum_{{n}=\mathrm{0}} ^{\propto} \left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} \:{e}^{−\left({n}+\mathrm{1}\right){x}} {dx} \\ $$$${and}\:{by}\:{the}\:{chsngement}\:\left({n}+\mathrm{1}\right){x}={t} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} {e}^{−\left({n}+\mathrm{1}\right){x}} {dx}\:=\:\int_{\mathrm{0}} ^{{n}+\mathrm{1}} \left(\frac{{t}}{{n}+\mathrm{1}}\right)^{{n}} {e}^{−{t}} \frac{{dt}}{{n}+\mathrm{1}} \\ $$$$=\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{{n}} \:{e}^{−{t}} {dt}=\:\frac{{A}_{{n}} }{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} } \\ $$$${and}\:{finally}\:?\:{I}=\:\sum_{{n}=\mathrm{0}} ^{{n}=\propto} \frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:{A}_{{n}} \\ $$$${with}\:{A}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{{n}} \:{e}^{−{t}} {dt}\:. \\ $$

Commented by abdo imad last updated on 01/Jan/18

∫_0 ^1 x^n e^(−(n+1)x) dx=(1/((n+1)^(n+1) )) ∫_0 ^(n+1) t^n e^(−t) dt = (A_n /((n+1)^(n+1) )) with A_n = ∫_0 ^(n+1) t^n e^(−t) dt.

$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {e}^{−\left({n}+\mathrm{1}\right){x}} {dx}=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:\int_{\mathrm{0}} ^{{n}+\mathrm{1}} {t}^{{n}} \:{e}^{−{t}} {dt} \\ $$$$=\:\frac{{A}_{{n}} }{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:{with}\:{A}_{{n}} \:=\:\int_{\mathrm{0}} ^{{n}+\mathrm{1}} {t}^{{n}} {e}^{−{t}} {dt}. \\ $$

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