Question Number 29971 by abdo imad last updated on 14/Feb/18
$${find}\:{J}\left({x}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{x}+{e}^{{t}} }\:\:\:\:?. \\ $$
Commented by abdo imad last updated on 16/Feb/18
![J(x)= ∫_0 ^∞ (dt/(e^t ( 1+x e^(−t) ))) =∫_0 ^∞ e^(−t) (Σ_(n=0) ^∞ x^n e^(−nt) )dt = Σ_(n=0) ^∞ x^n ∫_0 ^∞ e^(−(n+1)t) dt the (n+1)t=u give ∫_0 ^∞ e^(−(n+1)t) dt= ∫_0 ^∞ e^(−u) (du/(n+1)) =(1/(n+1)) [−e^(−u) ]_0 ^(+∞) =(1/(n+1)) J(x)= Σ_(n=0) ^∞ (x^n /(n+1)) ⇒ x J(x)=Σ_(n=0) ^∞ (x^(n+1) /(n+1)) let derivate J(x) +xJ^′ (x)= Σ_(n=0) ^∞ x^n = (1/(1−x)) so J is solution of the d.e y +xy^′ =(1/(1−x)) eh⇒ xy^′ =−y ⇒(y^′ /y) =((−1)/x) ⇒ ln∣y∣= −ln∣x∣+c ⇒y= (λ/x) mvc method give y^′ =((λ^′ x −λ)/x^2 ) ⇒(λ/x) +λ^′ −(λ/x) = (1/(1−x)) ⇒λ(x)=∫ (dx/(1−x)) +κ =−ln∣1−x∣ +k but k=λ(0)=0 y(x)=−(1/x)ln∣1−x∣ ⇒ J(x)=−(1/x)ln∣1−x∣ .](https://www.tinkutara.com/question/Q30110.png)
$${J}\left({x}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{{e}^{{t}} \left(\:\mathrm{1}+{x}\:{e}^{−{t}} \right)}\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} \:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} {e}^{−{nt}} \right){dt} \\ $$$$=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:{x}^{{n}} \:\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left({n}+\mathrm{1}\right){t}} {dt}\:\:{the}\:\left({n}+\mathrm{1}\right){t}={u}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left({n}+\mathrm{1}\right){t}} {dt}=\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}} \:\frac{{du}}{{n}+\mathrm{1}}\:=\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\left[−{e}^{−{u}} \right]_{\mathrm{0}} ^{+\infty} \:=\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$${J}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}} }{{n}+\mathrm{1}}\:\Rightarrow\:{x}\:{J}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:{let}\:{derivate} \\ $$$${J}\left({x}\right)\:+{xJ}^{'} \left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\:{so}\:{J}\:{is}\:{solution}\:{of}\:{the} \\ $$$${d}.{e}\:{y}\:+{xy}^{'} \:\:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\:{eh}\Rightarrow\:{xy}^{'} =−{y}\:\Rightarrow\frac{{y}^{'} }{{y}}\:=\frac{−\mathrm{1}}{{x}}\:\Rightarrow \\ $$$${ln}\mid{y}\mid=\:−{ln}\mid{x}\mid+{c}\:\Rightarrow{y}=\:\frac{\lambda}{{x}}\:\:{mvc}\:{method}\:{give} \\ $$$${y}^{'} \:=\frac{\lambda^{'} {x}\:−\lambda}{{x}^{\mathrm{2}} }\:\Rightarrow\frac{\lambda}{{x}}\:+\lambda^{'} \:−\frac{\lambda}{{x}}\:=\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\Rightarrow\lambda\left({x}\right)=\int\:\frac{{dx}}{\mathrm{1}−{x}}\:+\kappa \\ $$$$=−{ln}\mid\mathrm{1}−{x}\mid\:+{k}\:{but}\:{k}=\lambda\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${y}\left({x}\right)=−\frac{\mathrm{1}}{{x}}{ln}\mid\mathrm{1}−{x}\mid\:\Rightarrow\:{J}\left({x}\right)=−\frac{\mathrm{1}}{{x}}{ln}\mid\mathrm{1}−{x}\mid\:. \\ $$
Commented by abdo imad last updated on 16/Feb/18
$${the}\:{Q}.\:{is}\:{find}\:{J}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{x}\:+{e}^{{t}} }\:{with}\:\mid{x}\mid<\mathrm{1}. \\ $$