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Question Number 40587 by mondodotto@gmail.com last updated on 24/Jul/18
find ∫e^x lnx dx
$$\boldsymbol{\mathrm{find}}\:\int\boldsymbol{\mathrm{e}}^{\boldsymbol{{x}}} \boldsymbol{\mathrm{ln}{x}}\:\boldsymbol{\mathrm{d}{x}} \\ $$
Commented by prof Abdo imad last updated on 25/Jul/18
∫ e^x ln(x)dx = ∫(Σ_(n=0) ^∞ (x^n /(n!)))ln(x)dx =Σ_(n=0) ^∞ (1/(n!)) ∫ x^n ln(x)dx by parts A_n =∫ x^n ln(x)dx=(1/(n+1))x^(n+1) ln(x)−∫ (1/(n+1))x^n dx =(1/(n+1)) x^(n+1) ln(x) −(1/((n+1)^2 )) +λ⇒ ∫ e^x ln(x)dx=Σ_(n=0) ^∞ (1/(n!)){ (1/(n+1)) x^(n+1) ln(x)−(1/((n+1)^2 )) +λ} =ln(x)Σ_(n=0) ^∞ (x^(n+1) /((n+1)n!)) −Σ_(n=0) ^∞ (1/((n+1)^2 n!)) +λe let w(x)=Σ_(n=0) ^∞ (x^(n+1) /((n+1)n!)) we have w^′ (x) =Σ_(n=0) ^∞ (x^n /(n!)) =e^x ⇒w(x)=e^x +c w(0)=0=1+c ⇒c=−1 ⇒w(x)=e^x −1 ⇒ ∫ e^x ln(x)dx =(e^x −1)ln(x)−Σ_(n=0) ^∞ (1/((n+1)^2 n!)) +λe ...be continued...
$$\int\:{e}^{{x}} {ln}\left({x}\right){dx}\:=\:\int\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} }{{n}!}\right){ln}\left({x}\right){dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{n}!}\:\int\:\:\:{x}^{{n}} {ln}\left({x}\right){dx}\:{by}\:{parts} \\ $$$${A}_{{n}} =\int\:{x}^{{n}} {ln}\left({x}\right){dx}=\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} {ln}\left({x}\right)−\int\:\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}^{{n}} {dx} \\ $$$$=\frac{\mathrm{1}}{{n}+\mathrm{1}}\:{x}^{{n}+\mathrm{1}} {ln}\left({x}\right)\:−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:+\lambda\Rightarrow \\ $$$$\int\:{e}^{{x}} {ln}\left({x}\right){dx}=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{n}!}\left\{\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\:{x}^{{n}+\mathrm{1}} {ln}\left({x}\right)−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:+\lambda\right\} \\ $$$$={ln}\left({x}\right)\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right){n}!}\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} {n}!}\:+\lambda{e} \\ $$$${let}\:{w}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right){n}!}\:\:{we}\:{have} \\ $$$${w}^{'} \left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}} }{{n}!}\:={e}^{{x}} \:\Rightarrow{w}\left({x}\right)={e}^{{x}} \:+{c}\: \\ $$$${w}\left(\mathrm{0}\right)=\mathrm{0}=\mathrm{1}+{c}\:\Rightarrow{c}=−\mathrm{1}\:\Rightarrow{w}\left({x}\right)={e}^{{x}} −\mathrm{1}\:\Rightarrow \\ $$$$\int\:{e}^{{x}} {ln}\left({x}\right){dx}\:=\left({e}^{{x}} −\mathrm{1}\right){ln}\left({x}\right)−\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} {n}!} \\ $$$$+\lambda{e}\:…{be}\:{continued}… \\ $$

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