Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 53956 by maxmathsup by imad last updated on 27/Jan/19

give∫_0 ^1 e^(−x) ln(1−x)dx at form of serie

$${give}\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}} {ln}\left(\mathrm{1}−{x}\right){dx}\:\:{at}\:{form}\:{of}\:{serie} \\ $$

Commented by maxmathsup by imad last updated on 31/Jan/19

we have e^(−x) =Σ_(n.=0) ^∞ (((−1)^n x^n )/(n!)) ln^′ (1−x) =−(1/(1−x)) =−Σ_(n=0) ^∞ x^n ⇒ln(1−x) =−Σ_(n=0) ^∞ (x^(n+1) /(n+1)) =−Σ_(n=1) ^∞ (x^n /n) ⇒e^(−x) ln(1−x) =−(Σ_(n=0) ^∞ (((−1)^n )/(n!)) x^n )(Σ_(n=1) ^∞ (x^n /n)) =−( 1+ Σ_(n=1) ^∞ (((−1)^n )/(n!)) x^n )(Σ_(n=1) ^∞ (x^n /n)) =−Σ_(n=1) ^∞ (x^n /n) −(Σ_(n=1) ^∞ (((−1)^n )/(n!)) x^n )(Σ_(n=1) ^∞ (x^n /n)) but (Σ_(n=1) ^∞ (((−1)^n )/(n!)) x^n ).(Σ_(n=1) ^∞ (x^n /n))=Σ_(n=0) ^∞ C_n x^n wih C_n =Σ_(i+j =n ) a_i b_j =Σ_(i=1) ^n (((−1)^i )/(i!)) (1/(n−i+1)) ⇒ e^(−x) ln(1−x) =−Σ_(n=1) ^∞ (x^n /n) −Σ_(n=0) ^∞ ( Σ_(i=1) ^n (((−1)^i )/(i!(n−i+1))))x^n ⇒ ∫_0 ^1 e^(−x) ln(1−x)dx =−Σ_(n=1) ^∞ (x^(n+1) /(n(n+1))) −Σ_(n=0) ^∞ (1/(n+1))(Σ_(i=1) ^n (((−1)^i )/(i!(n−i+1))))x^(n+1) .

$${we}\:{have}\:{e}^{−{x}} \:=\sum_{{n}.=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \:{x}^{{n}} }{{n}!} \\ $$$${ln}^{'} \left(\mathrm{1}−{x}\right)\:=−\frac{\mathrm{1}}{\mathrm{1}−{x}}\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:\Rightarrow{ln}\left(\mathrm{1}−{x}\right)\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:\Rightarrow{e}^{−{x}} {ln}\left(\mathrm{1}−{x}\right)\:=−\left(\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:{x}^{{n}} \right)\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{x}^{{n}} }{{n}}\right) \\ $$$$=−\left(\:\mathrm{1}+\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:{x}^{{n}} \right)\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\right) \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:−\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:{x}^{{n}} \right)\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\right)\:\:\:{but}\: \\ $$$$\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:{x}^{{n}} \right).\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:{C}_{{n}} {x}^{{n}} \:\:{wih} \\ $$$${C}_{{n}} =\sum_{{i}+{j}\:={n}\:} \:{a}_{{i}} {b}_{{j}} =\sum_{{i}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{i}} }{{i}!}\:\frac{\mathrm{1}}{{n}−{i}+\mathrm{1}}\:\Rightarrow \\ $$$${e}^{−{x}} {ln}\left(\mathrm{1}−{x}\right)\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:−\sum_{{n}=\mathrm{0}} ^{\infty} \left(\:\sum_{{i}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{i}} }{{i}!\left({n}−{i}+\mathrm{1}\right)}\right){x}^{{n}} \:\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}} {ln}\left(\mathrm{1}−{x}\right){dx}\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}+\mathrm{1}} }{{n}\left({n}+\mathrm{1}\right)}\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{n}+\mathrm{1}}\left(\sum_{{i}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{i}} }{{i}!\left({n}−{i}+\mathrm{1}\right)}\right){x}^{{n}+\mathrm{1}} . \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com