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

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