Question Number 49941 by turbo msup by abdo last updated on 12/Dec/18
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}} {ln}\left(\mathrm{1}+{x}\right){dx} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 12/Dec/18
Commented by tanmay.chaudhury50@gmail.com last updated on 12/Dec/18
$$\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{x}} {ln}\left(\mathrm{1}+{x}\right){dx}\approx\mathrm{5}×\left(\mathrm{0}.\mathrm{2}×\mathrm{0}.\mathrm{2}\right) \\ $$
Commented by Abdo msup. last updated on 16/Dec/18
![for ∣x∣<1 we hsve ln^′ (1+x)=(1/(1+x)) =Σ_(n=0) ^∞ (−1)^n x^n ⇒ ln(1+x)=Σ_(n=0) ^∞ (((−1)^n )/(n+1))x^(n+1) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n)x^n ⇒ ∫_0 ^1 e^(−x) ln(1+x)dx =∫_0 ^1 e^(−x) (Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^n )dx =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) ∫_0 ^1 x^n e^(−x) dx by parts u =x^n and v^′ =e^(−x) A_n =∫_0 ^1 x^n e^(−x) dx =[−x^n e^(−x) ]_0 ^1 +∫_0 ^1 n x^(n−1) e^(−x) dx =−(1/e) +n A_(n−1) ⇒ A_n =nA_(n−1) −(1/e) ...be continued...](https://www.tinkutara.com/question/Q50334.png)
$${for}\:\mid{x}\mid<\mathrm{1}\:\:\:{we}\:{hsve}\:{ln}^{'} \left(\mathrm{1}+{x}\right)=\frac{\mathrm{1}}{\mathrm{1}+{x}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}{x}^{{n}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}} {ln}\left(\mathrm{1}+{x}\right){dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}} \left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{x}^{{n}} \right){dx} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} \:{e}^{−{x}} {dx}\:\:{by}\:{parts}\:\:\:{u}\:={x}^{{n}} \:{and}\:{v}^{'} \:={e}^{−{x}} \\ $$$${A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} \:{e}^{−{x}} {dx}\:=\left[−{x}^{{n}} \:{e}^{−{x}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:+\int_{\mathrm{0}} ^{\mathrm{1}} \:{n}\:{x}^{{n}−\mathrm{1}} \:{e}^{−{x}} {dx} \\ $$$$=−\frac{\mathrm{1}}{{e}}\:+{n}\:{A}_{{n}−\mathrm{1}} \:\Rightarrow\:{A}_{{n}} ={nA}_{{n}−\mathrm{1}} −\frac{\mathrm{1}}{{e}}\:\:\:…{be}\:{continued}… \\ $$