Question Number 73397 by mathmax by abdo last updated on 11/Nov/19
$${find}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{t}} {ln}\left(\mathrm{1}−{xt}^{\mathrm{2}} \right){dt}\:\:{with}\:\mid{x}\mid<\mathrm{1} \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{t}} {ln}\left(\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right){dt} \\ $$
Commented by abdomathmax last updated on 12/Nov/19
![1) we have ln^′ (1−u)=((−1)/(1−u)) =−Σ_(n=0) ^∞ u^n if ∣u∣<1 ⇒ ln(1−u) =−Σ_(n=0) ^∞ (u^(n+1) /(n+1)) +c (c=0) ⇒ ln(1−u) =−Σ_(n=1) ^∞ (u^n /n) we have ∣xt^2 ∣<1 ⇒ ln(1−xt^2 ) =−Σ_(n=1) ^∞ ((x^n t^(2n) )/n) ⇒ f(x)=−∫_0 ^1 e^(−t) (Σ_(n=1) ^∞ ((x^n t^(2n) )/n))dt =−Σ_(n=1) ^∞ (x^n /n) ∫_0 ^1 t^(2n) e^(−t) dt =−Σ_(n=1) ^∞ A_n (x^n /n) with A_n =∫_0 ^1 t^(2n) e^(−t) dt let determinate I_n =∫_0 ^1 t^n e^(−t) dt by parts u=t^n and v^′ =e^(−t) ⇒ I_n =[−e^(−t) t^n ]_0 ^1 +∫_0 ^1 nt^(n−1) e^(−t) dt =−e^(−1) +n I_(n−1) =nI_(n−1) −(1/e) (n>0) let W_n =(I_n /(n!)) ⇒W_(n+1) =(I_(n+1) /((n+1)!)) =(((n+1)I_n −(1/e))/((n+1)!)) =(I_n /(n!)) −(1/(e(n+1)!)) =W_n −(1/(e(n+1)!)) ⇒ W_(n+1) −W_n =−(1/(e(n+1)!)) ⇒ Σ_(k=0) ^(n−1) (W_(k+1) −W_k ) =−(1/e)Σ_(k=0) ^(n−1) (1/((k+1)!)) ⇒W_n =W_0 −(1/e) Σ_(k=1) ^n (1/(k!)) we have W_0 =I_0 =∫_0 ^1 e^(−t) dt =[−e^(−t) ]_0 ^1 =1−e^(−1) ⇒W_n =1−e^(−1) −(1/e)Σ_(k=1) ^n (1/(k!)) ⇒ I_n =n!W_n =n!{1−e^(−1) −e^(−1) Σ_(k=1) ^n (1/(k!))} ⇒ A_n =(2n)!{1−e^(−1) −e^(−1) Σ_(k=1) ^(2n) (1/(k!))} ⇒ f(x)=−Σ_(n=1) ^∞ (2n)!{1−e^(−1) −e^(−1) Σ_(k=1) ^(2n) (1/(k!))}(x^n /n) =(e^(−1) −1)Σ_(n=1) ^∞ (((2n)!)/n) x^n +e^(−1) Σ_(n=1) ^∞ (((2n)!)/n)(Σ_(k=1) ^(2n) (1/(k!)))x^n .....be continued....](https://www.tinkutara.com/question/Q73469.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{ln}^{'} \left(\mathrm{1}−{u}\right)=\frac{−\mathrm{1}}{\mathrm{1}−{u}}\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:{u}^{{n}} \:\:\:{if}\:\mid{u}\mid<\mathrm{1}\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}−{u}\right)\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{u}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:+{c}\:\:\left({c}=\mathrm{0}\right)\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}−{u}\right)\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{u}^{{n}} }{{n}}\:\:\:\:\:{we}\:{have}\:\mid{xt}^{\mathrm{2}} \mid<\mathrm{1}\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}−{xt}^{\mathrm{2}} \right)\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} \:{t}^{\mathrm{2}{n}} }{{n}}\:\Rightarrow \\ $$$${f}\left({x}\right)=−\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{t}} \:\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} \:{t}^{\mathrm{2}{n}} }{{n}}\right){dt} \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{n}} \:{e}^{−{t}} \:\:{dt}\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:{A}_{{n}} \frac{{x}^{{n}} }{{n}}\:\:{with} \\ $$$${A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{n}} \:{e}^{−{t}} \:{dt}\:\:\:{let}\:{determinate}\:\:{I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{{n}} \:{e}^{−{t}} \:{dt} \\ $$$${by}\:{parts}\:\:{u}={t}^{{n}} \:{and}\:{v}^{'} ={e}^{−{t}} \:\Rightarrow \\ $$$${I}_{{n}} =\left[−{e}^{−{t}} \:{t}^{{n}} \right]_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} {nt}^{{n}−\mathrm{1}} \:{e}^{−{t}} \:{dt} \\ $$$$=−{e}^{−\mathrm{1}} \:+{n}\:{I}_{{n}−\mathrm{1}} ={nI}_{{n}−\mathrm{1}} −\frac{\mathrm{1}}{{e}}\:\:\left({n}>\mathrm{0}\right) \\ $$$${let}\:{W}_{{n}} =\frac{{I}_{{n}} }{{n}!}\:\Rightarrow{W}_{{n}+\mathrm{1}} =\frac{{I}_{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)!}\:=\frac{\left({n}+\mathrm{1}\right){I}_{{n}} −\frac{\mathrm{1}}{{e}}}{\left({n}+\mathrm{1}\right)!} \\ $$$$=\frac{{I}_{{n}} }{{n}!}\:−\frac{\mathrm{1}}{{e}\left({n}+\mathrm{1}\right)!}\:={W}_{{n}} −\frac{\mathrm{1}}{{e}\left({n}+\mathrm{1}\right)!}\:\Rightarrow \\ $$$${W}_{{n}+\mathrm{1}} −{W}_{{n}} =−\frac{\mathrm{1}}{{e}\left({n}+\mathrm{1}\right)!}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({W}_{{k}+\mathrm{1}} −{W}_{{k}} \right)\:=−\frac{\mathrm{1}}{{e}}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)!} \\ $$$$\Rightarrow{W}_{{n}} ={W}_{\mathrm{0}} −\frac{\mathrm{1}}{{e}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}!}\:\:{we}\:{have}\:{W}_{\mathrm{0}} ={I}_{\mathrm{0}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{t}} \:{dt} \\ $$$$=\left[−{e}^{−{t}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:=\mathrm{1}−{e}^{−\mathrm{1}} \:\Rightarrow{W}_{{n}} =\mathrm{1}−{e}^{−\mathrm{1}} −\frac{\mathrm{1}}{{e}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}!}\:\Rightarrow \\ $$$${I}_{{n}} ={n}!{W}_{{n}} ={n}!\left\{\mathrm{1}−{e}^{−\mathrm{1}} −{e}^{−\mathrm{1}} \sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}!}\right\}\:\Rightarrow \\ $$$${A}_{{n}} =\left(\mathrm{2}{n}\right)!\left\{\mathrm{1}−{e}^{−\mathrm{1}} −{e}^{−\mathrm{1}} \sum_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \:\frac{\mathrm{1}}{{k}!}\right\}\:\Rightarrow \\ $$$${f}\left({x}\right)=−\sum_{{n}=\mathrm{1}} ^{\infty} \left(\mathrm{2}{n}\right)!\left\{\mathrm{1}−{e}^{−\mathrm{1}} −{e}^{−\mathrm{1}} \sum_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \:\frac{\mathrm{1}}{{k}!}\right\}\frac{{x}^{{n}} }{{n}} \\ $$$$=\left({e}^{−\mathrm{1}} −\mathrm{1}\right)\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(\mathrm{2}{n}\right)!}{{n}}\:{x}^{{n}} \:+{e}^{−\mathrm{1}} \sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(\mathrm{2}{n}\right)!}{{n}}\left(\sum_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \:\frac{\mathrm{1}}{{k}!}\right){x}^{{n}} \\ $$$$…..{be}\:{continued}…. \\ $$$$ \\ $$
Answered by mind is power last updated on 11/Nov/19
![f(x)=[−e^(−t) ln(1−xt^2 )]_0 ^1 +∫_0 ^1 ((e^(−t) .−2xt)/(1−xt^2 ))dt =−e^(−1) ln(1−x)−2x∫_0 ^1 .((te^(−t) dt)/((1−(√x)t)(1+(√x)t))) =−e^(−1) ln(1−x)−(√x)∫_0 ^1 ((e^(−t) .(1+t(√x)−(1−t(√x))))/((1−t(√x))(1+t(√x))))dt =−e^(−1) ln(1−x)−(√x)∫_0 ^1 ((e^(−t) dt)/(1−t(√x)))+(√x)∫_0 ^1 ((e^(−t) dt)/(1+t(√x))) let u=1−t(√x) in first and s=1+t(√x) in 2nd =−e^(−1) ln(1−x)+∫_1 ^(1−(√x)) (e^((u−1)/( (√x))) /u)du+∫_1 ^(1+(√x)) (e^((1−s)/( (√x))) /s)ds m=(u/( (√x))) in first m=(s/( (√x))) in]2nd =−e^(−1) ln(1−x)+∫^((1/( (√x)))−1) _(1/( (√x))) ((e^m .e^(−(1/( (√x)))) )/m).dm+∫_(1/( (√x))) ^(1+(1/( (√x)))) e^(1/( (√x))) .e^(−m) .(dm/m)346 We use Ei(w)=∫_(−∞) ^w (e^x /x)dx=∫_∞ ^(−w) (e^(−x) /x)dx Ei is exponential integral function we get −e^(−1) ln(1−x)+{Ei((1/( (√x)))−1)−Ei((1/( (√x))))}e^((−1)/( (√x))) +e^(1/( (√x))) {Ei(−1−(1/( (√x))))−Ei(−(1/( (√x))))} ∫_0 ^1 e^(−t) ln(1−(t^2 /2))dt put x=(1/2) =e^(−1) ln(2)+{Ei((1/( (√2)))−1)−Ei((1/( (√2))))}e^(−(1/( (√2)))) +e^(1/( (√2))) {Ei(−1−(1/( (√2))))−Ei((1/( (√2))))}](https://www.tinkutara.com/question/Q73414.png)
$${f}\left({x}\right)=\left[−{e}^{−{t}} {ln}\left(\mathrm{1}−{xt}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{e}^{−{t}} .−\mathrm{2}{xt}}{\mathrm{1}−{xt}^{\mathrm{2}} }{dt} \\ $$$$=−{e}^{−\mathrm{1}} {ln}\left(\mathrm{1}−{x}\right)−\mathrm{2}{x}\int_{\mathrm{0}} ^{\mathrm{1}} .\frac{{te}^{−{t}} {dt}}{\left(\mathrm{1}−\sqrt{{x}}{t}\right)\left(\mathrm{1}+\sqrt{{x}}{t}\right)} \\ $$$$=−{e}^{−\mathrm{1}} {ln}\left(\mathrm{1}−{x}\right)−\sqrt{{x}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{e}^{−{t}} .\left(\mathrm{1}+{t}\sqrt{{x}}−\left(\mathrm{1}−{t}\sqrt{{x}}\right)\right)}{\left(\mathrm{1}−{t}\sqrt{{x}}\right)\left(\mathrm{1}+{t}\sqrt{{x}}\right)}{dt} \\ $$$$=−{e}^{−\mathrm{1}} {ln}\left(\mathrm{1}−{x}\right)−\sqrt{{x}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{e}^{−{t}} {dt}}{\mathrm{1}−{t}\sqrt{{x}}}+\sqrt{{x}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{e}^{−{t}} {dt}}{\mathrm{1}+{t}\sqrt{{x}}} \\ $$$${let}\:{u}=\mathrm{1}−{t}\sqrt{{x}}\:{in}\:{first}\:{and}\:{s}=\mathrm{1}+{t}\sqrt{{x}}\:{in}\:\mathrm{2}{nd} \\ $$$$=−{e}^{−\mathrm{1}} {ln}\left(\mathrm{1}−{x}\right)+\int_{\mathrm{1}} ^{\mathrm{1}−\sqrt{{x}}} \frac{{e}^{\frac{{u}−\mathrm{1}}{\:\sqrt{{x}}}} }{{u}}{du}+\int_{\mathrm{1}} ^{\mathrm{1}+\sqrt{{x}}} \frac{{e}^{\frac{\mathrm{1}−{s}}{\:\sqrt{{x}}}} }{{s}}{ds} \\ $$$$\left.{m}=\frac{{u}}{\:\sqrt{{x}}}\:{in}\:{first}\:{m}=\frac{{s}}{\:\sqrt{{x}}}\:\:{in}\right]\mathrm{2}{nd} \\ $$$$=−{e}^{−\mathrm{1}} {ln}\left(\mathrm{1}−{x}\right)+\underset{\frac{\mathrm{1}}{\:\sqrt{{x}}}} {\int}^{\frac{\mathrm{1}}{\:\sqrt{{x}}}−\mathrm{1}} \frac{{e}^{{m}} .{e}^{−\frac{\mathrm{1}}{\:\sqrt{{x}}}} }{{m}}.{dm}+\int_{\frac{\mathrm{1}}{\:\sqrt{{x}}}} ^{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{x}}}} {e}^{\frac{\mathrm{1}}{\:\sqrt{{x}}}} .{e}^{−{m}} .\frac{{dm}}{{m}}\mathrm{346} \\ $$$${We}\:{use}\:\:{Ei}\left({w}\right)=\int_{−\infty} ^{{w}} \frac{{e}^{{x}} }{{x}}{dx}=\int_{\infty} ^{−{w}} \frac{{e}^{−{x}} }{{x}}{dx} \\ $$$${Ei}\:\:{is}\:{exponential}\:{integral}\:{function}\: \\ $$$${we}\:{get}\:−{e}^{−\mathrm{1}} {ln}\left(\mathrm{1}−{x}\right)+\left\{{Ei}\left(\frac{\mathrm{1}}{\:\sqrt{{x}}}−\mathrm{1}\right)−{Ei}\left(\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)\right\}{e}^{\frac{−\mathrm{1}}{\:\sqrt{{x}}}} +{e}^{\frac{\mathrm{1}}{\:\sqrt{{x}}}} \left\{{Ei}\left(−\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)−{Ei}\left(−\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)\right\} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{t}} {ln}\left(\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right){dt}\:{put}\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$={e}^{−\mathrm{1}} {ln}\left(\mathrm{2}\right)+\left\{{Ei}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\mathrm{1}\right)−{Ei}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\right\}{e}^{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} +{e}^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \left\{{Ei}\left(−\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)−{Ei}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\right\} \\ $$
Commented by mind is power last updated on 11/Nov/19
$${y}'{re}\:{welcom} \\ $$
Commented by mathmax by abdo last updated on 11/Nov/19
$${thank}\:{you}\:{sir}. \\ $$