Menu Close

find-f-x-0-1-e-t-ln-1-xt-2-dt-with-x-lt-1-2-calculate-0-1-e-t-ln-1-t-2-2-dt-

Tinku Tara 0 Comments
Pin




Question Number 73397 by mathmax by abdo last updated on 11/Nov/19
find f(x)=∫_0 ^1 e^(−t) ln(1−xt^2 )dt with ∣x∣<1 2)calculate ∫_0 ^1 e^(−t) ln(1−(t^2 /2))dt
findf(x)=∫01e−tln(1−xt2)dtwith∣x∣<12)calculate∫01e−tln(1−t22)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....
1)wehaveln′(1−u)=−11−u=−∑n=0∞unif∣u∣<1⇒ln(1−u)=−∑n=0∞un+1n+1+c(c=0)⇒ln(1−u)=−∑n=1∞unnwehave∣xt2∣<1⇒ln(1−xt2)=−∑n=1∞xnt2nn⇒f(x)=−∫01e−t(∑n=1∞xnt2nn)dt=−∑n=1∞xnn∫01t2ne−tdt=−∑n=1∞AnxnnwithAn=∫01t2ne−tdtletdeterminateIn=∫01tne−tdtbypartsu=tnandv′=e−t⇒In=[−e−ttn]01+∫01ntn−1e−tdt=−e−1+nIn−1=nIn−1−1e(n>0)letWn=Inn!⇒Wn+1=In+1(n+1)!=(n+1)In−1e(n+1)!=Inn!−1e(n+1)!=Wn−1e(n+1)!⇒Wn+1−Wn=−1e(n+1)!⇒∑k=0n−1(Wk+1−Wk)=−1e∑k=0n−11(k+1)!⇒Wn=W0−1e∑k=1n1k!wehaveW0=I0=∫01e−tdt=[−e−t]01=1−e−1⇒Wn=1−e−1−1e∑k=1n1k!⇒In=n!Wn=n!{1−e−1−e−1∑k=1n1k!}⇒An=(2n)!{1−e−1−e−1∑k=12n1k!}⇒f(x)=−∑n=1∞(2n)!{1−e−1−e−1∑k=12n1k!}xnn=(e−1−1)∑n=1∞(2n)!nxn+e−1∑n=1∞(2n)!n(∑k=12n1k!)xn…..becontinued….
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))))}
f(x)=[−e−tln(1−xt2)]01+∫01e−t.−2xt1−xt2dt=−e−1ln(1−x)−2x∫01.te−tdt(1−xt)(1+xt)=−e−1ln(1−x)−x∫01e−t.(1+tx−(1−tx))(1−tx)(1+tx)dt=−e−1ln(1−x)−x∫01e−tdt1−tx+x∫01e−tdt1+txletu=1−txinfirstands=1+txin2nd=−e−1ln(1−x)+∫11−xeu−1xudu+∫11+xe1−sxsdsm=uxinfirstm=sxin]2nd=−e−1ln(1−x)+∫1x1x−1em.e−1xm.dm+∫1x1+1xe1x.e−m.dmm346WeuseEi(w)=∫−∞wexxdx=∫∞−we−xxdxEiisexponentialintegralfunctionweget−e−1ln(1−x)+{Ei(1x−1)−Ei(1x)}e−1x+e1x{Ei(−1−1x)−Ei(−1x)}∫01e−tln(1−t22)dtputx=12=e−1ln(2)+{Ei(12−1)−Ei(12)}e−12+e12{Ei(−1−12)−Ei(12)}
Commented by mind is power last updated on 11/Nov/19
y′re welcom
y′rewelcom
Commented by mathmax by abdo last updated on 11/Nov/19
thank you sir.
thankyousir.

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *