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Question Number 52945 by Tawa1 last updated on 15/Jan/19

Commented by maxmathsup by imad last updated on 15/Jan/19

3) let A =∫_0 ^∞ ((x ln^2 (x))/(e^x −1)) dx ⇒A =∫_0 ^∞ ((x e^(−x) ln^2 (x))/(1−e^(−x) )) dx =∫_0 ^∞ x e^(−x) ln^2 (x)(Σ_(n=0) ^∞ e^(−nx) )dx =Σ_(n=0) ^∞ ∫_0 ^∞ x e^(−(n+1)x) ln^2 x dx =Σ_(n=0) ^∞ A_n with A_n =∫_0 ^∞ x e^(−(n+1)x) ln^2 (x)dx A_n =_((n+1)x=t) ∫_0 ^∞ (t/(n+1)) e^(−t) ln^2 ((t/(n+1)))(dt/(n+1)) =(1/((n+1)^2 ))∫_0 ^∞ t e^(−t) (ln(t)−ln(n+1))^2 dt ⇒(n+1)^2 A_n =∫_0 ^∞ t e^(−t) (ln^2 (t)−2ln(n+1)ln(t) +ln^2 (n+1))dt =∫_0 ^∞ t e^(−t) ln^2 (t)dt−2ln(n+1)∫_0 ^∞ t e^(−t) ln(t)dt +ln^2 (n+1)∫_0 ^∞ t e^(−t) dt but ∫_0 ^∞ t e^(−t) dt =[−t e^(−t) ]_0 ^(+∞) +∫_0 ^∞ e^(−t) dt =[−e^(−t) ]_0 ^(+∞) =1 by part u^′ =t e^(−t) and v=ln(t) ⇒u =∫ t e^(−t) dt =−t e^(−t) +∫ e^(−t) dt =−t e^(−t) −e^(−t) =−(t+1)e^(−t) +1 ⇒∫_0 ^∞ t e^(−t) ln(t)dt =[(−(t+1)e^(−t) +1)ln(t)]_0 ^(+∞) −∫_0 ^∞ t e^(−t) (dt/t) =−∫_0 ^∞ e^(−t) dt =[e^(−t) ]_0 ^(+∞) =−1 let find ∫_0 ^∞ t e^(−t) ln^2 (t)dt by parts u^′ =t e^(−t) and v=ln^2 (t) ⇒ u =∫ t e^(−t) dt =1−(t+1)e^(−t) ⇒∫_0 ^∞ t e^(−t) ln^2 (t)dt =[(1−(t+1)e^(−t) ln^2 t]_0 ^(+∞) −∫_0 ^∞ t e^(−t) ((2ln(t))/t)dt =−2 ∫_0 ^∞ e^(−t) ln(t)dt =−2(−γ)=2γ ( this result is proved )⇒ (n+1)^2 A_n =2γ +2ln(n+1) +ln^2 (n+1) ⇒ A_n =((2γ)/((n+1)^2 )) + 2((ln(n+1))/((n+1)^2 )) +((ln^2 (n+1))/((n+1)^2 )) ⇒ A =2γ Σ_(n=0) ^∞ (1/((n+1)^2 )) +2Σ_(n=0) ^∞ ((ln(n+1))/((n+1)^2 )) +Σ_(n=0) ^∞ ((ln^2 (n+1))/((n+1)^2 )) Σ_(n=0) ^∞ (1/((n+1)^2 )) =ξ(2)=(π^2 /6) ....becontinued...

3)letA=∫0∞xln2(x)ex−1dx⇒A=∫0∞xe−xln2(x)1−e−xdx=∫0∞xe−xln2(x)(∑n=0∞e−nx)dx=∑n=0∞∫0∞xe−(n+1)xln2xdx=∑n=0∞AnwithAn=∫0∞xe−(n+1)xln2(x)dxAn=(n+1)x=t∫0∞tn+1e−tln2(tn+1)dtn+1=1(n+1)2∫0∞te−t(ln(t)−ln(n+1))2dt⇒(n+1)2An=∫0∞te−t(ln2(t)−2ln(n+1)ln(t)+ln2(n+1))dt=∫0∞te−tln2(t)dt−2ln(n+1)∫0∞te−tln(t)dt+ln2(n+1)∫0∞te−tdtbut∫0∞te−tdt=[−te−t]0+∞+∫0∞e−tdt=[−e−t]0+∞=1bypartu′=te−tandv=ln(t)⇒u=∫te−tdt=−te−t+∫e−tdt=−te−t−e−t=−(t+1)e−t+1⇒∫0∞te−tln(t)dt=[(−(t+1)e−t+1)ln(t)]0+∞−∫0∞te−tdtt=−∫0∞e−tdt=[e−t]0+∞=−1letfind∫0∞te−tln2(t)dtbypartsu′=te−tandv=ln2(t)⇒u=∫te−tdt=1−(t+1)e−t⇒∫0∞te−tln2(t)dt=[(1−(t+1)e−tln2t]0+∞−∫0∞te−t2ln(t)tdt=−2∫0∞e−tln(t)dt=−2(−γ)=2γ(thisresultisproved)⇒(n+1)2An=2γ+2ln(n+1)+ln2(n+1)⇒An=2γ(n+1)2+2ln(n+1)(n+1)2+ln2(n+1)(n+1)2⇒A=2γ∑n=0∞1(n+1)2+2∑n=0∞ln(n+1)(n+1)2+∑n=0∞ln2(n+1)(n+1)2∑n=0∞1(n+1)2=ξ(2)=π26....becontinued...

Commented by Tawa1 last updated on 15/Jan/19

God bless you sir, waiting for the rest sir.

Godblessyousir,waitingfortherestsir.

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

1)y=cos^(−1) (((1−x^2 )/(1+x^2 ))) put x=tana y=cos^(−1) (((1−tan^2 a)/(1+tan^2 a)))=cos^(−1) (cos2a) y=2a=2tan^(−1) x (dy/dx)=(2/(1+x^2 )) (1+x^2 )(dy/dx)−2=0 proved

1)y=cos−1(1−x21+x2)putx=tanay=cos−1(1−tan2a1+tan2a)=cos−1(cos2a)y=2a=2tan−1xdydx=21+x2(1+x2)dydx−2=0proved

Commented by Tawa1 last updated on 15/Jan/19

God bless you sir.

Godblessyousir.

Commented by Tawa1 last updated on 15/Jan/19

Please sir, 3 and 5

Pleasesir,3and5

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

2)cos^(−1) x=a cosa=x sin[cot^(−1) {tana)}] =sin[cot^(−1) {cot((π/2)−a)}] =sin((π/2)−a) =cosa=x proved

2)cos−1x=acosa=xsin[cot−1{tana)}]=sin[cot−1{cot(π2−a)}]=sin(π2−a)=cosa=xproved

Commented by Tawa1 last updated on 15/Jan/19

God bless you sir

Godblessyousir

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

4)f(x)=x^(n−2) (x^2 −3x+2) =x^(n−2) (x−2)(x−1) quotient=x^(n−2) (x−1) remainder=0

4)f(x)=xn−2(x2−3x+2)=xn−2(x−2)(x−1)quotient=xn−2(x−1)remainder=0

Commented by Tawa1 last updated on 15/Jan/19

God bless you sir

Godblessyousir

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