Question Number 114045 by mnjuly1970 last updated on 17/Sep/20

... advanced calculus... i : prove that :: ∫_0 ^( 1) ((ln(1+ln(1−x)))/(ln(1−x))) dx =^? Σ_(n=1) ^∞ ((Γ(n+1))/n^2 ) ii: prove that :: Ω =∫_0 ^( 1) ((ln(1+x))/(x(1+x^2 )))dx =^? ((5π^2 )/(48)) m.n.july 1970#

Answered by Dwaipayan Shikari last updated on 16/Sep/20

∫_0 ^1 ((log(1+x))/(x(1+x)))=∫_0 ^1 ((log(1+x))/x)−((log(1+x))/((1+x)))dx =∫_0 ^1 (−1)^n Σ_(n=1) ^∞ (x^(n−1) /n)−∫_1 ^2 ((logu)/u)du =Σ_(n=1) ^∞ (−1)^n (1/n^2 )−(1/2)[(logu)^2 ]_1 ^2 =(π^2 /(12))−(1/2)(log2)^2

Answered by mathmax by abdo last updated on 16/Sep/20

I =∫_0 ^1 ((ln(1+x))/(x(1+x)))dx =∫_0 ^1 ((1/x)−(1/(x+1)))ln(1+x)dx =∫_0 ^1 ((ln(1+x))/x)dx−∫_0 ^1 ((ln(1+x))/(1+x))dx but ∫_0 ^1 ((ln(1+x))/(1+x)) dx =_(1+x=t) ∫_1 ^2 ((ln(t))/(x>t)) dt =[(1/2)ln^2 t]_1 ^2 =((ln^2 (2))/2) we have (d/dx)ln(1+x)=(1/(1+x)) =Σ_(n=0) ^∞ (−1)^n x^n ⇒ln(1+x)=Σ_(n=0) ^∞ (((−1)^n x^(n+1) )/(n+1)) =Σ_(n=1) ^∞ (((−1)^(n−1) x^n )/n) ⇒((ln(1+x))/x) =Σ_(n=1) ^∞ (((−1)^(n−1) x^(n−1) )/n) ⇒ ∫_0 ^1 ((ln(1+x))/x)dx =Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ) =−Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =−{ 2^(1−2) −1}ξ(2) =(π^2 /(12)) ⇒ I =(1/2)ln^2 (2)−(π^2 /(12))

Commented bymathmax by abdo last updated on 16/Sep/20

sorry I =(π^2 /(12))−(1/2)ln^2 (2)

Commented bymnjuly1970 last updated on 17/Sep/20

thank you sir that was my mistake denominator is x(1+x^2 ) i edited it.

Answered by mathmax by abdo last updated on 16/Sep/20

A =∫_0 ^1 ((ln(1+ln(1−x)))/(ln(1−x))) dx we do tbe changement ln(1−x)=−t ⇒ 1−x =e^(−t) ⇒x =1−e^(−t) ⇒ A =∫_0 ^(+∞) ((ln(1+t))/(−t)) (e^(−t) )dt =−∫_0 ^∞ ((e^(−t) ln(1+t))/t) dt =−∫_0 ^∞ e^(−t) (Σ_(n=1) ^(∞ ) (((−1)^(n−1) t^(n−1) )/n))dt =Σ_(n=1) ^(∞ ) (((−1)^n )/n) ∫_0 ^∞ t^(n−1) e^(−t) dt =Σ_(n=2) ^∞ (−1)^n ((Γ(n−1))/n) −1

Commented bymnjuly1970 last updated on 17/Sep/20

A=−∫_0 ^( ∞) ((ln(1−t)e^(−t) )/t)dt is correct.please check it .thank you so much for your effort .

Answered by mindispower last updated on 17/Sep/20

i) ∫_0 ^1 ((ln(1+ln(1−x)))/(ln(1−x)))dx=∫_0 ^1 ((ln(1+ln(x)))/(ln(x)))dx =∫_0 ^∞ ((ln(1−x))/(−x))e^(−x) dx somthing wrong ln(1−x)∈C x>1