∫_0 ^( ∞) ((x ln^2 (x))/(e^x − 1)) dx


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Question Number 52999 by Tawa1 last updated on 16/Jan/19

$\int_{0}^{\infty} \frac{x \ln^{2} (x)}{e^{x} - 1} dx$
Commented by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19

$t o u g h q u e s t i o n . . . w a i t . . .$
Commented by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19

Commented by Tawa1 last updated on 16/Jan/19

$Alright sir .$
Commented by Tawa1 last updated on 16/Jan/19

$I wait$
Commented by maxmathsup by imad last updated on 16/Jan/19
$let I =∫_0 ^∞ ((xln^2 (x))/(e^x −1)) dx ⇒ I =∫_0 ^∞ ((xe^(−x) ln^2 (x))/(1−e^(−x) )) =∫_0 ^∞ x e^(−x) ln^2 (x)(Σ_(n=0) ^∞ e^(−nx) ) =Σ_(n=0) ^∞ ∫_0 ^∞ x e^(−(n+1)x) ln^2 (x)dx =_((n+1)x=t) Σ_(n=0) ^∞ ∫_0 ^∞ (t/(n+1)) e^(−t) ln^2 ((t/(n+1)))(dt/(n+1)) =Σ_(n=0) ^∞ (1/((n+1)^2 ))∫_0 ^∞ t e^(−t) ln^2 ((t/(n+1)))dt =Σ_(n=0) ^∞ (A_n /((n+1)^2 )) with A_n =∫_0 ^∞ t e^(−t) ln^2 ((t/(n+1)))dt A_n =∫_0 ^∞ t e^(−t) (ln(t)−ln(n+1))^2 dt =∫_0 ^∞ t e^(−t) (ln^2 (t)−2ln(n+1)ln(t) +ln^2 (n+1))dt =∫_0 ^∞ t e^(−t) ln^2 (t)dt −2ln(n+1)∫_0 ^∞ t e^(−t) ln(t)dt +ln^2 (n+1)∫_0 ^∞ t e^(−t) dt but ∫_0 ^∞ t e^(−t) dt =[−t e^(−t) ]_0 ^(+∞) +∫_0 ^∞ e^(−t) dt =[−e^(−t) ]_0 ^(+∞) =1 let find I=∫_0 ^∞ t e^(−t) ln(t)dt ⇒I =∫_0 ^∞ e^(−t) (tln(t)) by psrts u^′ =e^(−t) and v=tln(t) ⇒v^′ =ln(t)+1 ⇒ I =[−e^(−t) tln(t)]_0 ^(+∞) +∫_0 ^∞ e^(−t) (1+ln(t))dt =∫_0 ^∞ e^(−t) dt +∫_0 ^∞ e^(−t) ln(t)dt =1−γ (euler costant) let find J =∫_0 ^∞ t e^(−t) ln^2 (t)dt ⇒J =∫_0 ^∞ e^(−t) (tln^2 t)dt by parts u^′ =e^(−t) and v =t ln^2 (t) ⇒v^′ =ln^2 t +2t ((ln(t))/t) =ln^2 (t)+2ln(t) ⇒ J = [−e^(−t) t ln^2 (t)]_0 ^(+∞) +∫_0 ^∞ e^(−t) {ln^2 (t)+2ln(t)}dt =∫_0 ^∞ e^(−t) ln^2 (t)dt +2 ∫_0 ^∞ e^(−t) ln(t)dt =−2γ +∫_0 ^∞ e^(−t) ln^2 (t)dt let find ∫_0 ^∞ e^(−t) ln^2 (t)dt ....be continued....$
$l e t I = \int_{0}^{\infty} \frac{{x l n}^{2} (x)}{e^{x} - 1} d x \Rightarrow I = \int_{0}^{\infty} \frac{{x e}^{- x} {l n}^{2} (x)}{1 - e^{- x}}$ $= \int_{0}^{\infty} x e^{- x} {l n}^{2} (x) (\sum_{n = 0}^{\infty} e^{- n x}) = \sum_{n = 0}^{\infty} \int_{0}^{\infty} x e^{- (n + 1) x} {l n}^{2} (x) d x$ $=_{(n + 1) x = t} \sum_{n = 0}^{\infty} \int_{0}^{\infty} \frac{t}{n + 1} e^{- t} {l n}^{2} (\frac{t}{n + 1}) \frac{d t}{n + 1}$ $= \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{2}} \int_{0}^{\infty} t e^{- t} {l n}^{2} (\frac{t}{n + 1}) d t = \sum_{n = 0}^{\infty} \frac{A_{n}}{{(n + 1)}^{2}} w i t h A_{n} = \int_{0}^{\infty} t e^{- t} {l n}^{2} (\frac{t}{n + 1}) d t$ $A_{n} = \int_{0}^{\infty} t e^{- t} {(l n (t) - l n (n + 1))}^{2} d t$ $= \int_{0}^{\infty} t e^{- t} ({l n}^{2} (t) - 2 l n (n + 1) l n (t) + {l n}^{2} (n + 1)) d t$ $= \int_{0}^{\infty} t e^{- t} {l n}^{2} (t) d t - 2 l n (n + 1) \int_{0}^{\infty} t e^{- t} l n (t) d t + {l n}^{2} (n + 1) \int_{0}^{\infty} t e^{- t} d t b u t$ $\int_{0}^{\infty} t e^{- t} d t = {[- t e^{- t}]}_{0}^{+ \infty} + \int_{0}^{\infty} e^{- t} d t = {[- e^{- t}]}_{0}^{+ \infty} = 1$ $l e t f i n d I = \int_{0}^{\infty} t e^{- t} l n (t) d t \Rightarrow I = \int_{0}^{\infty} e^{- t} (t l n (t)) b y p s r t s$ $u^{^{'}} = e^{- t} a n d v = t l n (t) \Rightarrow v^{^{'}} = l n (t) + 1 \Rightarrow$ $I = {[- e^{- t} t l n (t)]}_{0}^{+ \infty} + \int_{0}^{\infty} e^{- t} (1 + l n (t)) d t$ $= \int_{0}^{\infty} e^{- t} d t + \int_{0}^{\infty} e^{- t} l n (t) d t = 1 - γ (e u l e r c o s t a n t)$ $l e t f i n d J = \int_{0}^{\infty} t e^{- t} {l n}^{2} (t) d t \Rightarrow J = \int_{0}^{\infty} e^{- t} ({t l n}^{2} t) d t b y p a r t s$ $u^{^{'}} = e^{- t} a n d v = t {l n}^{2} (t) \Rightarrow v^{^{'}} = {l n}^{2} t + 2 t \frac{l n (t)}{t} = {l n}^{2} (t) + 2 l n (t) \Rightarrow$ $J = {[- e^{- t} t {l n}^{2} (t)]}_{0}^{+ \infty} + \int_{0}^{\infty} e^{- t} {{l n}^{2} (t) + 2 l n (t)} d t$ $= \int_{0}^{\infty} e^{- t} {l n}^{2} (t) d t + 2 \int_{0}^{\infty} e^{- t} l n (t) d t = - 2 γ + \int_{0}^{\infty} e^{- t} {l n}^{2} (t) d t$ $l e t f i n d \int_{0}^{\infty} e^{- t} {l n}^{2} (t) d t . . . . b e c o n t i n u e d . . . .$
Commented by Tawa1 last updated on 16/Jan/19

$God bless you sir . waiting . . .$
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