Question Number 52999 by Tawa1 last updated on 16/Jan/19
$$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{1}}\:\:\boldsymbol{\mathrm{dx}}\:\:\: \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19
$${tough}\:{question}...{wait}... \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19
Commented by Tawa1 last updated on 16/Jan/19
$$\mathrm{Alright}\:\mathrm{sir}. \\ $$
Commented by Tawa1 last updated on 16/Jan/19
$$\mathrm{I}\:\mathrm{wait} \\ $$
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....](Q53076.png)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{xln}^{\mathrm{2}} \left({x}\right)}{{e}^{{x}} −\mathrm{1}}\:{dx}\:\Rightarrow\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{xe}^{−{x}} {ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}−{e}^{−{x}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:{x}\:{e}^{−{x}} {ln}^{\mathrm{2}} \left({x}\right)\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{nx}} \right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\int_{\mathrm{0}} ^{\infty} \:{x}\:{e}^{−\left({n}+\mathrm{1}\right){x}} {ln}^{\mathrm{2}} \left({x}\right){dx} \\ $$$$=_{\left({n}+\mathrm{1}\right){x}={t}} \:\:\sum_{{n}=\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \:\frac{{t}}{{n}+\mathrm{1}}\:{e}^{−{t}} {ln}^{\mathrm{2}} \left(\frac{{t}}{{n}+\mathrm{1}}\right)\frac{{dt}}{{n}+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−{t}} {ln}^{\mathrm{2}} \left(\frac{{t}}{{n}+\mathrm{1}}\right){dt}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{A}_{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:{with}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−{t}} {ln}^{\mathrm{2}} \left(\frac{{t}}{{n}+\mathrm{1}}\right){dt} \\ $$$${A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−{t}} \left({ln}\left({t}\right)−{ln}\left({n}+\mathrm{1}\right)\right)^{\mathrm{2}} {dt}\: \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−{t}} \left({ln}^{\mathrm{2}} \left({t}\right)−\mathrm{2}{ln}\left({n}+\mathrm{1}\right){ln}\left({t}\right)\:+{ln}^{\mathrm{2}} \left({n}+\mathrm{1}\right)\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−{t}} {ln}^{\mathrm{2}} \left({t}\right){dt}\:−\mathrm{2}{ln}\left({n}+\mathrm{1}\right)\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−{t}} {ln}\left({t}\right){dt}\:+{ln}^{\mathrm{2}} \left({n}+\mathrm{1}\right)\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−{t}} \:{dt}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−{t}} {dt}\:=\left[−{t}\:{e}^{−{t}} \right]_{\mathrm{0}} ^{+\infty} \:+\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {dt}\:=\left[−{e}^{−{t}} \right]_{\mathrm{0}} ^{+\infty} =\mathrm{1} \\ $$$$\:{let}\:{find}\:{I}=\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−{t}} {ln}\left({t}\right){dt}\:\:\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} \left({tln}\left({t}\right)\right)\:{by}\:{psrts}\: \\ $$$${u}^{'} ={e}^{−{t}} \:{and}\:{v}={tln}\left({t}\right)\:\Rightarrow{v}^{'} ={ln}\left({t}\right)+\mathrm{1}\:\Rightarrow \\ $$$${I}\:=\left[−{e}^{−{t}} {tln}\left({t}\right)\right]_{\mathrm{0}} ^{+\infty} \:+\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−{t}} \left(\mathrm{1}+{ln}\left({t}\right)\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {dt}\:+\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {ln}\left({t}\right){dt}\:=\mathrm{1}−\gamma\:\:\:\:\left({euler}\:{costant}\right) \\ $$$${let}\:{find}\:{J}\:=\int_{\mathrm{0}} ^{\infty} \:\:{t}\:{e}^{−{t}} {ln}^{\mathrm{2}} \left({t}\right){dt}\:\:\Rightarrow{J}\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} \left({tln}^{\mathrm{2}} {t}\right){dt}\:\:{by}\:{parts} \\ $$$${u}^{'} ={e}^{−{t}} \:{and}\:{v}\:={t}\:{ln}^{\mathrm{2}} \left({t}\right)\:\Rightarrow{v}^{'} ={ln}^{\mathrm{2}} {t}\:\:+\mathrm{2}{t}\:\frac{{ln}\left({t}\right)}{{t}}\:={ln}^{\mathrm{2}} \left({t}\right)+\mathrm{2}{ln}\left({t}\right)\:\Rightarrow \\ $$$${J}\:=\:\left[−{e}^{−{t}} {t}\:{ln}^{\mathrm{2}} \left({t}\right)\right]_{\mathrm{0}} ^{+\infty} \:+\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} \left\{{ln}^{\mathrm{2}} \left({t}\right)+\mathrm{2}{ln}\left({t}\right)\right\}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {ln}^{\mathrm{2}} \left({t}\right){dt}\:+\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {ln}\left({t}\right){dt}\:=−\mathrm{2}\gamma\:+\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {ln}^{\mathrm{2}} \left({t}\right){dt} \\ $$$${let}\:{find}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}} {ln}^{\mathrm{2}} \left({t}\right){dt}\:....{be}\:{continued}.... \\ $$$$ \\ $$
Commented by Tawa1 last updated on 16/Jan/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{waiting}\:... \\ $$