Integration Questions

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

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{xln}^{\mathrm{2}} \left({x}\right)}{{e}^{{x}} −\mathrm{1}}\:{dx}\:\Rightarrow\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{xe}^{−{x}} {ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}−{e}^{−{x}} } \\$$$$=\int_{\mathrm{0}} ^{\infty} \:\:{x}\:{e}^{−{x}} {ln}^{\mathrm{2}} \left({x}\right)\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{nx}} \right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\int_{\mathrm{0}} ^{\infty} \:{x}\:{e}^{−\left({n}+\mathrm{1}\right){x}} {ln}^{\mathrm{2}} \left({x}\right){dx} \\$$$$=_{\left({n}+\mathrm{1}\right){x}={t}} \:\:\sum_{{n}=\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \:\frac{{t}}{{n}+\mathrm{1}}\:{e}^{−{t}} {ln}^{\mathrm{2}} \left(\frac{{t}}{{n}+\mathrm{1}}\right)\frac{{dt}}{{n}+\mathrm{1}} \\$$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−{t}} {ln}^{\mathrm{2}} \left(\frac{{t}}{{n}+\mathrm{1}}\right){dt}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{A}_{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:{with}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−{t}} {ln}^{\mathrm{2}} \left(\frac{{t}}{{n}+\mathrm{1}}\right){dt} \\$$$${A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−{t}} \left({ln}\left({t}\right)−{ln}\left({n}+\mathrm{1}\right)\right)^{\mathrm{2}} {dt}\: \\$$$$=\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−{t}} \left({ln}^{\mathrm{2}} \left({t}\right)−\mathrm{2}{ln}\left({n}+\mathrm{1}\right){ln}\left({t}\right)\:+{ln}^{\mathrm{2}} \left({n}+\mathrm{1}\right)\right){dt} \\$$$$=\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−{t}} {ln}^{\mathrm{2}} \left({t}\right){dt}\:−\mathrm{2}{ln}\left({n}+\mathrm{1}\right)\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−{t}} {ln}\left({t}\right){dt}\:+{ln}^{\mathrm{2}} \left({n}+\mathrm{1}\right)\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−{t}} \:{dt}\:{but} \\$$$$\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−{t}} {dt}\:=\left[−{t}\:{e}^{−{t}} \right]_{\mathrm{0}} ^{+\infty} \:+\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {dt}\:=\left[−{e}^{−{t}} \right]_{\mathrm{0}} ^{+\infty} =\mathrm{1} \\$$$$\:{let}\:{find}\:{I}=\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−{t}} {ln}\left({t}\right){dt}\:\:\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} \left({tln}\left({t}\right)\right)\:{by}\:{psrts}\: \\$$$${u}^{'} ={e}^{−{t}} \:{and}\:{v}={tln}\left({t}\right)\:\Rightarrow{v}^{'} ={ln}\left({t}\right)+\mathrm{1}\:\Rightarrow \\$$$${I}\:=\left[−{e}^{−{t}} {tln}\left({t}\right)\right]_{\mathrm{0}} ^{+\infty} \:+\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−{t}} \left(\mathrm{1}+{ln}\left({t}\right)\right){dt} \\$$$$=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {dt}\:+\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {ln}\left({t}\right){dt}\:=\mathrm{1}−\gamma\:\:\:\:\left({euler}\:{costant}\right) \\$$$${let}\:{find}\:{J}\:=\int_{\mathrm{0}} ^{\infty} \:\:{t}\:{e}^{−{t}} {ln}^{\mathrm{2}} \left({t}\right){dt}\:\:\Rightarrow{J}\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} \left({tln}^{\mathrm{2}} {t}\right){dt}\:\:{by}\:{parts} \\$$$${u}^{'} ={e}^{−{t}} \:{and}\:{v}\:={t}\:{ln}^{\mathrm{2}} \left({t}\right)\:\Rightarrow{v}^{'} ={ln}^{\mathrm{2}} {t}\:\:+\mathrm{2}{t}\:\frac{{ln}\left({t}\right)}{{t}}\:={ln}^{\mathrm{2}} \left({t}\right)+\mathrm{2}{ln}\left({t}\right)\:\Rightarrow \\$$$${J}\:=\:\left[−{e}^{−{t}} {t}\:{ln}^{\mathrm{2}} \left({t}\right)\right]_{\mathrm{0}} ^{+\infty} \:+\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} \left\{{ln}^{\mathrm{2}} \left({t}\right)+\mathrm{2}{ln}\left({t}\right)\right\}{dt} \\$$$$=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {ln}^{\mathrm{2}} \left({t}\right){dt}\:+\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {ln}\left({t}\right){dt}\:=−\mathrm{2}\gamma\:+\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {ln}^{\mathrm{2}} \left({t}\right){dt} \\$$$${let}\:{find}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}} {ln}^{\mathrm{2}} \left({t}\right){dt}\:....{be}\:{continued}.... \\$$$$\\$$

Commented by Tawa1 last updated on 16/Jan/19

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{waiting}\:... \\$$