Question Number 114959 by mathdave last updated on 22/Sep/20
$${long}\:{time}\:{question}\:{proposed}\:{by} \\ $$$${math}\:{abdo} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} }{dx} \\ $$$$ \\ $$
Answered by mathdave last updated on 22/Sep/20
$${solution} \\ $$$${let} \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} }{dx}+\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} }{dx}={A}+{B} \\ $$$${puttting}\:\:{x}=\frac{\mathrm{1}}{{x}}\:\:{into}\:{integral}\:{B} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} }{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} \mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}\:} }{dx}={A}−{B} \\ $$$${let} \\ $$$${A}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{x}}{\mathrm{1}−{x}^{\mathrm{6}} }{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} \mathrm{ln}{x}}{\mathrm{1}−{x}^{\mathrm{6}} }{dx} \\ $$$${A}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{6}{n}} .\mathrm{ln}{xdx}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} .{x}^{\mathrm{6}{n}} .\mathrm{ln}{xdx} \\ $$$${A}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{6}{n}} .{x}^{{a}−\mathrm{1}} {dx}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{6}{n}+\mathrm{2}} .{x}^{{a}−\mathrm{1}} {dx} \\ $$$${A}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{6}{n}+{a}}\right)−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{6}{n}+\mathrm{2}+{a}}\right) \\ $$$${A}=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{6}{n}+\mathrm{1}\right)^{\mathrm{2}} }+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{6}{n}+\mathrm{3}\right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{36}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{36}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$${A}=−\frac{\mathrm{1}}{\mathrm{36}}\psi^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{6}}\right)+\frac{\mathrm{1}}{\mathrm{36}}\psi^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)………..\left(\mathrm{1}\right)\:{and}\: \\ $$$${B}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} \mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} \mathrm{ln}{x}}{\mathrm{1}−{x}^{\mathrm{6}} }{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{4}} \mathrm{ln}{x}}{\mathrm{1}−{x}^{\mathrm{6}} }{dx} \\ $$$${B}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} .\mathrm{ln}{x}.{x}^{\mathrm{6}{n}} {dx}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{4}} .\mathrm{ln}{x}.{x}^{\mathrm{6}{n}} {dx} \\ $$$${B}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{6}{n}+\mathrm{2}} .{x}^{{a}−\mathrm{1}} {dx}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{6}{n}+\mathrm{4}} .{x}^{{a}−\mathrm{1}} {dx} \\ $$$${B}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{6}{n}+\mathrm{2}+{a}}\right)−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{6}{n}+\mathrm{4}+{a}}\right) \\ $$$${B}=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{6}{n}+\mathrm{3}\right)^{\mathrm{2}} }+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{6}{n}+\mathrm{5}\right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{36}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{36}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{5}}{\mathrm{6}}\right)^{\mathrm{2}} } \\ $$$${B}=−\frac{\mathrm{1}}{\mathrm{36}}\psi^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{36}}\psi^{\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{6}}\right)……..\left(\mathrm{2}\right) \\ $$$${but}\:\:{I}={A}−{B} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{36}}\psi^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{6}}\right)+\frac{\mathrm{1}}{\mathrm{36}}\psi^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\left(−\frac{\mathrm{1}}{\mathrm{36}}\psi^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{36}}\psi^{\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{6}}\right)\right) \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{18}}\psi^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{36}}\left[\psi^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{6}}\right)+\psi^{\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{6}}\right)\right] \\ $$$${using}\:\:\psi^{\mathrm{1}} \left({z}\right)+\psi^{\mathrm{1}} \left(\mathrm{1}−{z}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{sin}^{\mathrm{2}} \left(\pi{z}\right)} \\ $$$$\psi^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{6}}\right)+\psi^{\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{6}}\right)=\frac{\pi^{\mathrm{2}} }{\left[\mathrm{sin}\left(\frac{\pi}{\mathrm{6}}\right)\right]^{\mathrm{2}} }=\mathrm{4}\pi^{\mathrm{2}} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{18}}\psi^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\frac{\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{36}}=\frac{\mathrm{1}}{\mathrm{18}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\right)−\frac{\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{36}}=−\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{36}}=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\because\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} }{dx}=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$${b}\mathrm{8}{y}\:{nathdave}\left(\mathrm{22}/\mathrm{09}/\mathrm{2020}\right) \\ $$
Commented by I want to learn more last updated on 23/Sep/20
$$\mathrm{And}\:\mathrm{sir},\:\mathrm{does}\:\mathrm{the}\:\mathrm{identity}\:\mathrm{has}\:\mathrm{proof}? \\ $$
Commented by I want to learn more last updated on 23/Sep/20
$$\mathrm{Weldone}\:\mathrm{sir}. \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{with}\:\mathrm{this}\:\:\:\:\:\:\:\:\int_{\:\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{10}\:\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{x}}\:\mathrm{dx} \\ $$$$\mathrm{Also}\:\mathrm{without}\:\mathrm{the}\:\mathrm{limit}.\:\:\:\int\:\:\frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{x}}\:\mathrm{dx}. \\ $$$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$
Commented by mathdave last updated on 23/Sep/20
$${note}\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left({ax}\right)}{{x}^{{n}} }{dx}=\frac{\pi{a}^{{n}−\mathrm{1}} }{\mathrm{2}\Gamma\left({n}\right)\mathrm{sin}\left(\frac{\pi{n}}{\mathrm{2}}\right)} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left({x}\right)}{{x}}{dx}=\frac{\pi\left(\mathrm{1}\right)^{\mathrm{1}−\mathrm{1}} }{\mathrm{2}\Gamma\left(\mathrm{1}\right)\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}\right)}=\frac{\pi}{\mathrm{2}} \\ $$$${where}\:\Gamma\left(\mathrm{1}\right)=\mathrm{1},{a}=\mathrm{1}\:{and}\:{n}=\mathrm{1} \\ $$$$\because\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left({x}\right)}{{x}}{dx}=\frac{\pi}{\mathrm{2}} \\ $$
Commented by mathdave last updated on 23/Sep/20
$${u}\:{can}\:{using}\:{IBP}\:{for}\:{d}\:{one}\:{without}\:{d} \\ $$$${symmentric}\:{bondaries} \\ $$
Commented by I want to learn more last updated on 23/Sep/20
$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}. \\ $$
Commented by I want to learn more last updated on 23/Sep/20
$$\mathrm{I}\:\mathrm{will}\:\mathrm{love}\:\mathrm{to}\:\mathrm{see}\:\mathrm{your}\:\mathrm{work}\:\mathrm{sir} \\ $$
Commented by mathdave last updated on 23/Sep/20
$${yah}\:{it}\:{has}\:{proof} \\ $$
Commented by I want to learn more last updated on 23/Sep/20
$$\mathrm{Please}\:\mathrm{sir}\:\mathrm{help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{it}\:\mathrm{when}\:\mathrm{you}\:\mathrm{are}\:\mathrm{less}\:\mathrm{busy}.\:\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Commented by Tawa11 last updated on 06/Sep/21
$$\mathrm{great}\:\mathrm{sir} \\ $$
Answered by Bird last updated on 24/Sep/20
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{lnx}}{{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$${we}\:{use}\:\int_{\mathrm{0}} ^{\infty} \:{q}\left({x}\right){ln}\left({x}\right){dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{Re}\left(\Sigma\:{Res}\left({q}\left({z}\right){ln}^{\mathrm{2}} {z}\:,{a}_{{k}} \right)\right) \\ $$$${w}\left({z}\right)=\frac{{ln}^{\mathrm{2}} {z}}{{z}^{\mathrm{4}} \:+{z}^{\mathrm{2}} \:+\mathrm{1}}\:\:{pole}\:{of}\:{w}? \\ $$$${u}^{\mathrm{2}\:} \:+{u}\:+\mathrm{1}=\mathrm{0}\:\:\left({u}={z}^{\mathrm{2}} \right) \\ $$$$\Delta=−\mathrm{3}\:\Rightarrow{u}_{\mathrm{1}} =\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:{and} \\ $$$${u}_{\mathrm{2}} =\frac{−\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:={e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \\ $$$${w}\left({z}\right)\:=\frac{{ln}^{\mathrm{2}} {z}}{\left({z}^{\mathrm{2}} −{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)\left({z}^{\mathrm{2}} −{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)} \\ $$$$=\frac{{ln}^{\mathrm{2}} {z}}{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)} \\ $$$${Res}\left({w},{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)=\frac{\left(\frac{{i}\pi}{\mathrm{3}}\right)^{\mathrm{2}} }{\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{3}}} \left(\mathrm{2}{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{i}}×\frac{−\pi^{\mathrm{2}} }{\mathrm{9}}\:×\frac{{e}^{−\frac{{i}\pi}{\mathrm{3}}} }{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{18}{i}\sqrt{\mathrm{3}}}.{e}^{−\frac{{i}\pi}{\mathrm{3}}} \\ $$$$ \\ $$$${Res}\left({f},−{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\:=\frac{\left({i}\pi+\frac{{i}\pi}{\mathrm{3}}\right)^{\mathrm{2}} }{−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{3}}} \left(\mathrm{2}{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)} \\ $$$$=\frac{−\left(\frac{\mathrm{4}\pi}{\mathrm{3}}\right)^{\mathrm{2}} }{−\mathrm{4}{i}}×\frac{{e}^{\frac{{i}\pi}{\mathrm{3}}} }{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\frac{\mathrm{16}}{\mathrm{9}}\pi^{\mathrm{2}} ×\frac{{e}^{\frac{{i}\pi}{\mathrm{3}}} }{\mathrm{2}{i}\sqrt{\mathrm{3}}} \\ $$$${Res}\left({f},{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\:=\frac{\left(−\frac{{i}\pi}{\mathrm{3}}\right)^{\mathrm{2}} }{\left(−\mathrm{2}{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{3}}} } \\ $$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{9}}×\frac{{e}^{{i}\frac{\pi}{\mathrm{3}}} }{−\mathrm{4}{i}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{18}{i}\sqrt{\mathrm{3}}}.{e}^{\frac{{i}\pi}{\mathrm{3}}} \\ $$$${Res}\left({f},−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right) \\ $$$$=\frac{\left({i}\pi\:−\frac{{i}\pi}{\mathrm{3}}\right)^{\mathrm{2}} }{\left(−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\left(−\mathrm{2}{i}\:{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right.} \\ $$$$=\frac{−\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)^{\mathrm{2}} }{\mathrm{4}{i}\:×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}.{e}^{\frac{{i}\pi}{\mathrm{3}}} \:=−\frac{\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{9}}×\frac{{e}^{\frac{{i}\pi}{\mathrm{3}}} }{\mathrm{2}{i}\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\Sigma\:{Res}\left({f}\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{18}{i}\sqrt{\mathrm{3}}}{e}^{−\frac{{i}\pi}{\mathrm{3}}} \\ $$$$+\frac{\mathrm{16}}{\mathrm{9}}\pi^{\mathrm{2}} \:\frac{{e}^{\frac{{i}\pi}{\mathrm{3}}} }{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:\:+\frac{\pi^{\mathrm{2}} }{\mathrm{18}{i}\sqrt{\mathrm{3}}}\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \\ $$$$−\frac{\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{18}{i}\sqrt{\mathrm{3}}}×{e}^{\frac{{i}\pi}{\mathrm{3}}} \\ $$$$=\frac{{i}\pi^{\mathrm{2}} }{\mathrm{18}\sqrt{\mathrm{3}}}\left\{\frac{\mathrm{1}}{\mathrm{2}}−\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right\} \\ $$$$−\frac{\mathrm{16}{i}\pi^{\mathrm{2}} }{\mathrm{18}\sqrt{\mathrm{3}}}\left\{\frac{\mathrm{1}}{\mathrm{2}}+\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right\}−\frac{{i}\pi^{\mathrm{2}} }{\mathrm{18}\sqrt{\mathrm{3}}}\left\{\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right\} \\ $$$$+\frac{\mathrm{2}\pi^{\mathrm{2}} {i}}{\mathrm{9}\sqrt{\mathrm{3}}}\left\{\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right\}…{be}\:{continued}… \\ $$