Question Number 130307 by mnjuly1970 last updated on 24/Jan/21
$$\:\:\:\:\:\:\:\:\:\:\:\:…\:{nice}\:\:\:….\:\:\:\:\:{cslculus}… \\ $$$$\:\:\:{evaluate}\::::\:\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\:\left({x}\:\right)}{\:\sqrt{{x}}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:{dx}=? \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 24/Jan/21
$$\int_{\mathrm{0}} ^{\infty} \frac{{log}\left({x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{{x}}}{dx}\:\:\:\:\:\:\:\:\:\:\:\:{x}={u}^{\mathrm{2}} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\infty} \frac{{log}\left({u}\right)}{\mathrm{1}+{u}^{\mathrm{4}} }{du}=\mathrm{4}\left(\frac{−\pi^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}} }{cosec}\left(\frac{\pi}{\mathrm{4}}\right){cot}\left(\frac{\pi}{\mathrm{4}}\right)\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$
Commented by Dwaipayan Shikari last updated on 24/Jan/21
$${Q}\mathrm{130137} \\ $$
Commented by mnjuly1970 last updated on 24/Jan/21
$${thank}\:{you}\:{so}\:{much}\:{mr}\:{dwaypayan}. \\ $$$$\:\: \\ $$
Answered by mathmax by abdo last updated on 24/Jan/21
$$\mathrm{A}=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnx}}{\:\sqrt{\mathrm{x}}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}\:\:\Rightarrow\mathrm{A}=_{\sqrt{\mathrm{x}}=\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{ln}\left(\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{t}\left(\mathrm{1}+\mathrm{t}^{\mathrm{4}} \right)}\left(\mathrm{2t}\right)\mathrm{dt} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnt}}{\mathrm{t}^{\mathrm{4}} \:+\mathrm{1}}\mathrm{dt}\:\mathrm{wedo}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{t}=\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\Rightarrow \\ $$$$\mathrm{A}=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\mathrm{lnu}}{\mathrm{1}+\mathrm{u}}\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \:\mathrm{du}\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \mathrm{lnu}}{\mathrm{1}+\mathrm{u}}\mathrm{du} \\ $$$$\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{a}−\mathrm{1}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{e}^{\left(\mathrm{a}−\mathrm{1}\right)\mathrm{lnt}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:=\frac{\pi}{\mathrm{sin}\left(\pi\mathrm{a}\right)}\:\mathrm{with}\:\mathrm{0}<\mathrm{a}<\mathrm{1} \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\mathrm{a}−\mathrm{1}} \mathrm{lnt}}{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:\Rightarrow\mathrm{A}\:=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{f}^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)=−\frac{\pi^{\mathrm{2}} \mathrm{cos}\left(\pi\mathrm{a}\right)}{\mathrm{sin}^{\mathrm{2}} \left(\pi\mathrm{a}\right)}\:\Rightarrow\mathrm{f}^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)=−\pi^{\mathrm{2}} \:×\frac{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{2}}}=−\pi^{\mathrm{2}} \sqrt{\mathrm{2}}\:\:\Rightarrow \\ $$$$\mathrm{A}=−\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\sqrt{\mathrm{2}\:}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\pi^{\mathrm{2}} \\ $$