Question Number 198151 by sonukgindia last updated on 11/Oct/23
Answered by Mathspace last updated on 12/Oct/23
$${I}=\int_{\mathrm{0}} ^{\infty} \:\frac{{zlnz}}{\mathrm{1}+{z}^{\mathrm{3}} }{dz}\:\:\:\:\left({z}={t}^{\frac{\mathrm{1}}{\mathrm{3}}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{3}}} {lnt}}{\mathrm{1}+{t}}.\frac{\mathrm{1}}{\mathrm{3}}{t}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} {dt} \\ $$$$\Rightarrow\mathrm{9}{I}=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{1}} {lnt}}{\mathrm{1}+{t}}{dt} \\ $$$${let}\:{J}_{{a}} =\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt} \\ $$$${we}\:{have}\:{J}^{,} {a}=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{a}−\mathrm{1}} {lnt}}{\mathrm{1}+{t}}{dt} \\ $$$${and}\:{J}^{'} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)=\mathrm{9}{I}\:\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{9}}{J}^{'} \left(\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$$${J}_{{a}} =\frac{\pi}{{sin}\left(\pi{a}\right)}\:\Rightarrow \\ $$$${J}_{{a}} ^{'} =−\pi^{\mathrm{2}} \:\frac{{cos}\left(\pi{a}\right)}{{sin}^{\mathrm{2}} \left(\pi{a}\right)}\:{and} \\ $$$${J}^{'} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)=−\pi^{\mathrm{2}} \frac{{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)}{{sin}^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)} \\ $$$$=−\pi^{\mathrm{2}} .\frac{\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}.\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}=\frac{\pi^{\mathrm{2}} }{\:\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$${I}=\frac{\pi^{\mathrm{2}} }{\mathrm{9}\sqrt{\mathrm{3}}}=\frac{\pi^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{27}} \\ $$