Question Number 138967 by mnjuly1970 last updated on 20/Apr/21
$$\:\Theta=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({x}\right).{log}^{\mathrm{2}} \left({x}\right)}{{x}}{dx}=\frac{\pi}{\mathrm{24}}\left(\mathrm{12}\:\gamma^{\:\mathrm{2}} +\pi^{\mathrm{2}} \right)….\checkmark \\ $$
Answered by Dwaipayan Shikari last updated on 20/Apr/21
$$\chi\left(\mu\right)=\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}^{\mu} }{dx}=\frac{\pi}{\mathrm{2}\Gamma\left(\mu\right){sin}\left(\frac{\pi\mu}{\mathrm{2}}\right)} \\ $$$$\chi''\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}{log}^{\mathrm{2}} \left({x}\right){dx}=\frac{\pi}{\mathrm{2}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\gamma^{\mathrm{2}} \right) \\ $$