Question Number 144351 by mnjuly1970 last updated on 24/Jun/21
$$ \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:{Evaluate}\:\:\::: \\ $$$$\:\: \\ $$$$\boldsymbol{\phi}:=\int_{\mathrm{0}} ^{\:\infty} \frac{\left.\:{sin}\:\sqrt[{\:\mathrm{3}\:}]{{x}}\:\right){log}\:\left(\frac{\mathrm{1}}{{x}}\:\right)}{{x}}{dx}=? \\ $$$$\:\:\: \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 24/Jun/21
$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({u}\right)}{{u}^{\alpha} }=\frac{\pi}{\mathrm{2}\Gamma\left(\alpha\right){sin}\left(\frac{\pi}{\mathrm{2}}\alpha\right)} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({u}\right)}{{u}^{\alpha} }{log}\left(\frac{\mathrm{1}}{{u}}\right){du}=−\frac{\pi\Gamma'\left(\alpha\right)}{\mathrm{2}\Gamma\left(\alpha\right)^{\mathrm{2}} {sin}\left(\frac{\pi}{\mathrm{2}}\alpha\right)}−\frac{\pi{cosec}\left(\frac{\pi\alpha}{\mathrm{2}}\right){cot}\left(\frac{\pi\alpha}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\alpha\right)} \\ $$$${Here} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left(\sqrt[{\mathrm{3}}]{{x}}\right)}{{x}}{log}\left(\frac{\mathrm{1}}{{x}}\right){dx}\:\:{x}={u}^{\mathrm{3}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({u}\right)}{{u}}{log}\left(\frac{\mathrm{1}}{{u}}\right){du}=\frac{\pi\gamma}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 24/Jun/21
$$\:\:\:\:{thanks}\:{alot}… \\ $$