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Question Number 155013 by amin96 last updated on 24/Sep/21

$${\int }_0^{1}ln(-lnx)\frac{x^{\mu -1}}{\sqrt{-ln\left(x\right)}}dx=?$$

Answered by mnjuly1970 last updated on 24/Sep/21

$$-ln\left(x\right)=t^2\Rightarrow x=e^{-t^2}$$$${\int }_0^{\mathrm{\infty }}ln\left(t^2\right).\frac{e^{-t^2(\mu -1)}}{t}\left(2t\right)e^{-t^2}dt$$$$4{\underset{0}{\int }}^{\mathrm{\infty }}ln\left(t\right).e^{-\mu t^2}dt\stackrel{t\sqrt{\mu }=y}{=}\frac{4}{\sqrt{\mu }}{\int }_0^{\mathrm{\infty }}ln\left(\frac{y}{\sqrt{\mu }}\right)e^{-y^2}dy$$$$=\frac{4}{\sqrt{\mu }}{\int }_0^{\mathrm{\infty }}ln\left(y\right)e^{-y^2}dy-\frac{2ln\left(\mu \right)}{\sqrt{\mu }}{\int }_0^{\mathrm{\infty }}{e}^{-y^2}dy$$$$=\frac{4}{\sqrt{\mu }}{\int }_0^{\mathrm{\infty }}ln\left(y\right).e^{-y^2}dy-\sqrt{\frac{\pi }{\mu }}ln\left(\mu \right)$$$$\stackrel{y^2=z}{=}\frac{1}{\sqrt{\mu }}{\int }_0^{\mathrm{\infty }}ln\left(z\right).z^{\frac{-1}{2}}.e^{-z}dz-ln\left(\mu \right).\sqrt{\frac{\pi }{\mu }}$$$$=\frac{1}{\sqrt{\mu }}\psi \left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)-ln\left(\mu \right).\sqrt{\frac{\pi }{\mu }}$$$$=\sqrt{\frac{\pi }{\mu }}(\psi \left(\frac{1}{2}\right)-ln\left(\mu \right))$$$$=\sqrt{\frac{\pi }{\mu }}(-\gamma -2ln\left(2\right)-ln\left(\mu \right))$$$$=\sqrt{\frac{\pi }{\mu }}(-\gamma -ln\left(4\mu \right))◼$$

Answered by ArielVyny last updated on 24/Sep/21

$$-lnx=t\to \frac{1}{x}=e^t\to 1={xe}^t\to x=e^{-t}$$$$dx=-e^{-t}dt$$$${\int }_0^{\mathrm{\infty }}ln\left(t\right)\frac{e^{-t(\mu -1)}}{\sqrt{t}}{e}^{-t}dt={\int }_0^{\mathrm{\infty }}ln\left(t\right)e^{-t\mu }{t}^{\frac{-1}{2}}dt$$$$t\mu =u\to \mu dt=du$$$${\int }_0^{\mathrm{\infty }}ln\left(\frac{u}{\mu }\right)e^{-u}{\left(\frac{u}{\mu }\right)}^{-\frac{1}{2}}\frac{1}{\mu }du$$$${\mu }^{-\frac{1}{2}}{\int }_0^{\mathrm{\infty }}ln\left(u\right)e^{-u}{u}^{-\frac{1}{2}}du-{\mu }^{-\frac{1}{2}}{\int }_0^{\mathrm{\infty }}ln\left(\mu \right)e^{-u}{u}^{-\frac{1}{2}}du$$$${\mu }^{-\frac{1}{2}}d\left(\mathrm{\Gamma }\left(\frac{1}{2}\right)\right)-{\mu }^{-\frac{1}{2}}ln\left(\mu \right)\mathrm{\Gamma }\left(\frac{1}{2}\right)$$$$I={\mu }^{-\frac{1}{2}}\frac{{\mathrm{\Gamma }}^{\prime }\left(\frac{1}{2}\right)}{1}-{\mu }^{-\frac{1}{2}}ln\left(\mu \right)\mathrm{\Gamma }\left(\frac{1}{2}\right)$$$$\frac{I}{\mathrm{\Gamma }\left(\frac{1}{2}\right)}={\mu }^{-\frac{1}{2}}\psi \left(\frac{1}{2}\right)-{\mu }^{-\frac{1}{2}}ln\left(\mu \right)$$$${\int }_0^{1}ln(-lnx)\frac{x^{\mu -1}}{\sqrt{-lnx}}dx=I\frac{\sqrt{\pi }}{\pi }.={\mu }^{-\frac{1}{2}}\psi \left(\frac{1}{2}\right)-{\mu }^{-\frac{1}{2}}ln\left(\mu \right)$$$$I=\sqrt{\pi }{\mu }^{-\frac{1}{2}}\psi \left(\frac{1}{2}\right)-\sqrt{\pi }{\mu }^{-\frac{1}{2}}ln\left(\mu \right)$$

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