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Question Number 131991 by mnjuly1970 last updated on 10/Feb/21

$$...nicecalculus...$$$$provethat:$$$${\varphi }_1={\int }_0^1{li}_2(1-x^2)=\frac{{\pi }^2}{2}-4$$$${\varphi }_2={\int }_0^1\frac{log(1-t)}{t^{\frac{3}{4}}\sqrt{1-t}}dt={\pi }^{\frac{3}{2}}.\frac{\sqrt{2}}{{\mathrm{\Gamma }}^2\left(\frac{3}{4}\right)}(log\left(2\right)-\frac{\pi }{2})$$$$hint1:\psi \left(\frac{3}{4}\right)\underset{easy}{\stackrel{why??}{=}}-\gamma +\frac{\pi }{2}-3log\left(2\right)$$$$hint2:\psi \left(\frac{1}{2}\right)\underset{easy}{\stackrel{why??}{=}}-\gamma -2log\left(2\right)$$

Answered by Dwaipayan Shikari last updated on 10/Feb/21

$$\psi \left(\frac{1}{2}\right)=-\gamma +{\int }_0^1\frac{1-x^{-\frac{1}{2}}}{1-x}dx=-\gamma +2{\int }_0^1\frac{x-1}{1-x^2}dx$$$$=-\gamma -2log\left(2\right)$$

Answered by Dwaipayan Shikari last updated on 10/Feb/21

$$I\left(b\right)={\int }_0^1{t}^{a-1}{(1-t)}^{b-1}dt=\frac{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }(a+b)}$$$$I^{\prime }\left(b\right)={\int }_0^1{t}^{a-1}{(1-t)}^{b-1}log(1-t)dt$$$$=\frac{\mathrm{\Gamma }\left(a\right)(\mathrm{\Gamma }(a+b){\mathrm{\Gamma }}^{\prime }\left(b\right)-{\mathrm{\Gamma }}^{\prime }(a+b)\mathrm{\Gamma }\left(b\right))}{{\mathrm{\Gamma }}^2(a+b)}$$$$putb=\frac{1}{2}a=\frac{1}{4}$$

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