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Question Number 122330 by mnjuly1970 last updated on 15/Nov/20

$$...advancedcalculus...$$$$provethat:$$$$\mathcal{R}e\left({\int }_0^{\frac{\pi }{2}}{sin}^3\left(x\right)ln\left(ln\left(cos\left(x\right)\right)\right)dx\right)$$$$\stackrel{?}{=}\frac{ln\left(3\right)-2\gamma }{3}✓$$

Answered by mathmax by abdo last updated on 16/Nov/20

$$\mathrm{from}\mathrm{where}\mathrm{you}\mathrm{get}\mathrm{this}\mathrm{sir}\mathrm{mn}...$$

Commented by mathmax by abdo last updated on 16/Nov/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by mnjuly1970 last updated on 16/Nov/20

$$himrmax$$$$from$$$$Apppimterest$$$$and$$$$brilliantmathingoogle...$$

Answered by mindispower last updated on 16/Nov/20

$$letcos\left(x\right)=t\Rightarrow sin\left(x\right)dx=-dt$$$${\int }_0^{\frac{\pi }{2}}{sin}^2\left(x\right)ln\left(ln\left(cos\left(x\right)\right)\right)sin\left(x\right)dx$$$$={\int }_0^1(1-t^2)ln\left(ln\left(t\right)\right)dt$$$$-ln\left(t\right)=u\Rightarrow dt=-e^{-u}du$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}(1-e^{-2u})ln(-u)e^{-u}du$$$$ln(-u)=ln\left(u\right)+i\pi$$$$Re\left(\mathrm{\Omega }\right)={\int }_0^{\mathrm{\infty }}{e}^{-u}ln\left(u\right)du-{\int }_0^{\mathrm{\infty }}{e}^{-3u}ln\left(u\right)du...I$$$$3u=y\Rightarrow du=\frac{dy}{3}$$$$\iff I=\frac{2}{3}{\int }_0^{\mathrm{\infty }}{e}^{-u}ln\left(u\right)du+\frac{1}{3}{\int }_0^{\mathrm{\infty }}{e}^{-y}ln\left(3\right)dy$$$$\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt\Rightarrow \mathrm{\Gamma }\left(1\right)={\int }_0^{\mathrm{\infty }}{e}^{-t}dt$$$${\mathrm{\Gamma }}^{\prime }\left(x\right)={\int }_0^{\mathrm{\infty }}ln\left(t\right)t^{x-1}ln\left(t\right)dt$$$${\mathrm{\Gamma }}^{\prime }\left(1\right)={\int }_0^{\mathrm{\infty }}ln\left(t\right)e^{-t}dt$$$$Re\left(\mathrm{\Omega }\right)=\frac{2}{3}{\mathrm{\Gamma }}^{\prime }\left(1\right)+\frac{1}{3}\mathrm{\Gamma }\left(1\right)$$$${\mathrm{\Gamma }}^{\prime }\left(1\right)=\mathrm{\Psi }\left(1\right)=-\gamma$$$$\iff Re\left(\mathrm{\Omega }\right)=-\frac{2}{3}\gamma +\frac{1}{3}=\frac{1-2\gamma }{3}$$$$$$

Commented by mnjuly1970 last updated on 16/Nov/20

$$thankyousomuchsirmax$$$$$$

Commented by mnjuly1970 last updated on 16/Nov/20

$$gratefulsirmidspower.verynice$$$$solution....$$

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