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Question Number 121282 by mnjuly1970 last updated on 06/Nov/20

$$...\mathrm{advanced}\mathrm{mathematics}...$$$$$$$$\mathrm{prove}\mathrm{that}::$$$$\mathrm{\Psi }={\int }_0^1\mathrm{\Gamma }(2-x)\mathrm{\Gamma }(1+x)dx=\frac{7}{{\pi }^2}\zeta \left(3\right)$$$$...\mathrm{m}.\mathrm{n}.\mathrm{july}.1970...$$$$$$

Answered by mindispower last updated on 06/Nov/20

$$\mathrm{\Psi }={\int }_0^1(1-x)(-x)\mathrm{\Gamma }(-x)\mathrm{\Gamma }(1+x)dx$$$$\mathrm{\Gamma }(-x)\mathrm{\Gamma }(1+x)=\frac{\pi }{sin(-\pi x)}=-\frac{\pi }{sin\left(\pi x\right)}$$$$\mathrm{\Psi }=\pi {\int }_0^1\frac{x(x-1)}{sin\left(\pi x\right)}dx,WeuseintegrationBypart$$$$\Rightarrow \mathrm{\Psi }=\frac{1}{2}{\left[ln\left(\frac{1-cos\left(\pi x\right)}{1+cos\left(\pi x\right)}\right)x(x-1)\right]}_0^1-\frac{1}{2}\int (2x-1)ln\left(\frac{1-cos\left(\pi x\right)}{1+cos\left(\pi x\right)}\right)dx$$$$\mathrm{\Psi }=-\frac{1}{2}{\int }_0^1(2x-1)ln\left(\frac{2{sin}^2\left(\frac{\pi x}{2}\right)}{2{cos}^2\left(\frac{\pi x}{2}\right)}\right)dx$$$$=-{\int }_0^1(2x-1)ln\left(tg\left(\frac{\pi x}{2}\right)\right)dx$$$$letz=\frac{\pi x}{2}\Rightarrow dx=\frac{2}{\pi }dz$$$$=-\frac{2}{\pi }{\int }_0^{\frac{\pi }{2}}(\frac{4z}{\pi }-1)ln\left(tg\left(z\right)\right).dz$$$$=-\frac{8}{{\pi }^2}{\int }_0^{\frac{\pi }{2}}zln\left(tg\left(z\right)\right)dz$$$$\mathrm{\forall }x\in [0,\frac{\pi }{2}[wehave$$$$ln\left(tg\left(x\right)\right)=-2\mathrm{\Sigma }\frac{cos\left(2(2n-1)x\right)}{2n-1}$$$$\mathrm{\Psi }=-\frac{8}{{\pi }^2}\sum _{n⩾1}\frac{-2}{(2n-1)}\int xcos\left(2(2n-1)x\right)dxIBP$$$$-\frac{8}{{\pi }^2}.2\sum _{n⩾1}{\int }_0^{\frac{\pi }{2}}\frac{1}{2{(2n-1)}^2}sin\left(2(2n-1)\right)$$$$=-\frac{16}{{\pi }^2}\sum _{n⩾1}\left[\frac{cos\left((2n-1)\pi \right)-1}{4{(2n-1)}^3}\right]$$$$=.-\frac{16}{{\pi }^2}.\sum _{n⩾1}-\frac{1}{2{(2n-1)}^3}=-\frac{16}{{\pi }^2}.-\frac{1}{2}\left(\frac{7}{8}\zeta \left(3\right)\right)$$$$=\frac{7}{{\pi }^2}\zeta \left(3\right)$$$$$$$$$$$$$$

Commented by mnjuly1970 last updated on 06/Nov/20

$$bravosirmindspower$$$$goodforyou...$$

Commented by mnjuly1970 last updated on 06/Nov/20

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