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Question Number 140056 by Ndala last updated on 03/May/21

$$\mathrm{Prove}\mathrm{the}\mathrm{folowing}\mathrm{result}:$$$${\int }_0^{\frac{\pi }{2}}\cot \theta \cdot {\left(\log \sec \theta \right)}^{3}d\theta =\frac{{\pi }^4}{240}$$$$.$$$$\mathrm{I}\mathrm{need}\mathrm{your}\mathrm{help},\mathrm{if}\mathrm{possible}\mathrm{please}.$$

Answered by Ar Brandon last updated on 04/May/21

$${\int }_0^{\frac{\pi }{2}}\cot \theta {\left(\ln \sec \theta \right)}^3\mathrm{d}\theta =-{\int }_0^{\frac{\pi }{2}}{\sin }^{-1}\theta \cos \theta \cdot {\ln }^3\cos \theta \mathrm{d}\theta$$$$\beta (\mathrm{m},\mathrm{n})=2{\int }_0^{\frac{\pi }{2}}{\sin }^{2\mathrm{m}-1}\theta {\cos }^{2\mathrm{n}-1}\theta \mathrm{d}\theta =\frac{\mathrm{\Gamma }\left(\mathrm{m}\right)\mathrm{\Gamma }\left(\mathrm{n}\right)}{\mathrm{\Gamma }(\mathrm{m}+\mathrm{n})}$$$$\ln \beta (\mathrm{m},\mathrm{n})=\ln \mathrm{\Gamma }\left(\mathrm{m}\right)+\ln \mathrm{\Gamma }\left(\mathrm{n}\right)-\ln \mathrm{\Gamma }(\mathrm{m}+\mathrm{n})$$$$\frac{\beta {}^{\prime }(\mathrm{m},\mathrm{n})}{\beta (\mathrm{m},\mathrm{n})}=\psi \left(\mathrm{n}\right)-\psi (\mathrm{m}+\mathrm{n})$$$${\beta }^{″}(\mathrm{m},\mathrm{n})=\beta (\mathrm{m},\mathrm{n})[{\psi }^{\prime }\left(\mathrm{n}\right)-{\psi }^{\prime }(\mathrm{m}+\mathrm{n})]+{\beta }^{\prime }(\mathrm{m},\mathrm{n})[\psi \left(\mathrm{n}\right)-\psi (\mathrm{m}+\mathrm{n})]$$$${\beta }^{‴}(\mathrm{m},\mathrm{n})=\beta (\mathrm{m},\mathrm{n})[{\psi }^{″}\left(\mathrm{n}\right)-{\psi }^{″}(\mathrm{m}+\mathrm{n})]+{\beta }^{\prime }(\mathrm{m},\mathrm{n})[{\psi }^{\prime }\left(\mathrm{n}\right)-{\psi }^{\prime }(\mathrm{m}+\mathrm{n})]$$$$+{\beta }^{″}(\mathrm{m},\mathrm{n})[\psi \left(\mathrm{n}\right)-\psi (\mathrm{m}+\mathrm{n})]+{\beta }^{\prime }(\mathrm{m},\mathrm{n})[{\psi }^{\prime }\left(\mathrm{n}\right)-{\psi }^{\prime }(\mathrm{m}+\mathrm{n})]$$

Commented by Ndala last updated on 07/May/21

$$\mathrm{Can}\mathrm{you}\mathrm{complete}?$$

Answered by qaz last updated on 07/May/21

$${\int }_0^{\pi /2}\left(\cot \theta \right)\cdot {ln}^3\sec \theta d\theta ........\cos \theta \to x$$$$={\int }_0^1\frac{x}{1-x^2}{ln}^3\left(\frac{1}{x}\right)dx$$$$={\int }_0^{\mathrm{\infty }}\frac{e^{-2y}}{1-e^{-2y}}{y}^{3}dy................x\to e^{-y}$$$$={\int }_0^{\mathrm{\infty }}(\frac{1}{1-e^{-2y}}-1)y^{3}dy$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}{\int }_0^{\mathrm{\infty }}{y}^3{e}^{-2ny}dy$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{3!}{{\left(2n\right)}^4}$$$$=\frac{{\pi }^4}{240}$$

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