Question Number 140056 by Ndala last updated on 03/May/21
$$\mathrm{Prove}\:\mathrm{the}\:\mathrm{folowing}\:\mathrm{result}: \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cot}\:\theta\centerdot\left(\mathrm{log}\:\mathrm{sec}\:\theta\right)^{\mathrm{3}} {d}\theta=\frac{\pi^{\mathrm{4}} }{\mathrm{240}} \\ $$$$. \\ $$$$\mathrm{I}\:\mathrm{need}\:\mathrm{your}\:\mathrm{help},\:\mathrm{if}\:\mathrm{possible}\:\mathrm{please}. \\ $$
Answered by Ar Brandon last updated on 04/May/21
![∫_0 ^(π/2) cotθ(ln secθ)^3 dθ=−∫_0 ^(π/2) sin^(−1) θcosθ∙ln^3 cosθdθ β(m,n)=2∫_0 ^(π/2) sin^(2m−1) θcos^(2n−1) θdθ=((Γ(m)Γ(n))/(Γ(m+n))) lnβ(m,n)=lnΓ(m)+lnΓ(n)−lnΓ(m+n) ((β ′(m,n))/(β(m,n)))=ψ(n)−ψ(m+n) β′′(m,n)=β(m,n)[ψ′(n)−ψ′(m+n)]+β′(m,n)[ψ(n)−ψ(m+n)] β′′′(m,n)=β(m,n)[ψ′′(n)−ψ′′(m+n)]+β′(m,n)[ψ′(n)−ψ′(m+n)] +β′′(m,n)[ψ(n)−ψ(m+n)]+β′(m,n)[ψ′(n)−ψ′(m+n)]](https://www.tinkutara.com/question/Q140080.png)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cot}\theta\left(\mathrm{ln}\:\mathrm{sec}\theta\right)^{\mathrm{3}} \mathrm{d}\theta=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{−\mathrm{1}} \theta\mathrm{cos}\theta\centerdot\mathrm{ln}^{\mathrm{3}} \mathrm{cos}\theta\mathrm{d}\theta \\ $$$$\beta\left(\mathrm{m},\mathrm{n}\right)=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2m}−\mathrm{1}} \theta\mathrm{cos}^{\mathrm{2n}−\mathrm{1}} \theta\mathrm{d}\theta=\frac{\Gamma\left(\mathrm{m}\right)\Gamma\left(\mathrm{n}\right)}{\Gamma\left(\mathrm{m}+\mathrm{n}\right)} \\ $$$$\mathrm{ln}\beta\left(\mathrm{m},\mathrm{n}\right)=\mathrm{ln}\Gamma\left(\mathrm{m}\right)+\mathrm{ln}\Gamma\left(\mathrm{n}\right)−\mathrm{ln}\Gamma\left(\mathrm{m}+\mathrm{n}\right) \\ $$$$\frac{\beta\:'\left(\mathrm{m},\mathrm{n}\right)}{\beta\left(\mathrm{m},\mathrm{n}\right)}=\psi\left(\mathrm{n}\right)−\psi\left(\mathrm{m}+\mathrm{n}\right) \\ $$$$\beta''\left(\mathrm{m},\mathrm{n}\right)=\beta\left(\mathrm{m},\mathrm{n}\right)\left[\psi'\left(\mathrm{n}\right)−\psi'\left(\mathrm{m}+\mathrm{n}\right)\right]+\beta'\left(\mathrm{m},\mathrm{n}\right)\left[\psi\left(\mathrm{n}\right)−\psi\left(\mathrm{m}+\mathrm{n}\right)\right] \\ $$$$\beta'''\left(\mathrm{m},\mathrm{n}\right)=\beta\left(\mathrm{m},\mathrm{n}\right)\left[\psi''\left(\mathrm{n}\right)−\psi''\left(\mathrm{m}+\mathrm{n}\right)\right]+\beta'\left(\mathrm{m},\mathrm{n}\right)\left[\psi'\left(\mathrm{n}\right)−\psi'\left(\mathrm{m}+\mathrm{n}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\beta''\left(\mathrm{m},\mathrm{n}\right)\left[\psi\left(\mathrm{n}\right)−\psi\left(\mathrm{m}+\mathrm{n}\right)\right]+\beta'\left(\mathrm{m},\mathrm{n}\right)\left[\psi'\left(\mathrm{n}\right)−\psi'\left(\mathrm{m}+\mathrm{n}\right)\right] \\ $$
Commented by Ndala last updated on 07/May/21
$$\mathrm{Can}\:\:\mathrm{you}\:\mathrm{complete}? \\ $$
Answered by qaz last updated on 07/May/21
$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left(\mathrm{cot}\:\theta\right)\centerdot{ln}^{\mathrm{3}} \mathrm{sec}\:\theta{d}\theta……..\mathrm{cos}\:\theta\rightarrow{x} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\:\mathrm{1}−{x}^{\mathrm{2}} }{ln}^{\mathrm{3}} \left(\frac{\mathrm{1}}{{x}}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−\mathrm{2}{y}} }{\mathrm{1}−{e}^{−\mathrm{2}{y}} }{y}^{\mathrm{3}} {dy}…………….{x}\rightarrow{e}^{−{y}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{2}{y}} }−\mathrm{1}\right){y}^{\mathrm{3}} {dy} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\infty} {y}^{\mathrm{3}} {e}^{−\mathrm{2}{ny}} {dy} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{3}!}{\left(\mathrm{2}{n}\right)^{\mathrm{4}} } \\ $$$$=\frac{\pi^{\mathrm{4}} }{\mathrm{240}} \\ $$