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Question Number 125668 by mnjuly1970 last updated on 12/Dec/20

$$...advancedcalculus...$$$$evaluate:::$$$$\mathrm{\Phi }={\int }_0^{\frac{\pi }{4}}ln\left(ln\left(cot\left(x\right)\right)\right)dx=?$$$$$$

Answered by mindispower last updated on 13/Dec/20

$$\mathrm{\Phi }={\int }_1^{\mathrm{\infty }}\frac{ln\left(ln\left(x\right)\right)}{1+x^2}dx={\int }_0^{\mathrm{\infty }}\frac{ln\left(t\right)}{1+e^{2t}}{e}^{t}dt$$$$={\int }_0^{\mathrm{\infty }}\frac{ln\left(t\right)e^{-t}}{1+e^{-2t}}dt$$$$=\sum _{k⩾0}{\int }_0^{\mathrm{\infty }}{(-1)}^{k}ln\left(t\right)e^{-(1+2k)t}$$$$ln\left(t\right)={\partial }_s{t}^s\mid s=0$$$$g\left(s\right)=\sum _{k⩾0}{\int }_0^{\mathrm{\infty }}{(-1)}^k{t}^s{e}^{-(1+2k)t}dt$$$$=\sum _{k⩾0}{\int }_0^{\mathrm{\infty }}{(-1)}^k\frac{y^s{e}^{-y}dy}{{(1+2k)}^{s+1}}$$$$=\sum _{k⩾0}\frac{{(-1)}^k}{{(1+2k)}^{s+1}}{\int }_0^{\mathrm{\infty }}{y}^s{e}^{-y}dy$$$$=\sum _{k⩾0}\frac{{(-1)}^k}{{(1+2k)}^{s+1}}\mathrm{\Gamma }(s+1)$$$$=\mathrm{\Gamma }(s+1)\sum _{k⩾0}(\frac{1}{{(1+4k)}^{s+1}}-\frac{1}{{(3+4k)}^{s+1}})$$$$=\frac{\mathrm{\Gamma }(s+1)}{4^{s+1}}(\sum _{k⩾0}\frac{1}{{(k+\frac{1}{4})}^{s+1}}-\sum _{k⩾0}\frac{1}{{(k+\frac{3}{4})}^{s+1}})$$$$=\frac{\mathrm{\Gamma }(s+1)}{4^{s+1}}(\zeta (s+1,\frac{1}{4})-\zeta (s+1,\frac{3}{4})$$$$\zeta (s,q)=\sum _{n⩾0}\frac{1}{{(n+q)}^s},Zetshrwitzfunction$$$$wewant{g}^{\prime }\left(0\right)$$$$\frac{{\mathrm{\Gamma }}^{\prime }(s+1)4^{s+1}-4^{s+1}ln\left(4\right)\mathrm{\Gamma }(s+1)}{4^{2s+2}}{\mid }_{s=0}.\underset{s\to 0}{\lim }(\zeta (s+1,\frac{1}{4})-\zeta (s+1,\frac{3}{4}))$$$$+\underset{s\to 0}{\lim }{\partial }_s(\zeta (s+1,\frac{1}{4})-\zeta (s+1,\frac{3}{4})).\frac{\mathrm{\Gamma }(s+1)}{4^{s+1}}$$$$\zeta (s+1,q)=\frac{1}{s}+\mathrm{\Sigma }\frac{{(-1)}^n}{n!}{\gamma }_n\left(q\right)s^n,laurentserie$$$$expansion{\gamma }_n\left(q\right)Stieltjesconstante$$$$\zeta (s+1,\frac{1}{4})-\zeta (s+1,\frac{3}{4})=h\left(s\right)=\sum _{n⩾0}\frac{{(-1)}^n}{n!}({\gamma }_n\left(\frac{1}{4}\right)-{\gamma }_n\left(\frac{3}{4}\right))s^n$$$$\underset{s\to 0}{\lim }h\left(s\right)={\gamma }_0\left(\frac{1}{4}\right)-{\gamma }_0\left(\frac{3}{4}\right)$$$${\partial }_{s}h\left(s\right)=\sum _{n⩾1}\frac{{(-1)}^n}{(n-1)!}({\gamma }_n\left(\frac{1}{4}\right)-{\gamma }_n\left(\frac{3}{4}\right)s^{n-1}$$$$\underset{s\to 0}{\lim }{h}^{\prime }\left(s\right)={\gamma }_1\left(\frac{3}{4}\right)-{\gamma }_1\left(\frac{1}{4}\right)$$$${\gamma }_1\left(\frac{3}{4}\right)-{\gamma }_1(1-\frac{3}{4})=2\pi \underset{l=1}{\sum ^3}sin\left(\frac{6\pi l}{4}\right)ln\mathrm{\Gamma }\left(\frac{l}{4}\right)-\pi (\gamma +ln\left(8\pi \right))cot\left(\frac{3\pi }{4}\right)$$$$=-2\pi \mathrm{\Gamma }\left(\frac{1}{4}\right)+2\pi \mathrm{\Gamma }\left(\frac{3}{4}\right)+\pi (\gamma +ln\left(8\pi \right))$$$$=2\pi (\mathrm{\Gamma }\left(\frac{3}{4}\right)-\mathrm{\Gamma }\left(\frac{1}{4}\right))+\pi (\gamma +ln\left(8\pi \right))$$$${\gamma }_0\left(\frac{1}{4}\right)-{\gamma }_0\left(\frac{3}{4}\right)=\mathrm{\Psi }\left(\frac{3}{4}\right)-\mathrm{\Psi }\left(\frac{1}{4}\right)=-\pi cot\left(\frac{3\pi }{4}\right)$$$$=\pi$$$$g^{\prime }\left(0\right)=\frac{4{\mathrm{\Gamma }}^{\prime }\left(1\right)-4ln\left(4\right)\mathrm{\Gamma }\left(1\right)}{4^2}\pi +(2\pi (\mathrm{\Gamma }\left(\frac{3}{4}\right)-\mathrm{\Gamma }\left(\frac{1}{4}\right))+\pi (\gamma +ln\left(8\pi \right)).\frac{1}{4}$$$$$$$$$$$$=-\frac{\gamma }{4}\pi -\frac{ln\left(2\right)}{2}\pi +\frac{\pi }{2}(\mathrm{\Gamma }\left(\frac{3}{4}\right)-\mathrm{\Gamma }\left(\frac{1}{4}\right))+\frac{\gamma \pi }{4}+\frac{\pi }{4}ln\left(8\pi \right)$$$$=\frac{\pi }{4}ln\left(2\pi \right)+\frac{\pi }{2}(\mathrm{\Gamma }\left(\frac{3}{4}\right)-\mathrm{\Gamma }\left(\frac{1}{4}\right))$$

Commented by mindispower last updated on 13/Dec/20

$$wecan$$$$use\mathrm{\Gamma }\left(z\right)\mathrm{\Gamma }(z+\frac{1}{2})=2^{1-2z}\sqrt{\pi }\mathrm{\Gamma }\left(2z\right)$$$$\mathrm{\Gamma }\left(\frac{1}{4}\right).\mathrm{\Gamma }\left(\frac{3}{4}\right)=2^{\frac{1}{2}}\sqrt{\pi }.\mathrm{\Gamma }\left(\frac{1}{2}\right)=\pi \sqrt{2}$$$$\Rightarrow \mathrm{\Gamma }\left(\frac{3}{4}\right)=\pi \sqrt{2}.\frac{1}{\mathrm{\Gamma }\left(\frac{1}{4}\right)}...toexpressusing$$$$just\mathrm{\Gamma }\left(\frac{1}{4}\right)$$$$$$

Commented by mnjuly1970 last updated on 13/Dec/20

$$thanksalot$$$$mrpower......$$

Commented by mindispower last updated on 13/Dec/20

$$withepleasur$$

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