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Question Number 143546 by mathmax by abdo last updated on 15/Jun/21

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}\frac{{\log }^2\mathrm{x}}{{(8+{\mathrm{x}}^4)}^2}\mathrm{dx}$$

Answered by mathmax by abdo last updated on 16/Jun/21

$$\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{{\log }^2\mathrm{x}}{{\mathrm{a}}^4+{\mathrm{x}}^4}\mathrm{dx}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-{\int }_0^{\mathrm{\infty }}\frac{{4\mathrm{a}}^3{\log }^2\mathrm{x}}{{({\mathrm{a}}^4+{\mathrm{x}}^4)}^2}\mathrm{dx}\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(2^{\frac{3}{4}}\right)=-4{\left(2\right)}^{\frac{3}{4}}{\int }_0^{\mathrm{\infty }}\frac{{\log }^2\mathrm{x}}{{(8+{\mathrm{x}}^4)}^2}\mathrm{dx}=-2^{2+\frac{3}{4}}{\int }_0^{\mathrm{\infty }}\frac{{\log }^2\mathrm{x}}{{(8+{\mathrm{x}}^4)}^2}\mathrm{dx}$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{{\log }^2\mathrm{x}}{{(8+{\mathrm{x}}^4)}^2}\mathrm{dx}=-\frac{1}{2^{\frac{11}{4}}}{\mathrm{f}}^{{}^{\prime }}\left(2^{\frac{3}{4}}\right)$$$$\mathrm{f}\left(\mathrm{a}\right){=}_{\mathrm{x}=\mathrm{at}}{\int }_0^{\mathrm{\infty }}\frac{{\log }^2\left(\mathrm{at}\right)}{{\mathrm{a}}^4(1+{\mathrm{t}}^4)}\mathrm{adt}=\frac{1}{{\mathrm{a}}^3}{\int }_0^{\mathrm{\infty }}\frac{{\log }^2\mathrm{a}+2\mathrm{logalogt}+{\log }^2\mathrm{t}}{1+{\mathrm{t}}^4}\mathrm{dt}$$$$=\frac{{\log }^2\mathrm{a}}{{\mathrm{a}}^3}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{1+{\mathrm{t}}^4}+\frac{2\mathrm{loga}}{{\mathrm{a}}^3}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{logt}}{1+{\mathrm{t}}^4}\mathrm{dt}+\frac{1}{{\mathrm{a}}^3}{\int }_0^{\mathrm{\infty }}\frac{{\log }^2\left(\mathrm{t}\right)}{1+{\mathrm{t}}^4}\mathrm{dt}$$$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{1+{\mathrm{t}}^4}{=}_{{\mathrm{t}}^4=\mathrm{y}}\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{y}}^{\frac{1}{4}-1}}{1+\mathrm{y}}\mathrm{dy}=\frac{1}{4}\times \frac{\pi }{\sin \left(\frac{\pi }{4}\right)}=\frac{\pi }{4\times \frac{\sqrt{2}}{2}}$$$$=\frac{\pi }{2\sqrt{2}}$$$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{logt}}{1+{\mathrm{t}}^4}\mathrm{dt}{=}_{{\mathrm{t}}^4=\mathrm{y}}\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{logy}}{1+\mathrm{y}}\times \frac{1}{4}{\mathrm{y}}^{\frac{1}{4}-1}\mathrm{dy}$$$$=\frac{1}{16}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{y}}^{\frac{1}{4}-1}\mathrm{logy}}{1+\mathrm{y}}\mathrm{dy}=\frac{1}{16}{\mathrm{w}}^{{}^{\prime }}\left(\frac{1}{4}\right)\mathrm{with}$$$$\mathrm{w}\left(\lambda \right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{y}}^{\lambda -1}}{1+\mathrm{y}}\mathrm{dy}\Rightarrow {\mathrm{w}}^{{}^{\prime }}\left(\lambda \right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{y}}^{\lambda -1}\mathrm{logy}}{1+\mathrm{y}}\mathrm{dy}$$$$\mathrm{w}\left(\lambda \right)=\frac{\pi }{\sin \left(\pi \lambda \right)}\Rightarrow {\mathrm{w}}^{{}^{\prime }}\left(\lambda \right)=-\frac{{\pi }^2\cos \left(\pi \lambda \right)}{{\sin }^2\left(\pi \lambda \right)}\Rightarrow$$$${\mathrm{w}}^{{}^{\prime }}\left(\frac{1}{4}\right)=-{\pi }^2\times \frac{\frac{1}{\sqrt{2}}}{{\left(\frac{1}{\sqrt{2}}\right)}^2}=-2{\pi }^2.\frac{1}{\sqrt{2}}=-{\pi }^2\sqrt{2}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{logt}}{1+{\mathrm{t}}^4}\mathrm{dt}=-\frac{\sqrt{2}}{16}{\pi }^2$$$${\int }_0^{\mathrm{\infty }}\frac{{\log }^2\mathrm{t}}{1+{\mathrm{t}}^4}\mathrm{dt}{=}_{{\mathrm{t}}^4=\mathrm{y}}\frac{1}{16}{\int }_0^{\mathrm{\infty }}\frac{{\log }^2\mathrm{y}}{1+\mathrm{y}}\times \frac{1}{4}{\mathrm{y}}^{\frac{1}{4}-1}\mathrm{dt}$$$$=\frac{1}{64}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{y}}^{\frac{1}{4}-1}{\log }^2\mathrm{y}}{1+\mathrm{y}}\mathrm{dy}=\frac{1}{64}{\mathrm{w}}^{\left(2\right)}\left(\frac{1}{4}\right)$$$${\mathrm{w}}^{\left(2\right)}\left(\lambda \right)=-{\pi }^2\times \frac{-\pi \sin \left(\pi \lambda \right){\sin }^2\left(\pi \lambda \right)-2\pi \sin \left(\pi \lambda \right)\cos \left(\pi \lambda \right)}{{\sin }^4\left(\pi \lambda \right)}$$$$=-{\pi }^3\times \frac{{\sin }^2\left(\pi \lambda \right)-2\cos \left(\pi \lambda \right)}{{\sin }^3\left(\pi \lambda \right)}\Rightarrow$$$${\mathrm{w}}^{\left(2\right)}\left(\frac{1}{4}\right)=-{\pi }^3\times \frac{\frac{1}{2}-2\times \frac{1}{\sqrt{2}}}{{\left(\frac{1}{\sqrt{2}}\right)}^3}=-{\pi }^3\times \frac{\frac{1}{2}-\sqrt{2}}{\frac{1}{2\sqrt{2}}}$$$$=-2\sqrt{2}{\pi }^3\left(\frac{1-2\sqrt{2}}{2}\right)=\sqrt{2}{\pi }^3(2\sqrt{2}-1)\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{{\log }^2\mathrm{a}}{{\mathrm{a}}^3}\times \frac{\pi }{2\sqrt{2}}+\frac{2\mathrm{loga}}{{\mathrm{a}}^3}(-\frac{\sqrt{2}}{16}){\pi }^2+\frac{1}{{\mathrm{a}}^3}\times \frac{1}{64}\sqrt{2}{\pi }^3(2\sqrt{2}-1)$$$$\mathrm{\Phi }=-\frac{1}{2^{\frac{11}{4}}}{\mathrm{f}}^{{}^{\prime }}\left(2^{\frac{3}{4}}\right)....$$

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