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Question Number 142425 by Mathspace last updated on 31/May/21

$$calculate{\int }_0^{\mathrm{\infty }}\frac{{log}^{3}x}{1+x^3}dx$$

Answered by mathmax by abdo last updated on 02/Jun/21

$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{{\log }^3\mathrm{x}}{1+{\mathrm{x}}^3}\mathrm{dx}\mathrm{changement}{\mathrm{x}}^3=\mathrm{t}\mathrm{give}$$$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{{\left(\log \left({\mathrm{t}}^{\frac{1}{3}}\right)\right)}^3}{1+\mathrm{t}}\frac{1}{3}{\mathrm{t}}^{\frac{1}{3}-1}\mathrm{dt}=\frac{1}{3^4}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\frac{1}{3}-1}\frac{{\log }^3\mathrm{t}}{1+\mathrm{t}}\mathrm{dt}$$$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}-1}}{1+\mathrm{t}}\mathrm{dt}\mathrm{with}0<\mathrm{a}<1\mathrm{we}\mathrm{have}\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi }{\sin \left(\pi \mathrm{a}\right)}$$$$\mathrm{and}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{a}-1}\frac{\mathrm{logt}}{1+\mathrm{t}}\mathrm{dt}\Rightarrow {\mathrm{f}}^{,,}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}-1}{\log }^2\mathrm{t}}{1+\mathrm{t}}\mathrm{dt}\Rightarrow$$$${\mathrm{f}}^{\left(3\right)}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}-1}{\log }^3\mathrm{t}}{1+\mathrm{t}}\mathrm{dt}\Rightarrow {\mathrm{f}}^{\left(3\right)}\left(\frac{1}{3}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{3}-1}{\log }^3\mathrm{t}}{1+\mathrm{t}}\mathrm{dt}$$$$=3^4\mathrm{\Phi }\mathrm{but}\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi }{\sin \left(\pi \mathrm{a}\right)}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-{\pi }^2\frac{\cos \left(\pi \mathrm{a}\right)}{{\sin }^2\left(\pi \mathrm{a}\right)}\Rightarrow$$$${\mathrm{f}}^{\left(2\right)}\left(\mathrm{a}\right)=-{\pi }^2.\frac{-\pi \sin \left(\pi \mathrm{a}\right)-2\pi \sin \left(\pi \mathrm{a}\right)\cos \left(\pi \mathrm{a}\right)}{{\sin }^4\left(\pi \mathrm{a}\right)}$$$$={\pi }^3.\frac{1+2\cos \left(\pi \mathrm{a}\right)}{{\sin }^3\left(\pi \mathrm{a}\right)}\Rightarrow$$$${\mathrm{f}}^{\left(3\right)}\left(\mathrm{a}\right)={\pi }^3.\frac{-2\pi \sin \left(\pi \mathrm{a}\right).{\sin }^3\left(\pi \mathrm{a}\right)-3\pi {\sin }^2\left(\pi \mathrm{a}\right)\cos \left(\pi \mathrm{a}\right)}{{\sin }^6\left(\pi \mathrm{a}\right)}$$$$={\pi }^4.\frac{-{2\sin }^2\left(\pi \mathrm{a}\right)-3\cos \left(\pi \mathrm{a}\right)}{{\sin }^4\left(\pi \mathrm{a}\right)}\Rightarrow$$$${\mathrm{f}}^{\left(3\right)}\left(\frac{1}{3}\right)=-{\pi }^4\times \frac{2{\left(\frac{\sqrt{3}}{2}\right)}^2+3.\frac{1}{2}}{{\left(\frac{\sqrt{3}}{2}\right)}^4}=-{\pi }^4.\frac{\frac{3}{2}+\frac{3}{2}}{\frac{9}{16}}$$$$=(-{\pi }^4).\left(\frac{16}{9}\right)\left(3\right)=-\frac{16{\pi }^4}{3}\Rightarrow \mathrm{\Phi }=-\frac{16{\pi }^4}{3^5}$$$$$$

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