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Question Number 198151 by sonukgindia last updated on 11/Oct/23

Answered by Mathspace last updated on 12/Oct/23

$$I={\int }_0^{\mathrm{\infty }}\frac{zlnz}{1+z^3}dz(z=t^{\frac{1}{3}})$$$$=\frac{1}{3}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{3}}lnt}{1+t}.\frac{1}{3}{t}^{\frac{1}{3}-1}dt$$$$\Rightarrow 9I={\int }_0^{\mathrm{\infty }}\frac{t^{\frac{2}{3}-1}lnt}{1+t}dt$$$$let{J}_a={\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt$$$$wehave{J}^{,}a={\int }_0^{\mathrm{\infty }}\frac{t^{a-1}lnt}{1+t}dt$$$$and{J}^{{}^{\prime }}\left(\frac{2}{3}\right)=9I\Rightarrow I=\frac{1}{9}{J}^{{}^{\prime }}\left(\frac{2}{3}\right)$$$$J_a=\frac{\pi }{sin\left(\pi a\right)}\Rightarrow$$$$J_a^{{}^{\prime }}=-{\pi }^2\frac{cos\left(\pi a\right)}{{sin}^2\left(\pi a\right)}and$$$$J^{{}^{\prime }}\left(\frac{2}{3}\right)=-{\pi }^2\frac{cos\left(\frac{2\pi }{3}\right)}{{sin}^2\left(\frac{2\pi }{3}\right)}$$$$=-{\pi }^2.\frac{(-\frac{1}{2})}{\frac{\sqrt{3}}{2}}=\frac{{\pi }^2}{2}.\frac{2}{\sqrt{3}}=\frac{{\pi }^2}{\sqrt{3}}\Rightarrow$$$$I=\frac{{\pi }^2}{9\sqrt{3}}=\frac{{\pi }^2\sqrt{3}}{27}$$

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