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Question Number 42506 by maxmathsup by imad last updated on 26/Aug/18

$$findthevalueof{\int }_0^{\mathrm{\infty }}\frac{ln\left(t\right)}{\left({}^3\sqrt{t^2}\right)(1+t)}dt.$$

Commented by maxmathsup by imad last updated on 29/Aug/18

$$wehaveprovedhat{\int }_0^{\mathrm{\infty }}\frac{ln\left(t\right)t^{a-1}}{1+t}dt=-{\pi }^2\frac{cos\left(\pi a\right)}{{sin}^2\left(\pi a\right)}and$$$${\int }_0^{\mathrm{\infty }}\frac{ln\left(t\right)}{\left({}^3\sqrt{t^2}\right)(1+t)}dt={\int }_0^{\mathrm{\infty }}\frac{ln\left(t\right).t^{-\frac{2}{3}}}{1+t}dt={\int }_0^{\mathrm{\infty }}\frac{ln\left(t\right)t^{\frac{1}{3}-1}}{1+t}dt=-{\pi }^2\frac{cos\left(\frac{\pi }{3}\right)}{{sin}^2\left(\frac{\pi }{3}\right)}$$$$=-{\pi }^2\frac{1}{2{\left(\frac{\sqrt{3}}{2}\right)}^2}=\frac{-{\pi }^2}{\frac{\sqrt{3}}{2}}=\frac{-2{\pi }^2}{\sqrt{3}}.$$

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