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Question Number 73238 by mathmax by abdo last updated on 08/Nov/19

$$let0<a<1calculate{\int }_0^{\mathrm{\infty }}\frac{ln\left(t\right)t^{a-1}}{1+t}dtand{\int }_0^{\mathrm{\infty }}\frac{{ln}^2\left(t\right)t^{a-1}}{1+t}dt$$

Commented bymathmax by abdo last updated on 10/Nov/19

$$wehave{\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt=\frac{\pi }{sin\left(\pi a\right)}=f\left(a\right)if0<a<1(thisresultisprovedin$$ $$theplstform)\Rightarrow f^{{}^{\prime }}\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{\partial }{\partial a}\left(\frac{t^{a-1}}{1+t}\right)dt={\int }_0^{\mathrm{\infty }}\frac{\partial }{\partial a}\left(\frac{e^{(a-1)lnt}}{1+t}\right)dt$$ $$={\int }_0^{\mathrm{\infty }}\frac{lnt\times t^{a-1}}{1+t}but{f}^{{}^{\prime }}\left(a\right)=\pi \times \frac{-\pi cos\left(\pi a\right)}{{sin}^2\left(\pi a\right)}=-{\pi }^2\frac{cos\left(\pi a\right)}{{sin}^2\left(\pi a\right)}$$ $$also{f}^{\left(2\right)}\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{{ln}^2\left(t\right)t^{a-1}}{1+t}dtand{f}^{\left(2\right)}\left(a\right)=$$ $$-{\pi }^2\frac{-\pi sin\left(\pi a\right){sin}^2\left(\pi a\right)-2sin\left(\pi a\right)\pi cos\left(\pi a\right)}{{sin}^4\left(\pi a\right)}$$ $$={\pi }^3\times \frac{{sin}^2\left(\pi a\right)+2cos\left(\pi a\right)}{{sin}^3\left(\pi a\right)}\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{ln\left(t\right)t^{a-1}}{1+t}dt=-\frac{{\pi }^{2}cos\left(\pi a\right)}{{sin}^2\left(\pi a\right)}$$ $$and{\int }_0^{\mathrm{\infty }}\frac{{ln}^{2}t{t}^{a-1}}{1+t}dt={\pi }^3\times \frac{{sin}^2\left(\pi a\right)+2cos\left(\pi a\right)}{{sin}^3\left(\pi a\right)}.$$ $$$$

Answered by mind is power last updated on 09/Nov/19

Commented bymathmax by abdo last updated on 10/Nov/19

$$thankyousir.$$

Commented bymind is power last updated on 10/Nov/19

$${\mathrm{y}}^{\prime }\mathrm{re}\mathrm{welcom}$$

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