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Question Number 210572 by Spillover last updated on 12/Aug/24

Answered by mathmax last updated on 13/Aug/24

$$I={\int }_0^{\mathrm{\infty }}\frac{{ln}^{2}x}{1+x^4}dxchangementx=t^{\frac{1}{4}}give$$$$I=\frac{1}{16}{\int }_0^{\mathrm{\infty }}\frac{{ln}^{2}t}{1+t}\frac{1}{4}{t}^{\frac{1}{4}-1}dt\Rightarrow$$$$64I={\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{4}-1}}{1+t}{ln}^2\left(t\right)dt$$$$letf\left(\lambda \right)={\int }_0^{\mathrm{\infty }}\frac{t^{\lambda -1}}{1+t}dtwith0<\lambda <1$$$$f^{\left(2\right)}\left(\lambda \right)={\int }_0^{\mathrm{\infty }}\frac{{ln}^{2}t.t^{\lambda -1}}{1+t}dt\Rightarrow f^{\left(2\right)}\left(\frac{1}{4}\right)=64I$$$$f\left(\lambda \right)=\frac{\pi }{sin\left(\pi \lambda \right)}\Rightarrow f^{{}^{\prime }}\left(\lambda \right)=-\frac{{\pi }^{2}cos\left(\pi \lambda \right)}{{sin}^2\left(\pi \lambda \right)}$$$$f^{{}^{\prime }{}^{\prime }}\left(\lambda \right)=-{\pi }^2\times \frac{-\pi sin\left(\pi \lambda \right){sin}^2\left(\pi \lambda \right)-\pi cos\left(\pi \lambda \right)2sin\left(\pi \lambda \right)cos\left(\pi \lambda \right)}{{sin}^4\left(\pi \lambda \right)}$$$$=-{\pi }^3\times \frac{{sin}^2\left(\pi \lambda \right)+2{cos}^2\left(\pi \lambda \right)}{{sin}^3\left(\pi \lambda \right)}$$$$=-{\pi }^3\times \frac{1+{cos}^2\left(\pi \lambda \right)}{{sin}^3\left(\pi \lambda \right)}\Rightarrow$$$$f^{\left(2\right)}\left(\frac{1}{4}\right)=-{\pi }^3\times \frac{1+{cos}^2\left(\frac{\pi }{4}\right)}{{sin}^3\left(\frac{\pi }{4}\right)}$$$$=-{\pi }^3\times \frac{1+\frac{1}{2}}{\frac{1}{2\sqrt{2}}}=-{\pi }^3\left(2\sqrt{2}\right)\frac{3}{2}=-3\sqrt{2}{\pi }^3$$$$\Rightarrow 64I=-3\sqrt{2}{\pi }^3\Rightarrow I=-\frac{3\sqrt{2}}{64}{\pi }^3$$

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