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Question Number 201544 by sonukgindia last updated on 08/Dec/23

Answered by aleks041103 last updated on 09/Dec/23

$$J={\int }_0^1\frac{4{sin}^2\left(ln\left(x\right)\right)}{ln\left(1/x\right)}dx=-{\int }_0^1\frac{4{sin}^2(1.ln\left(x\right))}{ln\left(x\right)}dx$$$$I\left(s\right)=-{\int }_0^1\frac{4{sin}^2\left(slnx\right)}{ln\left(x\right)}dx,I\left(0\right)=0,J=I\left(1\right)$$$$I{}^{\prime }\left(s\right)=-{\int }_0^1\frac{8sin\left(slnx\right)cos\left(slnx\right)ln\left(x\right)}{ln\left(x\right)}dx=$$$$=-{\int }_0^{1}4sin\left(2slnx\right)dx=-4Im\left({\int }_0^1{e}^{2isln\left(x\right)}dx\right)=$$$$=-4Im\left({\int }_0^1{x}^{2is}dx\right)=-4Im\left(\frac{1}{1+2is}\right)=$$$$=-4Im\left(\frac{1-2is}{1+4s^2}\right)=\frac{8s}{1+4s^2}=\frac{4\left(2s\right)}{1+4s^2}=$$$$=\frac{{(1+4s^2)}^{\prime }}{1+4s^2}={\left(ln(1+4s^2)\right)}^{\prime }$$$$\Rightarrow I\left(s\right)=ln(1+4s^2)+c$$$$I\left(0\right)=0=ln\left(1\right)+c\Rightarrow c=0$$$$\Rightarrow I\left(s\right)=ln(1+4s^2)$$$$\Rightarrow J={\int }_0^1\frac{4{sin}^2\left(ln\left(x\right)\right)}{ln\left(1/x\right)}dx=I\left(1\right)=ln\left(5\right)$$

Answered by Mathspace last updated on 09/Dec/23

$$I=-4{\int }_0^1\frac{{sin}^2\left(lnx\right)}{lnx}dx(lnx=-t)$$$$=-4{\int }_{\mathrm{\infty }}^0\frac{{sin}^2\left(t\right)}{-t}\times (-e^{-t})dt$$$$={\int }_0^{\mathrm{\infty }}\frac{e^{-t}{sin}^{2}t}{t}dt$$$$f\left(\lambda \right)={\int }_0^{\mathrm{\infty }}\frac{e^{-\lambda t}{sin}^{2}t}{t}dt(\lambda >0)$$$$f^{{}^{\prime }}\left(\lambda \right)=-{\int }_0^{\mathrm{\infty }}{e}^{-\lambda t}{sin}^{2}tdt$$$$=-{\int }_0^{\mathrm{\infty }}{e}^{-\lambda t}\frac{1-cos\left(2t\right)}{2}dt$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{e}^{-\lambda t}cos\left(2t\right)-\frac{1}{2}{\int }_0^{\mathrm{\infty }}{e}^{-\lambda t}dt$$$$but$$$${\int }_0^{\mathrm{\infty }}{e}^{-\lambda t}dt={[-\frac{1}{\lambda }{e}^{-\lambda t}]}_0^{\mathrm{\infty }}=\frac{1}{\lambda }$$$${\int }_0^{\mathrm{\infty }}{e}^{-\lambda t}cos\left(2t\right)dt=Re\left({\int }_0^{\mathrm{\infty }}{e}^{(-\lambda +2i)t}dt\right)$$$$and{\int }_0^{\mathrm{\infty }}{e}^{(-\lambda +2i)t}dt$$$${=\frac{1}{-\lambda +2i}{e}^{(-\lambda +2i)t}]}_0^{\mathrm{\infty }}$$$$=\frac{-1}{-\lambda +2i}=\frac{1}{\lambda -2i}=\frac{\lambda +2i}{{\lambda }^2+4}$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}{e}^{-\lambda t}cos\left(2t\right)dt=\frac{\lambda }{{\lambda }^2+4}\Rightarrow$$$$f^{{}^{\prime }}\left(\lambda \right)=\frac{\lambda }{2({\lambda }^2+4)}-\frac{1}{2\lambda }\Rightarrow$$$$f\left(\lambda \right)=\frac{1}{4}ln({\lambda }^2+4)-\frac{1}{2}ln\lambda +\sigma$$$$=\frac{1}{2}ln\left(\sqrt{{\lambda }^2+4}\right)-\frac{1}{2}ln\lambda +\sigma$$$$=\frac{1}{2}ln\left(\frac{\sqrt{{\lambda }^2+4}}{\lambda }\right)+\sigma$$$${lim}_{\lambda \to \mathrm{\infty }}f\left(\lambda \right)=0=\sigma$$$${\int }_0^{\mathrm{\infty }}{e}^{-t}\frac{{sin}^{2}t}{t}dt=f\left(1\right)=\frac{1}{2}ln\left(\sqrt{5}\right)$$$$=\frac{ln5}{4}$$$$$$

Commented by Mathspace last updated on 09/Dec/23

$$sorryI=4{\int }_0^{\mathrm{\infty }}\frac{e^{-t}{sin}^{2}t}{t}dt\Rightarrow$$$$★I=\sqrt{5}★$$

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