Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 202393 by sonukgindia last updated on 26/Dec/23

Answered by Mathspace last updated on 26/Dec/23

$$I={\int }_0^{\mathrm{\infty }}\frac{lnx}{a+{bx}^2}dx(a>0,b>0)$$$$I=\frac{1}{a}{\int }_0^{\mathrm{\infty }}\frac{lnx}{1+\frac{b}{a}{x}^2}dx(\sqrt{\frac{b}{a}}x=t)$$$$=\frac{1}{a}{\int }_0^{\mathrm{\infty }}\frac{ln\left(\sqrt{\frac{a}{b}}t\right)}{1+t^2}\times \sqrt{\frac{a}{b}}dt$$$$=\frac{1}{\sqrt{ab}}{\int }_0^{\mathrm{\infty }}\frac{ln\left(\sqrt{\frac{a}{b}}\right)+lnt}{1+t^2}dt$$$$=\frac{1}{2\sqrt{ab}}(lna-lnb){\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^2}$$$$+\frac{1}{\sqrt{ab}}{\int }_0^{\mathrm{\infty }}\frac{lnt}{1+t^2}dtbut{\int }_0^{\mathrm{\infty }}\frac{lnt}{1+t^2}dt=0$$$$\Rightarrow I=\frac{lna-lnb}{2\sqrt{ab}}\times \frac{\pi }{2}$$$$=\frac{\pi }{4\sqrt{ab}}(lna-lnb)$$

Answered by Mathspace last updated on 27/Dec/23

$$J={\int }_0^{\mathrm{\infty }}\frac{{ln}^{2}x}{a+{bx}^2}dx\Rightarrow$$$$J=\frac{1}{a}{\int }_0^{\mathrm{\infty }}\frac{{ln}^{2}x}{1+\frac{b}{a}{x}^2}dx(\sqrt{\frac{b}{a}}x=t)$$$$=\frac{1}{a}{\int }_0^{\mathrm{\infty }}\frac{{ln}^2\left(\sqrt{\frac{a}{b}}t\right)}{1+t^2}\sqrt{\frac{a}{b}}dt$$$$=\frac{1}{\sqrt{ab}}{\int }_0^{\mathrm{\infty }}\frac{(ln{(\sqrt{\frac{a}{b}}+lnt)}^2}{1+t^2}dt$$$$\sqrt{ab}J={\int }_0^{\mathrm{\infty }}\frac{{ln}^2\left(\sqrt{\frac{a}{b}}\right)+2ln\left(\sqrt{\frac{a}{b}}\right)lnt+{ln}^{2}t}{1+t^2}dt$$$$=\frac{\pi }{2}{ln}^2\left(\sqrt{\frac{a}{b}}\right)+2ln\left(\sqrt{\frac{a}{b}}\right){\int }_0^{\mathrm{\infty }}\frac{lnt}{1+t^2}dt(\to 0)$$$$+{\int }_0^{\mathrm{\infty }}\frac{{ln}^{2}t}{1+t^2}dt$$$$=\frac{\pi }{2}{ln}^2\left(\sqrt{\frac{a}{b}}\right)+{\int }_0^{\mathrm{\infty }}\frac{{ln}^{2}t}{1+t^2}dt$$$$K={\int }_0^{\mathrm{\infty }}\frac{{ln}^{2}t}{1+t^2}dt(t=z^{\frac{1}{2}})$$$$K=\frac{1}{8}{\int }_0^{\mathrm{\infty }}\frac{{ln}^{2}z}{1+z}{z}^{\frac{1}{2}-1}dz$$$$$$$$$$

Commented by Mathspace last updated on 27/Dec/23

$$f\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{z^{a-1}}{1+z}dto<a<1$$$$f^{\left(2\right)}\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{t^{a-1}{ln}^{2}z}{1+z}dt$$$$f^{\left(2\right)}\left(\frac{1}{2}\right)={\int }_0^{\mathrm{\infty }}\frac{z^{\frac{1}{2}-1}{ln}^{2}z}{1+z}dz=8K$$$$f\left(a\right)=\frac{\pi }{sin\left(\pi a\right)}\Rightarrow$$$$f^{{}^{\prime }}\left(a\right)=-\pi \frac{\pi cos\left(\pi a\right)}{{sin}^2\left(\pi a\right)}$$$$=-{\pi }^2\frac{cos\left(\pi a\right)}{{sin}^2\left(\pi a\right)}$$$$f^{\left(2\right)}\left(a\right)=-{\pi }^2\frac{-\pi sin\left(\pi a\right){sin}^2\left(\pi a\right)-cos\left(\pi a\right)2\pi sin\left(\pi a\right)cos\left(\pi a\right)}{{sin}^4\left(\pi a\right)}$$$$={\pi }^3\frac{{sin}^2\left(\pi a\right)+2{cos}^2\left(\pi a\right)}{{sin}^3\left(\pi a\right)}$$$$={\pi }^3\frac{1+{cos}^2\left(\pi a\right)}{{sin}^3\left(\pi a\right)}$$$$f^{\left(2\right)}\left(\frac{1}{2}\right)={\pi }^3\times \frac{1+0}{1^3}={\pi }^3\Rightarrow$$$$8K={\pi }^3\Rightarrow K=\frac{{\pi }^3}{8}\Rightarrow$$$$\sqrt{ab}J=\frac{\pi }{2}{ln}^2\left(\sqrt{\frac{a}{b}}\right)+\frac{{\pi }^3}{8}\Rightarrow$$$$J=\frac{\pi }{2\sqrt{ab}}{ln}^2\left(\sqrt{\frac{a}{b}}\right)+\frac{{\pi }^3}{8\sqrt{ab}}$$$$by\left(mathsupbyabdo\right)$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com