Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 41762 by math khazana by abdo last updated on 12/Aug/18

$$letf\left(x\right)={\int }_0^1\frac{ln(1+{xt}^2)}{1+t^2}dt$$$$1)findasimpleformoff\left(x\right)$$$$2)calculate{\int }_0^1\frac{ln(1+t^2)}{1+t^2}dt$$$$3)calculate{\int }_0^1\frac{ln(1+2t^2)}{1+t^2}dt$$

Commented by math khazana by abdo last updated on 13/Aug/18

$$wehave{f}^{{}^{\prime }}\left(x\right)={\int }_0^1\frac{t^2}{(1+{xt}^2)(1+t^2)}dt$$$$letdecomposeF\left(t\right)=\frac{t^2}{(1+{xt}^2)(1+t^2)}$$$$F\left(t\right)=\frac{at+b}{t^2+1}+\frac{ct+d}{{xt}^2+1}$$$$F(-t)=F\left(t\right)\Rightarrow \frac{-at+b}{t^2+1}+\frac{-ct+d}{{xt}^2+1}=F\left(t\right)\Rightarrow$$$$a=c=0\Rightarrow F\left(t\right)=\frac{b}{t^2+1}+\frac{d}{{xt}^2+1}$$$${lim}_{t\to +\mathrm{\infty }}{t}^{2}F\left(t\right)=\frac{1}{x}=b+\frac{d}{x}\Rightarrow 1=bx+d\Rightarrow$$$$d=1-bx\Rightarrow F\left(t\right)=\frac{b}{t^2+1}+\frac{1-bx}{{xt}^2+1}$$$$F\left(0\right)=0=b+1-bx=(1-x)b+1\Rightarrow b=\frac{1}{x-1}$$$$d=1-\frac{x}{x-1}=\frac{-1}{x-1}\Rightarrow$$$$F\left(t\right)=\frac{1}{(x-1)(t^2+1)}-\frac{1}{(x-1)({xt}^2+1)}\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)=\frac{1}{x-1}{\int }_0^1\frac{dt}{1+t^2}-\frac{1}{x-1}{\int }_0^1\frac{dx}{1+{xt}^2}$$$$=\frac{\pi }{4(x-1)}-\frac{1}{x-1}{\int }_0^1\frac{dt}{1+{xt}^2}$$$$case1x>0changementt\sqrt{x}=ugive$$$${\int }_0^1\frac{dt}{1+{xt}^2}={\int }_0^{\sqrt{x}}\frac{1}{1+u^2}\frac{du}{\sqrt{x}}=\frac{1}{\sqrt{x}}arctan\left(\sqrt{x}\right)\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)=\frac{\pi }{4(x-1)}-\frac{arctan\left(\sqrt{x}\right)}{(x-1)\sqrt{x}}\Rightarrow$$$$f\left(x\right)=\frac{\pi }{4}{\int }_0^x\frac{dt}{t-1}-{\int }_0^x\frac{arctan\left(\sqrt{t}\right)}{(t-1)\sqrt{t}}+c$$$$c=f\left(0\right)=0\Rightarrow f\left(x\right)=\frac{\pi }{4}ln\mid x-1\mid -{\int }_0^x\frac{arctan\left(\sqrt{t}\right)}{(t-1)\sqrt{t}}dt$$$$but{\int }_0^x\frac{arctan\left(\sqrt{t}\right)}{(t-1)\sqrt{t}}dt{=}_{\sqrt{t}=u}2{\int }_0^{\sqrt{x}}\frac{arctan\left(u\right)}{(u^2-1)u}udu$$$$=2{\int }_0^{\sqrt{x}}\frac{arctan\left(u\right)}{u^2-1}du\Rightarrow$$$$f\left(x\right)=\frac{\pi }{4}ln\mid x-1\mid +2{\int }_0^{\sqrt{x}}\frac{arctanu}{1-u^2}du$$

Commented by math khazana by abdo last updated on 13/Aug/18

$$case2x<0{\int }_0^1\frac{dx}{1+{xt}^2}={\int }_0^1\frac{dt}{1-{\left(\sqrt{-x}t\right)}^2}$$$${=}_{\sqrt{-x}t=u}{\int }_0^{\sqrt{-x}}\frac{1}{1-u^2}\frac{du}{\sqrt{-x}}$$$$=\frac{1}{\sqrt{-x}}{\int }_0^{\sqrt{-x}}\frac{du}{1-u^2}=\frac{1}{2\sqrt{-x}}{\int }_0^{\sqrt{-x}}(\frac{1}{1-u}+\frac{1}{1+u})du$$$$=\frac{1}{2\sqrt{-x}}{[ln\mid \frac{1+u}{1-u}\mid ]}_0^{\sqrt{-x}}=\frac{1}{2\sqrt{-x}}ln\mid \frac{1+\sqrt{-x}}{1-\sqrt{-x}}\mid \Rightarrow$$$$f^{{}^{\prime }}\left(x\right)=\frac{\pi }{4(x-1)}-\frac{1}{2(x-1)\sqrt{-x}}ln\mid \frac{1+\sqrt{-x}}{1-\sqrt{-x}}\mid \Rightarrow$$$$f\left(x\right)=\frac{\pi }{4}ln\mid x-1\mid -\int \frac{1}{2(x-1)\sqrt{-x}}ln\mid \frac{1+\sqrt{-x}}{1-\sqrt{-x}}\mid dx+c$$$$$$

Commented by math khazana by abdo last updated on 13/Aug/18

$$2)letA={\int }_0^1\frac{ln(1+t^2)}{1+t^2}dt$$$$A{=}_{t=tan\theta }{\int }_0^{\frac{\pi }{4}}\frac{ln(1+{tan}^2\theta )}{1+{tan}^2\theta }(1+{tan}^2\theta )d\theta$$$$={\int }_0^{\frac{\pi }{4}}ln\left(\frac{1}{{cos}^2\theta }\right)d\theta =-2{\int }_0^{\frac{\pi }{4}}ln\left(cos\theta \right)d\theta$$$$letH={\int }_0^{\frac{\pi }{4}}ln\left(cos\theta \right)d\theta andK={\int }_0^{\frac{\pi }{4}}ln\left(sin\theta \right)d\theta$$$$wehaveH{=}_{\theta =\frac{\pi }{2}-t}-{\int }_{\frac{\pi }{2}}^{\frac{\pi }{4}}ln\left(sint\right)dt$$$$={\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}ln\left(sint\right)dt=-{\int }_0^{\frac{\pi }{4}}ln\left(sint\right)dt+{\int }_0^{\frac{\pi }{2}}ln\left(sint\right)dt$$$$=-K-\frac{\pi }{2}ln\left(2\right)\Rightarrow H+K=-\frac{\pi }{2}ln\left(2\right)$$$$K-H={\int }_0^{\frac{\pi }{4}}ln\left(sin\theta \right)d\theta -{\int }_0^{\frac{\pi }{4}}ln\left(cos\theta \right)d\theta$$$$={\int }_0^{\frac{\pi }{4}}ln\left(tan\theta \right)d\theta {=}_{tan\theta =u}{\int }_0^1\frac{ln\left(u\right)}{1+u^2}du$$$$={\int }_0^1\left(\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{u}^{2n}\right)ln\left(u\right)du$$$$=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\int }_0^1{u}^{2n}ln\left(u\right)dubyparts$$$$A_n={\int }_0^1{u}^{2n}ln\left(u\right)du={\left[\frac{1}{2n+1}{u}^{2n+1}ln\left(u\right)\right]}_0^1$$$$-{\int }_0^1\frac{1}{(2n+1)}{u}^{2n}du=-\frac{1}{{(2n+1)}^2}\Rightarrow$$$$K-H=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^{n+1}}{{(2n+1)}^2}={\lambda }_0\Rightarrow$$$$2K=-\frac{\pi }{2}ln\left(2\right)+{\lambda }_0\Rightarrow K=-\frac{\pi }{4}ln\left(2\right)+\frac{{\lambda }_0}{2}$$$$H=-\frac{\pi }{2}ln\left(2\right)-K=-\frac{\pi }{4}ln\left(2\right)-\frac{{\lambda }_0}{2}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com