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Question Number 31502 by abdo imad last updated on 09/Mar/18

$$findf\left(x\right)={\int }_0^{1}ln(1+{xt}^2)dtwithx>0.$$ $$2)givethevalueof{\int }_0^{1}ln(1+t^2)dtand{\int }_0^{1}ln(1+2t^2)dt.$$

Commented byabdo imad last updated on 15/Mar/18

$$wehave{f}^{{}^{\prime }}\left(x\right)={\int }_0^1\frac{t^2}{1+{xt}^2}dt=\frac{1}{x}{\int }_0^1\frac{{xt}^2}{1+{xt}^2}dt$$ $$=\frac{1}{x}{\int }_0^1\frac{1+{xt}^2-1}{1+{xt}^2}dt=\frac{1}{x}-\frac{1}{x}{\int }_0^1\frac{dt}{1+{xt}^2}ch.\sqrt{x}t=ugive$$ $${\int }_0^1\frac{dt}{1+{xt}^2}={\int }_0^{\sqrt{x}}\frac{1}{1+u^2}\frac{du}{\sqrt{x}}=\frac{1}{\sqrt{x}}{\int }_0^{\sqrt{x}}\frac{du}{1+u^2}$$ $$=\frac{1}{\sqrt{x}}artan\left(\sqrt{x}\right)\Rightarrow f^{{}^{\prime }}\left(x\right)=\frac{1}{x}-\frac{arctan(\sqrt{x)}}{x\sqrt{x}}\Rightarrow$$ $$f\left(x\right)=ln\mid x\mid -{\int }_1^x\frac{arctan\left(\sqrt{t}\right)}{t\sqrt{t}}dt+\lambda$$ $$\lambda =f\left(1\right)={\int }_0^{1}ln(1+t^2)dtso$$ $$f\left(x\right)=ln\left(x\right)-{\int }_1^x\frac{arctan(\sqrt{t)}}{t\sqrt{t}}dt+{\int }_0^{1}ln(1+t^2)dtch.\sqrt{t}=u$$ $$give{\int }_1^x\frac{arctan\left(\sqrt{t}\right)}{t\sqrt{t}}dt={\int }_1^{\sqrt{x}}\frac{arctanu}{u^3}\left(2u\right)du$$ $$=2{\int }_1^{\sqrt{x}}\frac{arctanu}{u^2}du$$ $$=2({[-\frac{1}{u}arctanu]}_1^{\sqrt{x}}+{\int }_1^{\sqrt{x}}\frac{du}{u(1+u^2)})$$ $$=2(\frac{\pi }{4}-\frac{arctan\left(\sqrt{x}\right)}{\sqrt{x}})+2{\int }_1^{\sqrt{x}}\frac{du}{u(1+u^2)}but$$ $${\int }_1^{\sqrt{x}}\frac{du}{u(1+u^2)}={\int }_1^{\sqrt{x}}(\frac{1}{u}-\frac{u}{1+u^2})du$$ $$={[ln\left(u\right)-\frac{1}{2}ln(1+u^2)]}_1^{\sqrt{x}}={\left[ln\left(\frac{u}{\sqrt{1+u^2}}\right)\right]}_1^{\sqrt{x}}$$ $$=ln\left(\frac{\sqrt{x}}{\sqrt{1+x}}\right)-ln\left(\frac{1}{\sqrt{2}}\right)=ln\left(\frac{\sqrt{x}}{\sqrt{1+x}}\right)+ln(\sqrt{2)}finally$$ $$f\left(x\right)=ln\left(x\right)-\frac{\pi }{2}+\frac{arctan\left(\sqrt{x}\right)}{2\sqrt{x}}-2ln\left(\frac{\sqrt{x}}{\sqrt{1+x}}\right)-ln\left(2\right)$$ $$+{\int }_0^{1}ln(1+t^2)dt.$$ $$$$

Commented byabdo imad last updated on 15/Mar/18

$$2)letputI={\int }_0^{1}ln(1+t^2)dtwehavefor\mid x\mid <1$$ $${ln}^{{}^{\prime }}(1+x)=\frac{1}{1+x}=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{x}^n\Rightarrow$$ $$ln(1+x)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+1}{x}^{n+1}=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n-1}\frac{x^n}{n}\Rightarrow$$ $$I={\int }_0^1\left(\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n-1}\frac{t^{2n}}{n}\right)dt$$ $$=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}\frac{1}{2n+1}=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n(2n+1)}$$ $$\frac{1}{2}I=\sum _{n=1}^{\mathrm{\infty }}(\frac{1}{2n}-\frac{1}{2n+1}){(-1)}^{n-1}$$ $$=\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}+\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{2n+1}$$ $$=\frac{1}{2}ln\left(2\right)+\frac{\pi }{4}-1\Rightarrow I=ln\left(2\right)+\frac{\pi }{2}-2$$ $$$$

Commented byabdo imad last updated on 15/Mar/18

$$wehaveprovedthat{\int }_0^{1}ln(1+t^2)dt=ln\left(2\right)+\frac{\pi }{2}-2\Rightarrow$$ $$f\left(x\right)={\int }_0^{1}ln(1+{xt}^2)dt=ln\left(x\right)-2+\frac{actan\left(\sqrt{x}\right)}{2\sqrt{x}}-2ln\left(\frac{\sqrt{x}}{\sqrt{1+x}}\right)$$ $$lettakex=2weobtain$$ $${\int }_0^{1}ln(1+2t^2)dt=ln\left(2\right)+\frac{arctan\left(\sqrt{2}\right)}{2\sqrt{2}}-2ln\left(\frac{\sqrt{2}}{\sqrt{3}}\right).$$ $$$$

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