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Question Number 79627 by mathmax by abdo last updated on 26/Jan/20

$$1)expicitef\left(x\right)={\int }_0^1\frac{ln(1+{xt}^2)}{1+t^2}dtwithx⩾0$$$$2)calculate{\int }_0^1\frac{ln(1+t^2)}{1+t^2}dtand{\int }_0^1\frac{ln(1+2t^2)}{1+t^2}dt$$

Commented by mathmax by abdo last updated on 30/Jan/20

$$1)wehave{f}^{{}^{\prime }}\left(x\right)={\int }_0^1\frac{t^2}{(1+{xt}^2)(t^2+1)}dt$$$$={\int }_0^1\frac{t^2+1-1}{(t^2+1)({xt}^2+1)}dt={\int }_0^1\frac{dt}{{xt}^2+1}-{\int }_0^1\frac{dt}{(t^2+1)({xt}^2+1)}$$$${\int }_0^1\frac{dt}{{xt}^2+1}{=}_{t\sqrt{x}=u}{\int }_0^{\sqrt{x}}\frac{du}{\sqrt{x}(1+u^2)}=\frac{1}{\sqrt{x}}arctan\left(\sqrt{x}\right)and$$$$letdecomposeF\left(t\right)=\frac{1}{(t^2+1)({xt}^2+1)}$$$$F\left(t\right)=\frac{at+b}{t^2+1}+\frac{ct+d}{{xt}^2+1}wehaveF(-t)=F\left(t\right)\Rightarrow$$$$\frac{-at+b}{t^2+1}+\frac{-ct+d}{{xt}^2+1}=F\left(t\right)\Rightarrow a=c=0\Rightarrow F\left(t\right)=\frac{b}{t^2+1}+\frac{d}{{xt}^2+1}$$$${lim}_{t\to +\mathrm{\infty }}{t}^{2}F\left(t\right)=0=b+\frac{d}{x}\Rightarrow d+bx=0\Rightarrow d=-bx$$$$F\left(0\right)=1=b+d\Rightarrow 1=b-bx=(1-x)b\Rightarrow b=\frac{1}{1-x}andd=\frac{-x}{1-x}\Rightarrow$$$$F\left(t\right)=\frac{1}{(1-x)(t^2+1)}-\frac{x}{(1-x)({xt}^2+1)}(x\ne 1)\Rightarrow$$$${\int }_0^{1}F\left(t\right)dt=\frac{1}{(1-x)}{\int }_0^1\frac{dt}{t^2+1}-\frac{x}{1-x}{\int }_0^1\frac{dt}{{xt}^2+1}$$$$=\frac{\pi }{4(1-x)}-\frac{x}{1-x}\times \frac{1}{\sqrt{x}}arctan\left(\sqrt{x}\right)\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)=\frac{1}{\sqrt{x}}arctan\left(\sqrt{x}\right)-\frac{\pi }{4(1-x)}+\frac{\sqrt{x}}{1-x}arctan\left(\sqrt{x}\right)$$$$=(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{1-x})arctan\left(\sqrt{x}\right)-\frac{\pi }{4(1-x)}$$$$=\frac{1}{(1-x)\sqrt{x}}arctan\left(\sqrt{x}\right)-\frac{\pi }{4(1-x)}\Rightarrow$$$$f\left(x\right)={\int }_0^x\frac{arctan\left(\sqrt{u}\right)}{(1-u)\sqrt{u}}du-\frac{\pi }{4}ln(\mid 1-x\mid )+c$$$$c=f\left(0\right)=0\Rightarrow f\left(x\right)={\int }_0^x\frac{arctan\left(\sqrt{u}\right)}{(1-u)\sqrt{u}}du-\frac{\pi }{4}ln\mid 1-x\mid$$$$\sqrt{u}=z\Rightarrow {\int }_0^x\frac{arctan\left(\sqrt{u}\right)}{(1-u)\sqrt{u}}du={\int }_0^{\sqrt{x}}\frac{arctan\left(z\right)}{(1-z^2)z}\left(2z\right)dz$$$$=2{\int }_0^{\sqrt{x}}\frac{arctan\left(z\right)}{1-z^2}dz\Rightarrow f\left(x\right)=2{\int }_0^{\sqrt{x}}\frac{arctan\left(z\right)}{1-z^2}dz-\frac{\pi }{4}ln\mid 1-x\mid (x\ne 1)$$$$$$

Commented by mathmax by abdo last updated on 30/Jan/20

$$2)letA={\int }_0^1\frac{ln(1+t^2)}{1+t^2}dt\Rightarrow A={\int }_0^1\frac{ln(1+it)+ln(1-it)}{1+t^2}dt$$$$={\int }_0^1\frac{ln(1+it)}{1+t^2}+conj\left({\int }_0^1\frac{ln(1+it)}{1+t^2}dt\right)=2Re\left({\int }_0^1\frac{ln(1+it)}{1+t^2}dt\right)$$$$letf\left(z\right)={\int }_0^1\frac{ln(1+zt)}{1+t^2}dtwehave{f}^{{}^{\prime }}\left(z\right)={\int }_0^1\frac{t}{(1+zt)(t^2+1)}dt$$$$decomposittionofF\left(t\right)=\frac{t}{(1+zt)(t^2+1)}$$$$=\frac{a}{zt+1}+\frac{bt+c}{t^2+1}$$$$a=(zt+1)F\left(t\right){\mid }_{t=-\frac{1}{z}}=-\frac{1}{z(\frac{1}{z^2}+1)}=-\frac{1}{z\left(\frac{1+z^2}{z^2}\right)}=-\frac{z}{1+z^2}$$$${lim}_{t\to +\mathrm{\infty }}tF\left(t\right)=0=\frac{a}{z}+b\Rightarrow b=-\frac{a}{z}=\frac{1}{1+z^2}$$$$F\left(0\right)=0=a+c\Rightarrow c=-a=\frac{z}{1+z^2}\Rightarrow$$$$F\left(t\right)=-\frac{z}{(1+z^2)(zt+1)}+\frac{1}{1+z^2}\times \frac{t+z}{t^2+1}\Rightarrow$$$$f^{{}^{\prime }}\left(z\right)={\int }_0^{1}F\left(t\right)dt=-\frac{z}{1+z^2}{\int }_0^1\frac{dt}{zt+1}+\frac{1}{2(1+z^2)}{\int }_0^1\frac{2t}{t^2+1}dt+\frac{z}{1+z^2}{\int }_0^1\frac{dt}{1+t^2}$$$$=-\frac{1}{1+z^2}ln(z+1)+\frac{1}{2(1+z^2)}ln\left(2\right)+\frac{z}{1+z^2}\times \frac{\pi }{4}\Rightarrow$$$$f\left(z\right)=-{\int }_0^z\frac{ln(1+u)}{u^2+1}du+\frac{ln\left(2\right)}{2}{\int }_0^z\frac{du}{1+u^2}+\frac{\pi }{8}{\int }_0^z\frac{2u}{1+u^2}du+c$$$$=\frac{ln\left(2\right)}{2}arctan\left(z\right)+\frac{\pi }{8}ln(1+z^2)-{\int }_0^z\frac{ln(1+u)}{u^2+1}du(c=f\left(0\right)=0)$$$$f\left(1\right)={\int }_0^1\frac{ln(1+t)}{1+t^2}dt=\frac{ln\left(2\right)}{2}\times \frac{\pi }{4}+\frac{\pi }{8}ln\left(2\right)-{\int }_0^1\frac{ln(1+u)}{u^2+1}du\Rightarrow$$$$2{\int }_0^1\frac{ln(1+u)}{1+u^2}du=\frac{\pi }{4}ln\left(2\right)\Rightarrow {\int }_0^1\frac{ln(1+u)}{1+u^2}du=\frac{\pi }{8}ln\left(2\right)$$$$theexplicitformoff\left(x\right)dontgivef\left(i\right)..solettryanotherway$$$$....becontinued...$$$$$$

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