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Question Number 52459 by Abdo msup. last updated on 08/Jan/19

$$letf\left(\alpha \right)={\int }_0^1\frac{ln(1+i\alpha x)}{1+x^2}dx$$$$1)determineaexplicitformoff\left(\alpha \right)$$$$2)calculate{\int }_0^1\frac{ln(1+ix)}{1+x^2}dxand{\int }_0^1\frac{ln(1+2ix)}{1+x^2}dx.$$

Commented by maxmathsup by imad last updated on 08/Jan/19

$$1)wehave{f}^{{}^{\prime }}\left(\alpha \right)={\int }_0^1\frac{ix}{(1+i\alpha x)(1+x^2)}dxsofor\alpha \ne 0$$$$f^{{}^{\prime }}\left(\alpha \right)=\frac{1}{\alpha }{\int }_0^1\frac{1+i\alpha x-1}{(1+i\alpha x)(x^2+1)}dx=\frac{1}{\alpha }-\frac{1}{\alpha }{\int }_0^1\frac{dx}{(i\alpha x+1)(x^2+1)}dxlet$$$$decomposeF\left(x\right)=\frac{1}{(i\alpha x+1)(x^2+1)}\Rightarrow$$$$F\left(x\right)=\frac{a}{i\alpha x+1}+\frac{bx+c}{x^2+1}$$$$a={lim}_{x\to \frac{-1}{i\alpha }}(i\alpha x+1)F\left(x\right)=\frac{1}{({\left(\frac{1}{i\alpha }\right)}^2+1)}=\frac{1}{\frac{1}{-{\alpha }^2}+1}=-\frac{1}{1-{\alpha }^2}{\alpha }^2=\frac{{\alpha }^2}{{\alpha }^2-1}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=\frac{a}{i\alpha }+b\Rightarrow b=-\frac{a}{i\alpha }=\frac{ia}{\alpha }\Rightarrow$$$$F\left(x\right)=\frac{{\alpha }^2}{{\alpha }^2-1}\frac{1}{i\alpha x+1}+\frac{\frac{ia}{\alpha }x+c}{x^2+1}(wesuppose\alpha \ne \stackrel{-}{+}1)$$$$F\left(o\right)=1=\frac{{\alpha }^2}{{\alpha }^2-1}+c\Rightarrow c=1-\frac{{\alpha }^2}{{\alpha }^2-1}=\frac{-1}{{\alpha }^2-1}\Rightarrow$$$$F\left(x\right)=\frac{{\alpha }^2}{({\alpha }^2-1)(1+i\alpha x)}+\frac{\frac{i}{\alpha }\frac{{\alpha }^2}{{\alpha }^2-1}x-\frac{1}{{\alpha }^2-1}}{x^2+1}$$$$=\frac{{\alpha }^2}{({\alpha }^2-1)(1+i\alpha x)}+\frac{1}{{\alpha }^2-1}\frac{i\alpha x-1}{x^2+1}\Rightarrow$$$$f^{{}^{\prime }}\left(\alpha \right)=\frac{1}{\alpha }-\frac{1}{\alpha }{\int }_0^1\frac{{\alpha }^2}{({\alpha }^2-1)(1+i\alpha x)}dx-\frac{1}{\alpha ({\alpha }^2-1)}{\int }_0^1\frac{i\alpha x-1}{x^2+1}dx$$$$=\frac{1}{\alpha }-\frac{\alpha }{{\alpha }^2-1}\frac{1}{i\alpha }{\int }_0^1\frac{i\alpha dx}{1+i\alpha x}dx-\frac{i}{2({\alpha }^2-1)}{\left[ln(x^2+1)\right]}_0^1+\frac{\pi }{4\alpha ({\alpha }^2-1)}$$$$=\frac{1}{\alpha }+\frac{i}{{\alpha }^2-1}ln(1+i\alpha )-\frac{iln\left(2\right)}{2({\alpha }^2-1)}+\frac{\pi }{4\alpha ({\alpha }^2-1)}\Rightarrow$$$$f\left(\alpha \right)=ln\mid \alpha \mid +i\int \frac{ln(1+i\alpha )}{{\alpha }^2-1}d\alpha +i\frac{ln\left(2\right)}{2}\int \frac{d\alpha }{{\alpha }^2-1}+\frac{\pi }{4}\int \frac{d\alpha }{\alpha ({\alpha }^2-1)}+cbut$$$$\int \frac{d\alpha }{{\alpha }^2-1}=\frac{1}{2}\int (\frac{1}{\alpha -1}-\frac{1}{\alpha +1})d\alpha =\frac{1}{2}ln\mid \frac{\alpha -1}{\alpha +1}\mid letdecompose$$$$w\left(\alpha \right)=\frac{1}{\alpha ({\alpha }^2-1)}=\frac{1}{\alpha (\alpha -1)(\alpha +1)}$$$$w\left(\alpha \right)=\frac{a}{\alpha }+\frac{b}{\alpha -1}+\frac{c}{\alpha +1}\Rightarrow a=-1$$$$b=\frac{1}{2}andc=\frac{1}{2}\Rightarrow w\left(\alpha \right)=-\frac{1}{\alpha }+\frac{1}{2(\alpha -1)}+\frac{1}{2(\alpha +1)}\Rightarrow$$$$\int \frac{d\alpha }{\alpha ({\alpha }^2-1)}=-ln\mid \alpha \mid +\frac{1}{2}ln\mid {\alpha }^2-1\mid \Rightarrow$$$$f\left(\alpha \right)=i\int \frac{ln(1+i\alpha )}{{\alpha }^2-1}d\alpha +\frac{iln\left(2\right)}{4}ln\mid \frac{\alpha -1}{\alpha +1}\mid +(1-\frac{\pi }{4})ln\mid \alpha \mid +\frac{\pi }{8}ln\mid {\alpha }^2-1\mid +c$$$$$$

Commented by Abdo msup. last updated on 08/Jan/19

$$2)wehave1+ix=\sqrt{1+x^2}(\frac{1}{\sqrt{1+x^2}}+i\frac{x}{\sqrt{1+x^2}})=r{e}^{i\theta }\Rightarrow$$$$r=\sqrt{1+x^2}and\theta =arctanx\Rightarrow$$$$ln(1+ix)=\frac{1}{2}ln(1+x^2)+iarctanx\Rightarrow$$$${\int }_0^1\frac{ln(1+ix)}{1+x^2}=\frac{1}{2}{\int }_0^1\frac{ln(1+x^2)}{1+x^2}dx+i{\int }_0^1\frac{arctanx}{1+x^2}dx$$$$bypartsI={\int }_0^1\frac{arctanx}{1+x^2}dx={\left[{arctan}^{2}x\right]}_0^1$$$$-{\int }_0^1\frac{arctanx}{1+x^2}=\frac{{\pi }^2}{16}-I\Rightarrow 2I=\frac{{\pi }^2}{16}\Rightarrow I=\frac{{\pi }^2}{32}$$$$letfind{\int }_0^1\frac{ln(1+x^2)}{1+x^2}dxlet$$$$f\left(t\right)={\int }_0^1\frac{ln(1+{tx}^2)}{1+x^2}dxwitht>0wehave$$$$f^{{}^{\prime }}\left(t\right)={\int }_0^1\frac{x^2}{(1+{tx}^2)(1+x^2)}dx$$$${=}_{\sqrt{t}x=u}{\int }_0^{\sqrt{t}}\frac{u^2}{t(1+u^2)(1+\frac{u^2}{t})}\frac{du}{\sqrt{t}}$$$$=\frac{1}{\sqrt{t}}{\int }_0^{\sqrt{t}}\frac{u^2}{(u^2+1)(u^2+t)}duletdecompose$$$$F\left(u\right)=\frac{u^2}{(u^2+1)(u^2+t)}\Rightarrow$$$$F\left(u\right)=\frac{au+b}{u^2+1}+\frac{cu+d}{u^2+t}$$$$F(-u)=F\left(u\right)\Rightarrow \frac{-au+b}{u^2+1}+\frac{-cu+d}{u^2+t}$$$$=F\left(u\right)\Rightarrow a=0andc=0\Rightarrow$$$$F\left(u\right)=\frac{b}{u^2+1}+\frac{d}{u^2+t}weseethat$$$$F\left(u\right)=\frac{1}{t-1}(\frac{1}{u^2+1}-\frac{1}{u^2+t})\Rightarrow$$$${\int }_0^{\sqrt{t}}F\left(u\right)du=\frac{1}{t-1}{\int }_0^{\sqrt{t}}\frac{du}{u^2+1}-\frac{1}{t-1}{\int }_0^{\sqrt{t}}\frac{du}{u^2+t}$$$$but{\int }_0^{\sqrt{t}}\frac{du}{1+u^2}=arctan\left(\sqrt{t}\right)$$$${\int }_0^{\sqrt{t}}\frac{du}{u^2+t}{=}_{u=\sqrt{t}\alpha }{\int }_0^1\frac{\sqrt{t}d\alpha }{t(1+{\alpha }^2)}=\frac{1}{\sqrt{t}}\frac{\pi }{4}\Rightarrow$$$$f^{{}^{\prime }}\left(t\right)=\frac{1}{\sqrt{t}}\{\frac{arctan\left(\sqrt{t}\right)}{t-1}-\frac{\pi }{4\sqrt{t}(t-1)}\}\Rightarrow$$$$f\left(t\right)=\int \frac{arctan\left(\sqrt{t}\right)}{\sqrt{t}(t-1)}dy-\frac{\pi }{4}\int \frac{dt}{t(t-1)}+C$$$$\int \frac{dt}{t(t-1)}=\int (\frac{1}{t-1}-\frac{1}{t})dt=ln\mid \frac{t-1}{t}\mid +c_1$$$$\int \frac{arctan\left(\sqrt{t}\right)}{\sqrt{t}(t-1)}dt{=}_{\sqrt{t}=u}\int \frac{arctan\left(u\right)}{u(u^2-1)}\left(2u\right)du$$$$=2\int \frac{arctan\left(u\right)}{u^2-1}du....becontinued...$$$$$$$$$$$$$$

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