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Question Number 51834 by Abdo msup. last updated on 31/Dec/18

$$calculatef\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{ln(1+{at}^2)}{1+t^4}dtwitha>0.$$ $$2)findthevalueof{\int }_0^{\mathrm{\infty }}\frac{ln(3+t^2)}{1+t^4}dt.$$

Commented byAbdo msup. last updated on 31/Dec/18

$$1)wehave{f}^{{}^{\prime }}\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{t^2}{(1+{at}^2)(t^4+1)}dt$$ $$=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{t^2}{({at}^2+1)(t^4+1)}dtlet$$ $$\phi \left(z\right)=\frac{z^2}{({az}^2+1)(z^4+1)}wehave$$ $$\phi \left(z\right)=\frac{z^2}{(\sqrt{a}z-i)(\sqrt{a}z+i)(z^2-i)(z^2+i)}$$ $$=\frac{z^2}{a(z-\frac{i}{\sqrt{a}})(z+\frac{i}{\sqrt{a}})(z-e^{\frac{i\pi }{4}})(z+e^{\frac{i\pi }{4}})(z-e^{-\frac{i\pi }{4}})(z+e^{-\frac{i\pi }{4}})}$$ $$thepolesof\phi are\stackrel{-}{+}\frac{i}{\sqrt{a}}and\stackrel{-}{+}{e}^{\frac{i\pi }{4}}and\stackrel{-}{+}{e}^{-\frac{i\pi }{4}}\left(allsimples\right)$$ $$residustbeoremgive$$ $${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \{Res(\phi ,\frac{i}{\sqrt{a}})+Res(\phi ,e^{\frac{i\pi }{4}})+Res(\phi ,-e^{-\frac{i\pi }{4}})\}$$ $$Res(\phi ,\frac{i}{\sqrt{a}})=\frac{-1}{a^2\frac{2i}{\sqrt{a}}(\frac{1}{a^2}+1)}=\frac{-\sqrt{a}}{2i(1+a^2)}=\frac{i\sqrt{a}}{2(1+a^2)}$$ $$Res(\phi ,e^{\frac{i\pi }{4}})=\frac{i}{(ai+1)\left(2e^{\frac{i\pi }{4}}\right)\left(2i\right)}=\frac{e^{-\frac{i\pi }{4}}}{4(1+ai)}$$ $$Res(\phi ,-e^{-\frac{i\pi }{4}})=\frac{-i}{(1-ai)(-2e^{-\frac{i\pi }{4}})(-2i)}$$ $$=-\frac{e^{\frac{i\pi }{4}}}{4(1-ai)}\Rightarrow$$ $${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \{\frac{i\sqrt{a}}{2(1+a^2)}-\frac{1}{4}\left(2iIm\left(\frac{e^{\frac{i\pi }{4}}}{1-ai}\right)\right\}$$ $$=\frac{-\pi \sqrt{a}}{1+a^2}-\pi Im\left(\frac{e^{\frac{i\pi }{4}}}{1-ai}\right)but$$ $$\frac{e^{\frac{i\pi }{4}}}{1-ai}=\frac{e^{\frac{i\pi }{4}}(1+ai)}{1+a^2}=\frac{(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})(1+ai)}{1+a^2}$$ $$=\frac{1}{\sqrt{2}}\frac{(1+i)(1+ai)}{1+a^2}=\frac{1+ai+i-a}{\sqrt{2}(1+a^2)}=\frac{1-a+i(1+a)}{\sqrt{2}(1+a^2)}\Rightarrow$$ $${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=\frac{-\pi \sqrt{a}}{1+a^2}-\pi \frac{1+a}{\sqrt{2}(1+a^2)}\Rightarrow$$ $$f^{{}^{\prime }}\left(a\right)=\frac{-\pi \sqrt{a}}{2(1+a^2)}-\frac{\pi (1+a)}{2\sqrt{2}(1+a^2)}\Rightarrow$$ $$f\left(a\right)=-\frac{\pi }{2}\int \frac{\sqrt{a}da}{1+a^2}-\frac{\pi }{2\sqrt{2}}\int \frac{1+a}{1+a^2}da+c$$

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