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Question Number 68600 by Abdo msup. last updated on 14/Sep/19

$$1)calculatef\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{cos\left(arctanx\right)}{a+x^2}dxwitha>0$$ $$2)?calculsteg\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{cos\left(arctanx\right)}{{(a+x^2)}^2}$$ $$3)findthevalueifintegrals$$ $${\int }_0^{\mathrm{\infty }}\frac{cos\left(arctanx\right)}{2+x^2}and{\int }_0^{\mathrm{\infty }}\frac{cos\left(arctanx\right)}{{(1+x^2)}^2}$$

Commented bymathmax by abdo last updated on 14/Sep/19

$$1)wehave2f\left(a\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(arctanx\right)}{x^2+a}dx=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{iarctan\left(x\right)}}{x^2+a}dx\right)$$ $$let\phi \left(z\right)=\frac{e^{iarctan\left(z\right)}}{z^2+a}\Rightarrow \phi \left(z\right)=\frac{e^{iarctan\left(z\right)}}{(z-i\sqrt{a})(z+i\sqrt{a})}residustheorem$$ $$give{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,i\sqrt{a})$$ $$Res(\phi ,i\sqrt{a})=\frac{e^{iarctan\left(i\sqrt{a}\right)}}{2i\sqrt{a}}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \times \frac{e^{iarctan\left(iz\right)}}{2i\sqrt{a}}$$ $$=\frac{\pi }{\sqrt{a}}\{cos\left(arctan\left(iz\right)\right)+isin\left(artan\left(iz\right)\right)\}$$ $$resttofindarctan\left(iz\right).....becontinued....$$

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