Question Number 68600 by Abdo msup. last updated on 14/Sep/19
$$\left.\mathrm{1}\right){calculatef}\left({a}\right)=\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({arctanx}\right)}{{a}+{x}^{\mathrm{2}} }{dx}\:{with}\:{a}>\mathrm{0} \\ $$ $$\left.\mathrm{2}\right)?{calculste}\:\:{g}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left({arctanx}\right)}{\left({a}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$ $$\left.\mathrm{3}\right){find}\:{the}\:{value}\:{if}\:{integrals} \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({arctanx}\right)}{\mathrm{2}+{x}^{\mathrm{2}} }\:{and}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({arctanx}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$
Commented bymathmax by abdo last updated on 14/Sep/19
$$\left.\mathrm{1}\right)\:{we}\:{have}\:\mathrm{2}{f}\left({a}\right)=\int_{−\infty} ^{+\infty} \:\frac{{cos}\left({arctanx}\right)}{{x}^{\mathrm{2}} \:+{a}}{dx}\:={Re}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{iarctan}\left({x}\right)} }{{x}^{\mathrm{2}} \:+{a}}{dx}\right) \\ $$ $${let}\:\varphi\left({z}\right)\:=\frac{{e}^{{i}\:{arctan}\left({z}\right)} }{{z}^{\mathrm{2}} +{a}}\:\Rightarrow\varphi\left({z}\right)\:=\frac{{e}^{{iarctan}\left({z}\right)} }{\left({z}−{i}\sqrt{{a}}\right)\left({z}+{i}\sqrt{{a}}\right)}\:{residus}\:{theorem} \\ $$ $${give}\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi{Res}\left(\varphi,{i}\sqrt{{a}}\right) \\ $$ $${Res}\left(\varphi,{i}\sqrt{{a}}\right)\:=\frac{{e}^{{iarctan}\left({i}\sqrt{{a}}\right)} }{\mathrm{2}{i}\sqrt{{a}}}\:\Rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi×\frac{{e}^{{iarctan}\left({iz}\right)} }{\mathrm{2}{i}\sqrt{{a}}} \\ $$ $$=\frac{\pi}{\sqrt{{a}}}\:\left\{{cos}\left({arctan}\left({iz}\right)\right)+{isin}\left({artan}\left({iz}\right)\right)\right\} \\ $$ $${rest}\:{to}\:{find}\:{arctan}\left({iz}\right)\:\:.....{be}\:{continued}.... \\ $$