Question Number 66466 by mathmax by abdo last updated on 15/Aug/19
$${find}\:\:{f}\left({a},{b}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{cos}\left({ax}\right){cos}\left({bx}\right)}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right)}{dx}\:\:{with}\:{a}>\mathrm{0}\:{and}\:{b}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left({x}\right){cos}\left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)}{dx} \\ $$
Commented by mathmax by abdo last updated on 17/Aug/19
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({a},{b}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({a}+{b}\right){x}+{cos}\left({a}−{b}\right){x}}{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} \:+{b}^{\left.\mathrm{2}\right)} \right.}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{−\infty} ^{+\infty} \:\frac{{cos}\left({a}+{b}\right){x}}{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right)}{dx}\:+\frac{\mathrm{1}}{\mathrm{4}}\int_{−\infty} ^{+\infty} \:\frac{{cos}\left({a}−{b}\right){x}}{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right)}{dx}\:\Rightarrow \\ $$$$\mathrm{4}{f}\left(,{b}\right)\:={H}\:+{K} \\ $$$${H}\:={Re}\left(\:\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\left({a}+{b}\right){x}} }{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right)}{dx}\right){let}\:\varphi\left({z}\right)=\frac{{e}^{{i}\left({a}+{b}\right){z}} }{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right)} \\ $$$$\varphi\left({z}\right)\:=\frac{{e}^{{i}\left({a}+{b}\right){z}} }{\left({x}−{ia}\right)\left({x}+{ia}\right)\left({x}−{ib}\right)\left({x}+{ib}\right)}\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\: \\ $$$$\overset{−} {+}{ia}\:{and}\:\overset{−} {+}{ib}\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{{Res}\left(\varphi,{ia}\right)+{Res}\left(\varphi,{ib}\right)\right\} \\ $$$${Res}\left(\varphi,{ia}\right)\:=\frac{{e}^{{i}\left({a}+{b}\right){ia}} }{\left(\mathrm{2}{ia}\right)\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}\:\:\:\:\:\:\:\:\:\:\left({ifa}\neq{b}\right) \\ $$$$=\frac{{e}^{−{a}\left({a}+{b}\right)} }{\left(\mathrm{2}{ia}\right)\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)} \\ $$$${Res}\left(\varphi,{ib}\right)\:=\:\frac{{e}^{{i}\left({a}+{b}\right){ib}} }{\left(\mathrm{2}{ib}\right)\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}\:=\frac{{e}^{−{b}\left({a}+{b}\right)} }{\mathrm{2}{ib}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:\:\frac{{e}^{−{a}^{\mathrm{2}} −{ab}} }{\mathrm{2}{ia}\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}\:+\frac{{e}^{−{b}^{\mathrm{2}} −{ab}} }{\mathrm{2}{ib}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}\right\} \\ $$$$=\frac{\pi}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\left\{\:\:\frac{{e}^{−{b}^{\mathrm{2}} −{ab}} }{{b}}−\frac{{e}^{−{a}^{\mathrm{2}} −{ab}} }{{a}}\right\}\:={H}\:\:\left({the}\:{integral}\:{is}\:{real}\right)\:{for}\:{k} \\ $$$${we}\:{change}\:{b}\:{by}\:−{b}\:{we}\:{get}\:\:{K}\:=\frac{\pi}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\left\{\frac{{e}^{−{b}^{\mathrm{2}} +{ab}} }{−{b}}−\frac{{e}^{−{a}^{\mathrm{2}} +{ab}} }{{a}}\right\}\:\Rightarrow \\ $$$${f}\left({a},{b}\right)\:=\frac{\pi}{\mathrm{4}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}\left\{\:\frac{{e}^{−{b}^{\mathrm{2}} −{ab}} −{e}^{−{b}^{\mathrm{2}} +{ab}} }{{b}}−\frac{{e}^{−{a}^{\mathrm{2}} −{ab}} +{e}^{−{a}^{\mathrm{2}} \:+{ab}} }{{a}}\right\} \\ $$$${and}\:{we}\:{must}\:{study}\:{the}\:{case}\:{a}={b}… \\ $$$$\left.\mathrm{2}\right)\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cosx}\:{cos}\left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx}\:={f}\left(\mathrm{1},\mathrm{2}\right) \\ $$$$=\frac{\pi}{\mathrm{4}\left(−\mathrm{3}\right)}\left\{\:\:\frac{{e}^{−\mathrm{6}} \:−{e}^{−\mathrm{2}} }{\mathrm{2}}\:−\frac{{e}^{−\mathrm{3}} \:+{e}^{\mathrm{1}} }{\mathrm{1}}\right\}\:=−\frac{\pi}{\mathrm{12}}\left\{\frac{{e}^{−\mathrm{6}} −{e}^{−\mathrm{2}} −\mathrm{2}{e}^{−\mathrm{3}} −\mathrm{2}{e}}{\mathrm{2}}\right\} \\ $$$$=\frac{\pi}{\mathrm{24}}\left\{\mathrm{2}{e}^{−\mathrm{3}} \:+{e}^{−\mathrm{2}} +\mathrm{2}{e}\:−{e}^{−\mathrm{6}} \right\}\:. \\ $$$$ \\ $$