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Question Number 66466 by mathmax by abdo last updated on 15/Aug/19

find f(a,b) =∫_0 ^∞ ((cos(ax)cos(bx))/((x^2 +a^2 )(x^2 +b^2 )))dx with a>0 and b>0 2)calculate ∫_0 ^∞ ((cos(x)cos(2x))/((x^2 +1)(x^2 +4)))dx

$${find}\:\:{f}\left({a},{b}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{cos}\left({ax}\right){cos}\left({bx}\right)}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right)}{dx}\:\:{with}\:{a}>\mathrm{0}\:{and}\:{b}>\mathrm{0} \\ $$ $$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left({x}\right){cos}\left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)}{dx} \\ $$

Commented bymathmax by abdo last updated on 17/Aug/19

1) we have f(a,b) =(1/2)∫_0 ^∞ ((cos(a+b)x+cos(a−b)x)/((x^2 +a^2 )(x^2 +b^(2)) ))dx =(1/4)∫_(−∞) ^(+∞) ((cos(a+b)x)/((x^2 +a^2 )(x^2 +b^2 )))dx +(1/4)∫_(−∞) ^(+∞) ((cos(a−b)x)/((x^2 +a^2 )(x^2 +b^2 )))dx ⇒ 4f(,b) =H +K H =Re( ∫_(−∞) ^(+∞) (e^(i(a+b)x) /((x^2 +a^2 )(x^2 +b^2 )))dx)let ϕ(z)=(e^(i(a+b)z) /((x^2 +a^2 )(x^2 +b^2 ))) ϕ(z) =(e^(i(a+b)z) /((x−ia)(x+ia)(x−ib)(x+ib))) so the poles of ϕ are +^− ia and +^− ib residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ {Res(ϕ,ia)+Res(ϕ,ib)} Res(ϕ,ia) =(e^(i(a+b)ia) /((2ia)(b^2 −a^2 ))) (ifa≠b) =(e^(−a(a+b)) /((2ia)(b^2 −a^2 ))) Res(ϕ,ib) = (e^(i(a+b)ib) /((2ib)(a^2 −b^2 ))) =(e^(−b(a+b)) /(2ib(a^2 −b^2 ))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ (e^(−a^2 −ab) /(2ia(b^2 −a^2 ))) +(e^(−b^2 −ab) /(2ib(a^2 −b^2 )))} =(π/(a^2 −b^2 )){ (e^(−b^2 −ab) /b)−(e^(−a^2 −ab) /a)} =H (the integral is real) for k we change b by −b we get K =(π/(a^2 −b^2 )){(e^(−b^2 +ab) /(−b))−(e^(−a^2 +ab) /a)} ⇒ f(a,b) =(π/(4(a^2 −b^2 ))){ ((e^(−b^2 −ab) −e^(−b^2 +ab) )/b)−((e^(−a^2 −ab) +e^(−a^2 +ab) )/a)} and we must study the case a=b... 2)∫_0 ^∞ ((cosx cos(2x))/((x^2 +1)(x^2 +4)))dx =f(1,2) =(π/(4(−3))){ ((e^(−6) −e^(−2) )/2) −((e^(−3) +e^1 )/1)} =−(π/(12)){((e^(−6) −e^(−2) −2e^(−3) −2e)/2)} =(π/(24)){2e^(−3) +e^(−2) +2e −e^(−6) } .

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({a},{b}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({a}+{b}\right){x}+{cos}\left({a}−{b}\right){x}}{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} \:+{b}^{\left.\mathrm{2}\right)} \right.}{dx} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{4}}\int_{−\infty} ^{+\infty} \:\frac{{cos}\left({a}+{b}\right){x}}{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right)}{dx}\:+\frac{\mathrm{1}}{\mathrm{4}}\int_{−\infty} ^{+\infty} \:\frac{{cos}\left({a}−{b}\right){x}}{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right)}{dx}\:\Rightarrow \\ $$ $$\mathrm{4}{f}\left(,{b}\right)\:={H}\:+{K} \\ $$ $${H}\:={Re}\left(\:\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\left({a}+{b}\right){x}} }{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right)}{dx}\right){let}\:\varphi\left({z}\right)=\frac{{e}^{{i}\left({a}+{b}\right){z}} }{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right)} \\ $$ $$\varphi\left({z}\right)\:=\frac{{e}^{{i}\left({a}+{b}\right){z}} }{\left({x}−{ia}\right)\left({x}+{ia}\right)\left({x}−{ib}\right)\left({x}+{ib}\right)}\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\: \\ $$ $$\overset{−} {+}{ia}\:{and}\:\overset{−} {+}{ib}\:\:{residus}\:{theorem}\:{give} \\ $$ $$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{{Res}\left(\varphi,{ia}\right)+{Res}\left(\varphi,{ib}\right)\right\} \\ $$ $${Res}\left(\varphi,{ia}\right)\:=\frac{{e}^{{i}\left({a}+{b}\right){ia}} }{\left(\mathrm{2}{ia}\right)\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}\:\:\:\:\:\:\:\:\:\:\left({ifa}\neq{b}\right) \\ $$ $$=\frac{{e}^{−{a}\left({a}+{b}\right)} }{\left(\mathrm{2}{ia}\right)\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)} \\ $$ $${Res}\left(\varphi,{ib}\right)\:=\:\frac{{e}^{{i}\left({a}+{b}\right){ib}} }{\left(\mathrm{2}{ib}\right)\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}\:=\frac{{e}^{−{b}\left({a}+{b}\right)} }{\mathrm{2}{ib}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$ $$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:\:\frac{{e}^{−{a}^{\mathrm{2}} −{ab}} }{\mathrm{2}{ia}\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}\:+\frac{{e}^{−{b}^{\mathrm{2}} −{ab}} }{\mathrm{2}{ib}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}\right\} \\ $$ $$=\frac{\pi}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\left\{\:\:\frac{{e}^{−{b}^{\mathrm{2}} −{ab}} }{{b}}−\frac{{e}^{−{a}^{\mathrm{2}} −{ab}} }{{a}}\right\}\:={H}\:\:\left({the}\:{integral}\:{is}\:{real}\right)\:{for}\:{k} \\ $$ $${we}\:{change}\:{b}\:{by}\:−{b}\:{we}\:{get}\:\:{K}\:=\frac{\pi}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\left\{\frac{{e}^{−{b}^{\mathrm{2}} +{ab}} }{−{b}}−\frac{{e}^{−{a}^{\mathrm{2}} +{ab}} }{{a}}\right\}\:\Rightarrow \\ $$ $${f}\left({a},{b}\right)\:=\frac{\pi}{\mathrm{4}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}\left\{\:\frac{{e}^{−{b}^{\mathrm{2}} −{ab}} −{e}^{−{b}^{\mathrm{2}} +{ab}} }{{b}}−\frac{{e}^{−{a}^{\mathrm{2}} −{ab}} +{e}^{−{a}^{\mathrm{2}} \:+{ab}} }{{a}}\right\} \\ $$ $${and}\:{we}\:{must}\:{study}\:{the}\:{case}\:{a}={b}... \\ $$ $$\left.\mathrm{2}\right)\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cosx}\:{cos}\left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx}\:={f}\left(\mathrm{1},\mathrm{2}\right) \\ $$ $$=\frac{\pi}{\mathrm{4}\left(−\mathrm{3}\right)}\left\{\:\:\frac{{e}^{−\mathrm{6}} \:−{e}^{−\mathrm{2}} }{\mathrm{2}}\:−\frac{{e}^{−\mathrm{3}} \:+{e}^{\mathrm{1}} }{\mathrm{1}}\right\}\:=−\frac{\pi}{\mathrm{12}}\left\{\frac{{e}^{−\mathrm{6}} −{e}^{−\mathrm{2}} −\mathrm{2}{e}^{−\mathrm{3}} −\mathrm{2}{e}}{\mathrm{2}}\right\} \\ $$ $$=\frac{\pi}{\mathrm{24}}\left\{\mathrm{2}{e}^{−\mathrm{3}} \:+{e}^{−\mathrm{2}} +\mathrm{2}{e}\:−{e}^{−\mathrm{6}} \right\}\:. \\ $$ $$ \\ $$

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