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Question Number 44706 by maxmathsup by imad last updated on 03/Oct/18
let f_α (x) = ((cos(αx))/(1+x^2 )) 1) calculate f^((n)) (x) and f^((n)) (0) 2) developp f at integr serie 3) give ∫_0 ^x f_α (t) dt at form of serie 4) developp ∫_0 ^∞ f_α (t)dt at integr serie .
$${let}\:{f}_{\alpha} \left({x}\right)\:=\:\frac{{cos}\left(\alpha{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie} \\ $$$$\left.\mathrm{3}\right)\:{give}\:\int_{\mathrm{0}} ^{{x}} \:{f}_{\alpha} \left({t}\right)\:{dt}\:\:{at}\:{form}\:{of}\:{serie}\: \\ $$$$\left.\mathrm{4}\right)\:{developp}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:{f}_{\alpha} \left({t}\right){dt}\:\:{at}\:\:{integr}\:{serie}\:. \\ $$
Commented by maxmathsup by imad last updated on 04/Oct/18
1) we have f_α (x) =((cos(αx))/((x−i)(x+i))) =((cos(αx))/(2i)){(1/(x−i)) −(1/(x+i))} =(1/(2i)){ ((cos(αx))/(x−i)) −((cos(αx))/(x+i))} ⇒f^((n)) (x) =(1/(2i)){ (((cos(αx))/(x−i)))^((n)) −(((cos(αx))/(x+i)))^((n)) } but (((cos(αx))/((x−i))))^((n)) =Σ_(k=0) ^n C_n ^k (cos(αx)^ )^((k)) ((1/(x−i)))^((n−k)) (leibniz formulae) =Σ_(k=0) ^n C_n ^k cos(αx +((kπ)/2)) (((−1)^(n−k) (n−k)!)/((x−i)^(n−k +1) )) also (((cos(αx))/(x+i)))^((n)) =Σ_(k=0) ^n C_n ^k cos(αx+(π/2))(((−1)^(n−k) (n−k)!)/((x+i)^(n−k+1) )) ⇒ f^((n)) (x)=(1/(2i)) Σ_(k=0) ^n C_n ^k (−1)^(n−k) (n−k)! cos(αx+((kπ)/2)){(1/((x−i)^(n−k+1) )) −(1/((x+i)^(n−k+1) ))} f^((n)) (0) =(1/(2i)) Σ_(k=0) ^n (−1)^(n−k) ((n!)/(k!)) cos(((kπ)/2)){i^(n−k+1) −(−i)^(n−k+1) } =Σ_(k=0) ^n (−1)^(n−k) ((n!)/(k!)) cos(((kπ)/2))sin(n−k+1)(π/2)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}_{\alpha} \left({x}\right)\:=\frac{{cos}\left(\alpha{x}\right)}{\left({x}−{i}\right)\left({x}+{i}\right)}\:=\frac{{cos}\left(\alpha{x}\right)}{\mathrm{2}{i}}\left\{\frac{\mathrm{1}}{{x}−{i}}\:−\frac{\mathrm{1}}{{x}+{i}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:\frac{{cos}\left(\alpha{x}\right)}{{x}−{i}}\:−\frac{{cos}\left(\alpha{x}\right)}{{x}+{i}}\right\}\:\Rightarrow{f}^{\left({n}\right)} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:\left(\frac{{cos}\left(\alpha{x}\right)}{{x}−{i}}\right)^{\left({n}\right)} −\left(\frac{{cos}\left(\alpha{x}\right)}{{x}+{i}}\right)^{\left({n}\right)} \right\}\:{but} \\ $$$$\left(\frac{{cos}\left(\alpha{x}\right)}{\left({x}−{i}\right)}\right)^{\left({n}\right)} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\left({cos}\left(\alpha{x}\right)^{} \right)^{\left({k}\right)} \:\left(\frac{\mathrm{1}}{{x}−{i}}\right)^{\left({n}−{k}\right)} \:\:\:\:\left({leibniz}\:{formulae}\right) \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:{cos}\left(\alpha{x}\:+\frac{{k}\pi}{\mathrm{2}}\right)\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}−\boldsymbol{{k}}} \left(\boldsymbol{{n}}−\boldsymbol{{k}}\right)!}{\left({x}−{i}\right)^{{n}−{k}\:+\mathrm{1}} }\:\:{also} \\ $$$$\left(\frac{{cos}\left(\alpha{x}\right)}{{x}+{i}}\right)^{\left({n}\right)} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:{cos}\left(\alpha{x}+\frac{\pi}{\mathrm{2}}\right)\frac{\left(−\mathrm{1}\right)^{{n}−{k}} \left({n}−{k}\right)!}{\left({x}+{i}\right)^{{n}−{k}+\mathrm{1}} }\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\:\left(−\mathrm{1}\right)^{{n}−{k}} \left({n}−{k}\right)!\:{cos}\left(\alpha{x}+\frac{{k}\pi}{\mathrm{2}}\right)\left\{\frac{\mathrm{1}}{\left({x}−{i}\right)^{{n}−{k}+\mathrm{1}} }\:−\frac{\mathrm{1}}{\left({x}+{i}\right)^{{n}−{k}+\mathrm{1}} }\right\} \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{1}\right)^{{n}−{k}} \:\frac{{n}!}{{k}!}\:{cos}\left(\frac{{k}\pi}{\mathrm{2}}\right)\left\{{i}^{{n}−{k}+\mathrm{1}} \:−\left(−{i}\right)^{{n}−{k}+\mathrm{1}} \right\} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{1}\right)^{{n}−{k}} \:\frac{{n}!}{{k}!}\:{cos}\left(\frac{{k}\pi}{\mathrm{2}}\right){sin}\left({n}−{k}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}} \\ $$
Commented by maxmathsup by imad last updated on 05/Oct/18
2) we have f(x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n =Σ_(n=0) ^∞ ( Σ_(k=0) ^n (((−1)^(n−k) )/(k!)) cos(((kπ)/2))sin(n−k+1)(π/2))x^n =Σ_(n=0) ^∞ (Σ_(k=0) ^n (((−1)^(n−k) )/(k!)) cos(((kπ)/2))cos((n−k)(π/2)) x^n .
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\left(\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−{k}} }{{k}!}\:{cos}\left(\frac{{k}\pi}{\mathrm{2}}\right){sin}\left({n}−{k}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}}\right){x}^{{n}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−{k}} }{{k}!}\:{cos}\left(\frac{{k}\pi}{\mathrm{2}}\right){cos}\left(\left({n}−{k}\right)\frac{\pi}{\mathrm{2}}\right)\:{x}^{{n}} \:.\right. \\ $$
Commented by maxmathsup by imad last updated on 05/Oct/18
3) ∫_0 ^x f_α (t)dt =Σ_(n=0) ^∞ (Σ_(k=0) ^n (((−1)^(n−k) )/(k!)) cos(((kπ)/2))cos((n−k)(π/2)) (x^(n+1) /(n+1)) .
$$\left.\mathrm{3}\right)\:\int_{\mathrm{0}} ^{{x}} \:{f}_{\alpha} \left({t}\right){dt}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−{k}} }{{k}!}\:{cos}\left(\frac{{k}\pi}{\mathrm{2}}\right){cos}\left(\left({n}−{k}\right)\frac{\pi}{\mathrm{2}}\right)\:\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:.\right. \\ $$
Commented by maxmathsup by imad last updated on 05/Oct/18
4) let calculate first ∫_0 ^∞ f_α (t)dt =I_α I_α = ∫_0 ^∞ ((cos(αt))/(1+t^2 ))dt ⇒2I_α = ∫_(−∞) ^(+∞) ((cos(αt))/(1+t^2 ))dt =Re(∫_(−∞) ^(+∞) (e^(iαt) /(1+t^2 ))dt) let ϕ(z) =(e^(iαz) /(1+z^2 )) the poles of ϕ are i and −i residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπRes(ϕ,i) =2iπ (e^(−α) /(2i)) =π e^(−α) ⇒ I_α =(π/2) e^(−α) ⇒∫_0 ^∞ f_α (t)dt =(π/2) e^(−α) =(π/2) Σ_(n=0) ^∞ (((−α)^n )/(n!)) =(π/2) Σ_(n=0) ^∞ (((−1)^n )/(n!)) α^n .
$$\left.\mathrm{4}\right)\:{let}\:{calculate}\:{first}\:\int_{\mathrm{0}} ^{\infty} \:\:{f}_{\alpha} \left({t}\right){dt}\:\:={I}_{\alpha} \\ $$$${I}_{\alpha} =\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\alpha{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\Rightarrow\mathrm{2}{I}_{\alpha} \:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\alpha{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:={Re}\left(\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{i}\alpha{t}} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\right) \\ $$$${let}\:\varphi\left({z}\right)\:=\frac{{e}^{{i}\alpha{z}} }{\mathrm{1}+{z}^{\mathrm{2}} }\:\:{the}\:{poles}\:{of}\:\varphi\:{are}\:{i}\:{and}\:−{i}\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi{Res}\left(\varphi,{i}\right)\:=\mathrm{2}{i}\pi\:\:\frac{{e}^{−\alpha} }{\mathrm{2}{i}}\:=\pi\:{e}^{−\alpha} \:\Rightarrow\:{I}_{\alpha} =\frac{\pi}{\mathrm{2}}\:{e}^{−\alpha} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:{f}_{\alpha} \left({t}\right){dt}\:=\frac{\pi}{\mathrm{2}}\:{e}^{−\alpha} \:=\frac{\pi}{\mathrm{2}}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\alpha\right)^{{n}} }{{n}!}\:=\frac{\pi}{\mathrm{2}}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\alpha^{{n}} \:. \\ $$$$ \\ $$

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