Question Number 207099 by tri26112004 last updated on 06/May/24
Answered by Berbere last updated on 06/May/24
![=∫_(−∞) ^∞ (e^(iπax) /((x^2 +β^2 )^(n+1) ))dx;a∈R_+ if Imx≥0 ∣e^(iπax) ∣=∣e^(iaπ(Rcos(β)+iRsin(β))) ∣=e^(−aπRsin(β)) ≤1 ∀β∈[0,π] applie Residue Theorem over [−R,R]∪Re^(iθ) ;θ∈[0,π] and Tacklim_(R→∞) ∫_(−∞) ^∞ (e^(iπax) /((x^2 +β)^(n+1) ))dx=∫_(−∞) ^∞ (e^(iaπx) /((x+iβ)^(n+1) (x−iβ)^(n+1) ))dx=f(a) =lim_(x→β) .2iπ.(1/(n!))∂_n (x−iβ)^(n+1) (e^(2iπax) /((x−iβ)^(n+1) (x+iβ)^(n+1) ))∣_(x=iβ) =((2iπ)/n)∂^n (e^(2iaπx) /((x+iβ)^(n+1) ))∣_(x=iβ) ;Libneiz formula ∂^n .e^(iaπx) .(1/((x+iβ)^(n+1) ))=Σ_(k=0) ^n ((n),(k) )(e^(iaπx) )^((n−k)) .((1/((x+iβ)^(n+1) )))^((k)) ∣_(x=iβ) =Σ_(k=0) ^n ((n),(k) )(iaπ)^((n−k)) e^(iaπx) .(((−1)^k (n+k)!)/(n!(x+iβ)^(n+1+k) ))∣_(x=iβ) =(e^(−aπβ) /(n!))Σ_(k=0) ^n ((n),(k) )(iaπ)^((n−k)) .(((n+k)!)/((2iβ)^(n+1−k) )) f(a)=((2iπ)/(n!)).(e^(−aπβ) /(n!))Σ_(k=0) ^n (iaπ)^((n−k)) (((n+k)!)/((2iβ)^(n+1−k) )) =((π/β))^(n+1) ((a/2))^n (e^(−aπβ) /((n!)^2 ))Σ_(k=0) ^n ((n),(k) ).(((n+k)!)/((2aβπ)^k ))](https://www.tinkutara.com/question/Q207111.png)
$$=\int_{−\infty} ^{\infty} \frac{{e}^{{i}\pi{ax}} }{\left({x}^{\mathrm{2}} +\beta^{\mathrm{2}} \right)^{{n}+\mathrm{1}} }{dx};{a}\in\mathbb{R}_{+} \\ $$$${if}\:{Imx}\geqslant\mathrm{0}\:\mid{e}^{{i}\pi{ax}} \mid=\mid{e}^{{ia}\pi\left({Rcos}\left(\beta\right)+{iRsin}\left(\beta\right)\right)} \mid={e}^{−{a}\pi{Rsin}\left(\beta\right)} \leqslant\mathrm{1} \\ $$$$\forall\beta\in\left[\mathrm{0},\pi\right] \\ $$$${applie}\:{Residue}\:{Theorem}\:{over}\:\left[−{R},{R}\right]\cup{Re}^{{i}\theta} ;\theta\in\left[\mathrm{0},\pi\right]\:{and}\:{Tack}\underset{{R}\rightarrow\infty} {\mathrm{lim}} \\ $$$$\int_{−\infty} ^{\infty} \frac{{e}^{{i}\pi{ax}} }{\left({x}^{\mathrm{2}} +\beta\right)^{{n}+\mathrm{1}} }{dx}=\int_{−\infty} ^{\infty} \frac{{e}^{{ia}\pi{x}} }{\left({x}+{i}\beta\right)^{{n}+\mathrm{1}} \left({x}−{i}\beta\right)^{{n}+\mathrm{1}} }{dx}={f}\left({a}\right) \\ $$$$=\underset{{x}\rightarrow\beta} {\mathrm{lim}}.\mathrm{2}{i}\pi.\frac{\mathrm{1}}{{n}!}\partial_{{n}} \left({x}−{i}\beta\right)^{{n}+\mathrm{1}} \frac{{e}^{\mathrm{2}{i}\pi{ax}} }{\left({x}−{i}\beta\right)^{{n}+\mathrm{1}} \left({x}+{i}\beta\right)^{{n}+\mathrm{1}} }\mid_{{x}={i}\beta} \\ $$$$=\frac{\mathrm{2}{i}\pi}{{n}}\partial^{{n}} \frac{{e}^{\mathrm{2}{ia}\pi{x}} }{\left({x}+{i}\beta\right)^{{n}+\mathrm{1}} }\mid_{{x}={i}\beta} ;{Libneiz}\:{formula} \\ $$$$\partial^{{n}} .{e}^{{ia}\pi{x}} .\frac{\mathrm{1}}{\left({x}+{i}\beta\right)^{{n}+\mathrm{1}} }=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\left({e}^{{ia}\pi{x}} \right)^{\left({n}−{k}\right)} .\left(\frac{\mathrm{1}}{\left({x}+{i}\beta\right)^{{n}+\mathrm{1}} }\right)^{\left({k}\right)} \mid_{{x}={i}\beta} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\left({ia}\pi\right)^{\left({n}−{k}\right)} {e}^{{ia}\pi{x}} .\frac{\left(−\mathrm{1}\right)^{{k}} \left({n}+{k}\right)!}{{n}!\left({x}+{i}\beta\right)^{{n}+\mathrm{1}+{k}} }\mid_{{x}={i}\beta} \\ $$$$=\frac{{e}^{−{a}\pi\beta} }{{n}!}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\left({ia}\pi\right)^{\left({n}−{k}\right)} .\frac{\left({n}+{k}\right)!}{\left(\mathrm{2}{i}\beta\right)^{{n}+\mathrm{1}−{k}} } \\ $$$${f}\left({a}\right)=\frac{\mathrm{2}{i}\pi}{{n}!}.\frac{{e}^{−{a}\pi\beta} }{{n}!}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({ia}\pi\right)^{\left({n}−{k}\right)} \frac{\left({n}+{k}\right)!}{\left(\mathrm{2}{i}\beta\right)^{{n}+\mathrm{1}−{k}} } \\ $$$$=\left(\frac{\pi}{\beta}\right)^{{n}+\mathrm{1}} \left(\frac{{a}}{\mathrm{2}}\right)^{{n}} \frac{{e}^{−{a}\pi\beta} }{\left({n}!\right)^{\mathrm{2}} }\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}.\frac{\left({n}+{k}\right)!}{\left(\mathrm{2}{a}\beta\pi\right)^{{k}} } \\ $$$$ \\ $$