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Question Number 207099 by tri26112004 last updated on 06/May/24

Answered by Berbere last updated on 06/May/24

$$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\pi ax}}{{(x^2+{\beta }^2)}^{n+1}}dx;a\in {\mathbb{R}}_{+}$$$$ifImx⩾0\mid e^{i\pi ax}\mid =\mid e^{ia\pi (Rcos\left(\beta \right)+iRsin\left(\beta \right))}\mid =e^{-a\pi Rsin\left(\beta \right)}⩽1$$$$\mathrm{\forall }\beta \in [0,\pi ]$$$$applieResidueTheoremover[-R,R]\cup {Re}^{i\theta };\theta \in [0,\pi ]andTack\underset{R\to \mathrm{\infty }}{\lim }$$$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\pi ax}}{{(x^2+\beta )}^{n+1}}dx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{ia\pi x}}{{(x+i\beta )}^{n+1}{(x-i\beta )}^{n+1}}dx=f\left(a\right)$$$$=\underset{x\to \beta }{\lim }.2i\pi .\frac{1}{n!}{\partial }_n{(x-i\beta )}^{n+1}\frac{e^{2i\pi ax}}{{(x-i\beta )}^{n+1}{(x+i\beta )}^{n+1}}{\mid }_{x=i\beta }$$$$=\frac{2i\pi }{n}{\partial }^n\frac{e^{2ia\pi x}}{{(x+i\beta )}^{n+1}}{\mid }_{x=i\beta };Libneizformula$$$${\partial }^n.e^{ia\pi x}.\frac{1}{{(x+i\beta )}^{n+1}}=\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right){\left(e^{ia\pi x}\right)}^{(n-k)}.{\left(\frac{1}{{(x+i\beta )}^{n+1}}\right)}^{\left(k\right)}{\mid }_{x=i\beta }$$$$=\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right){\left(ia\pi \right)}^{(n-k)}{e}^{ia\pi x}.\frac{{(-1)}^k(n+k)!}{n!{(x+i\beta )}^{n+1+k}}{\mid }_{x=i\beta }$$$$=\frac{e^{-a\pi \beta }}{n!}\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right){\left(ia\pi \right)}^{(n-k)}.\frac{(n+k)!}{{\left(2i\beta \right)}^{n+1-k}}$$$$f\left(a\right)=\frac{2i\pi }{n!}.\frac{e^{-a\pi \beta }}{n!}\underset{k=0}{\sum ^n}{\left(ia\pi \right)}^{(n-k)}\frac{(n+k)!}{{\left(2i\beta \right)}^{n+1-k}}$$$$={\left(\frac{\pi }{\beta }\right)}^{n+1}{\left(\frac{a}{2}\right)}^n\frac{e^{-a\pi \beta }}{{(n!)}^2}\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right).\frac{(n+k)!}{{\left(2a\beta \pi \right)}^k}$$$$$$

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