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Question Number 44706 by maxmathsup by imad last updated on 03/Oct/18

$$let{f}_{\alpha }\left(x\right)=\frac{cos\left(\alpha x\right)}{1+x^2}$$$$1)calculate{f}^{\left(n\right)}\left(x\right)and{f}^{\left(n\right)}\left(0\right)$$$$2)developpfatintegrserie$$$$3)give{\int }_0^x{f}_{\alpha }\left(t\right)dtatformofserie$$$$4)developp{\int }_0^{\mathrm{\infty }}{f}_{\alpha }\left(t\right)dtatintegrserie.$$

Commented by maxmathsup by imad last updated on 04/Oct/18

$$1)wehave{f}_{\alpha }\left(x\right)=\frac{cos\left(\alpha x\right)}{(x-i)(x+i)}=\frac{cos\left(\alpha x\right)}{2i}\{\frac{1}{x-i}-\frac{1}{x+i}\}$$$$=\frac{1}{2i}\{\frac{cos\left(\alpha x\right)}{x-i}-\frac{cos\left(\alpha x\right)}{x+i}\}\Rightarrow f^{\left(n\right)}\left(x\right)=\frac{1}{2i}\{{\left(\frac{cos\left(\alpha x\right)}{x-i}\right)}^{\left(n\right)}-{\left(\frac{cos\left(\alpha x\right)}{x+i}\right)}^{\left(n\right)}\}but$$$${\left(\frac{cos\left(\alpha x\right)}{(x-i)}\right)}^{\left(n\right)}=\sum _{k=0}^n{C}_n^k{\left(cos{\left(\alpha x\right)}^{}\right)}^{\left(k\right)}{\left(\frac{1}{x-i}\right)}^{(n-k)}\left(leibnizformulae\right)$$$$=\sum _{k=0}^n{C}_n^{k}cos(\alpha x+\frac{k\pi }{2})\frac{{(-1)}^{\mathit{n}-\mathit{k}}(\mathit{n}-\mathit{k})!}{{(x-i)}^{n-k+1}}also$$$${\left(\frac{cos\left(\alpha x\right)}{x+i}\right)}^{\left(n\right)}=\sum _{k=0}^n{C}_n^{k}cos(\alpha x+\frac{\pi }{2})\frac{{(-1)}^{n-k}(n-k)!}{{(x+i)}^{n-k+1}}\Rightarrow$$$$f^{\left(n\right)}\left(x\right)=\frac{1}{2i}\sum _{k=0}^n{C}_n^k{(-1)}^{n-k}(n-k)!cos(\alpha x+\frac{k\pi }{2})\{\frac{1}{{(x-i)}^{n-k+1}}-\frac{1}{{(x+i)}^{n-k+1}}\}$$$$f^{\left(n\right)}\left(0\right)=\frac{1}{2i}\sum _{k=0}^n{(-1)}^{n-k}\frac{n!}{k!}cos\left(\frac{k\pi }{2}\right)\{i^{n-k+1}-{(-i)}^{n-k+1}\}$$$$=\sum _{k=0}^n{(-1)}^{n-k}\frac{n!}{k!}cos\left(\frac{k\pi }{2}\right)sin(n-k+1)\frac{\pi }{2}$$

Commented by maxmathsup by imad last updated on 05/Oct/18

$$2)wehavef\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{f^{\left(n\right)}\left(0\right)}{n!}{x}^n$$$$=\sum _{n=0}^{\mathrm{\infty }}\left(\sum _{k=0}^n\frac{{(-1)}^{n-k}}{k!}cos\left(\frac{k\pi }{2}\right)sin(n-k+1)\frac{\pi }{2}\right)x^n$$$$=\sum _{n=0}^{\mathrm{\infty }}(\sum _{k=0}^n\frac{{(-1)}^{n-k}}{k!}cos\left(\frac{k\pi }{2}\right)cos\left((n-k)\frac{\pi }{2}\right)x^n.$$

Commented by maxmathsup by imad last updated on 05/Oct/18

$$3){\int }_0^x{f}_{\alpha }\left(t\right)dt=\sum _{n=0}^{\mathrm{\infty }}(\sum _{k=0}^n\frac{{(-1)}^{n-k}}{k!}cos\left(\frac{k\pi }{2}\right)cos\left((n-k)\frac{\pi }{2}\right)\frac{x^{n+1}}{n+1}.$$

Commented by maxmathsup by imad last updated on 05/Oct/18

$$4)letcalculatefirst{\int }_0^{\mathrm{\infty }}{f}_{\alpha }\left(t\right)dt=I_{\alpha }$$$$I_{\alpha }={\int }_0^{\mathrm{\infty }}\frac{cos\left(\alpha t\right)}{1+t^2}dt\Rightarrow 2I_{\alpha }={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(\alpha t\right)}{1+t^2}dt=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{i\alpha t}}{1+t^2}dt\right)$$$$let\phi \left(z\right)=\frac{e^{i\alpha z}}{1+z^2}thepolesof\phi areiand-iresidustheoremgive$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,i)=2i\pi \frac{e^{-\alpha }}{2i}=\pi e^{-\alpha }\Rightarrow I_{\alpha }=\frac{\pi }{2}{e}^{-\alpha }$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}{f}_{\alpha }\left(t\right)dt=\frac{\pi }{2}{e}^{-\alpha }=\frac{\pi }{2}\sum _{n=0}^{\mathrm{\infty }}\frac{{(-\alpha )}^n}{n!}=\frac{\pi }{2}\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n!}{\alpha }^n.$$$$$$

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