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Question Number 37898 by abdo mathsup 649 cc last updated on 19/Jun/18

$$calculatef\left(\lambda \right)={\int }_0^{+\mathrm{\infty }}{e}^{-\lambda x}cos\left(\pi \left[x\right]\right)dx$$$$with\lambda >0$$

Commented by prof Abdo imad last updated on 19/Jun/18

$$f\left(\lambda \right)=\sum _{n=0}^{\mathrm{\infty }}{\int }_n^{n+1}{e}^{-\lambda x}cos\left(n\pi \right)dx$$$$=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\int }_n^{n+1}{e}^{-\lambda x}dx$$$$=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{[-\frac{1}{\lambda }{e}^{-\lambda x}]}_n^{n+1}$$$$=\frac{1}{\lambda }\sum _{n=0}^{\mathrm{\infty }}{(-1)}^{n+1}(e^{-\lambda (n+1)}-e^{-\lambda n})$$$$=\frac{1}{\lambda }\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{e}^{-\lambda n}-\frac{1}{\lambda }\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{e}^{-\lambda (n+1)}$$$$=\frac{1}{\lambda }\sum _{n=0}^{\mathrm{\infty }}{(-e^{-\lambda })}^n-\frac{e^{-\lambda }}{\lambda }\sum _{n=0}^{\mathrm{\infty }}{(-e^{-\lambda })}^n$$$$=\frac{1-e^{-\lambda }}{\lambda }\frac{1}{1+e^{-\lambda }}\Rightarrow$$$$f\left(\lambda \right)=\frac{1-e^{-\lambda }}{\lambda (1+e^{-\lambda })}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18

$$\left[x\right]=01>x⩾0$$$$\left[x\right]=12>x⩾1$$$$\left[x\right]=23>x⩾2$$$$$$$$thusthevalueof\mathrm{\Pi }\left[x\right]isintregalmultipleof$$$$\mathrm{\Pi }...hencevalueofcos\left(\mathrm{\Pi }\left[x\right]\right)oscillates$$$$eitther+1or-1$$$$sothegivenintregalhastwovalue$$$$1){\int }_0^{+\mathrm{\infty }}{e}^{-\lambda x}\times 1\times dx$$$$=\mid \frac{e^{-\lambda x}}{-\lambda }{\mid }_0^{\mathrm{\infty }}$$$$=\frac{-1}{\lambda }(\frac{1}{e^{\mathrm{\infty }}}-\frac{1}{e^0})=\frac{1}{\lambda }$$$$2){\int }_0^{+\mathrm{\infty }}{e}^{-\lambda x}\times -1\times dx$$$$=-1\times \mid \frac{e^{-\lambda x}}{-\lambda }{\mid }_0^{\mathrm{\infty }}=\frac{1}{\lambda }(\frac{1}{e^{\mathrm{\infty }}}-\frac{1}{e^0})=-\frac{1}{\lambda }$$$$$$

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