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Old June 27th, 2009, 11:56 PM
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Default Integral-limit and ?!!....

Calculate this integral :

deduct :
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Old June 28th, 2009, 12:28 AM
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Quote:
Originally Posted by dhiab View Post
Calculate this integral :

deduct :
\int_{2n\pi}^{(2n+1)\pi} e^{-x}\cos(x)\ dx = \text{Re}\int_{2n\pi}^{(2n+1)\pi} e^{-x}e^{ix}\ dx= \text{Re}\int_{2n\pi}^{(2n+1)\pi} e^{(-1+i)x}\ dx = \text{Re} \left[\frac{e^{(-1+i)x}}{-1+i} \right]_{2n\pi}^{(2n+1)\pi}

and if you do things right you will find this is proportional to e^{-2n\pi}, which will immediate answer the next part, and then the series is geometric.

CB

Last edited by CaptainBlack; June 28th, 2009 at 12:49 AM.
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