Fourier Integral for Cosine Function with Infinite Limits

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SUMMARY

The discussion centers on evaluating the Fourier integral of the cosine function, specifically the integral \(\int \cos(wv) \, dv\) from \(-\infty\) to \(\infty\). The result presented is \(\frac{\sin(wv)}{\pi w}\), raising questions about the appearance of \(\pi\) and the divergence of the integral in standard terms. Participants emphasize the need for context, suggesting that if the integral is part of a Fourier series, it should be limited to one period, while for Fourier transforms, the cosine should be expressed in exponential form as \(\cos(wv) = \frac{e^{wv} + e^{-wv}}{2}\).

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kasse
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Working with Fourier integrals, I need to find the integral

[tex]\int cos (wv) dv[/tex] between -[tex]\infty[/tex] and [tex]\infty[/tex]. Is it possible to find this integral?

What I get is sin(wv)/(pi*w) with the infinite limits for v.
 
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Where did the pi come from? In the sense of a ordinary integral it makes no sense, it's divergent. In the sense of a Fourier integral it can make some sense, if you are also integrating over w. But I think you need to give us the complete context.
 
If you are working with Fourier series, why are you integrating from negative infinity to infinity? The integral should be over one period.

If you are working with the Fourier transform, why are you integrating cosine? Write everything in terms of exponentials: cos(wv)= (ewv+ e-wv)/2.

As Dick said, please give us the entire problem.
 

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