Understanding the Improper Integral ∫eiωtdω = √(2\pi) from -∞ to ∞

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SUMMARY

The improper integral ∫eiωtdω from -∞ to ∞ equals √(2π) is confirmed through the derivation of the Fourier transform. The discussion highlights that by multiplying the integral by itself, resulting in I^2, and solving yields 2π, which upon taking the square root, confirms the original statement. There is a mention of a potential confusion regarding the equality to 1/√(2π), which requires clarification.

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zezima1
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Is ∫eiωtdω = √(2\pi) (from -∞ to ∞)

It should be so if my books derivation of the Fourier transform is correct, but they don't really explain why the above equality is correct?

can anyone show me how they get to that result for that integral or does anyone perhaps have a good link?
 
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zezima1 said:
Is ∫eiωtdω = √(2\pi) (from -∞ to ∞)

It should be so if my books derivation of the Fourier transform is correct, but they don't really explain why the above equality is correct?

can anyone show me how they get to that result for that integral or does anyone perhaps have a good link?

If you set I = integral and multiplied by the same integral, you would have I^2. When you solve that you get 2pi but then you take the square root.

However, isn't is supposed to be equal to \frac{1}{\sqrt{2\pi}
 

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