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I happened to stumble on the following simplification of an integral:

$$ \frac{1}{\pi} \int_{0}^{\infty} dx \ e^{-ax} \cdot \cos(kx) = \frac{1}{\pi} Re \left[ \int_{0}^{\infty} dx \ e^{x (ik - a)} \right] $$

According to my reasoning: $ \cos(kx) = \frac{1}{2} \left( e^{ikx} + e^{-ikx} \right) $, so the term on the exponent on the right side comes somehow from the first term of the complex form of $\cos(kx)$, but the second term gets cancelled, why?

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Consider first $e^{ikx} = \cos(kx) + i \sin(kx)$ then you can easily deduce that $$ Re(e^{ikx}) = \cos(kx). $$

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