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Normalization of integral_without_prefactor #24

Answered by gudrunhe
vsht asked this question in Q&A
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Hi Vladyslav,

thanks for the comment, indeed the factor of 1/(i\pi^2) per loop is always included, we consider it as part of the integration measure.
In the documentation this is explained in section 6.6 (FAQs).

Cheers, Gudrun

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