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## Main Question or Discussion Point

is this trick valid at least in the 'regularization' sense ?? for example

[tex] \int_{-\infty}^{\infty} \frac{dx}{x^{2}-a^{2}} [/tex]

then we replace thi integral above by [tex] \int_{-\infty}^{\infty} \frac{dx}{x^{2}+ie-a^{2}} [/tex] for 'e' tending to 0

using Cauchy residue theorem i get [tex] \int_{-\infty}^{\infty} \frac{dx}{x^{2}+ie-a^{2}}=(i\pi )/a [/tex]

the same trick applied to [tex] \int_{-\infty}^{\infty} \frac{dx}{x+a}= 2\pi i [/tex]

however we have obtained a complex regularization to a Real valued integral, i have seen this trick whenever dealing integrals in Quantum Physics but i want to hear the opinion of a Mathematician.

[tex] \int_{-\infty}^{\infty} \frac{dx}{x^{2}-a^{2}} [/tex]

then we replace thi integral above by [tex] \int_{-\infty}^{\infty} \frac{dx}{x^{2}+ie-a^{2}} [/tex] for 'e' tending to 0

using Cauchy residue theorem i get [tex] \int_{-\infty}^{\infty} \frac{dx}{x^{2}+ie-a^{2}}=(i\pi )/a [/tex]

the same trick applied to [tex] \int_{-\infty}^{\infty} \frac{dx}{x+a}= 2\pi i [/tex]

however we have obtained a complex regularization to a Real valued integral, i have seen this trick whenever dealing integrals in Quantum Physics but i want to hear the opinion of a Mathematician.