I've tried to perform the following integral with mathematica (student edition):

Integrate[(I/Pi)/(-x + I/(2 t)), {x, -Infinity, Infinity}, Assumptions -> t> 0]

Mathematica says that this integral doesn't converge but it is wrong, the right result is 1. Is something wrong with the string?

  • $\begingroup$ By numerics: NIntegrate[(I/Pi)/(-x + I/(2)), {x, -Infinity, -I, Infinity}, Method -> "PrincipalValue"] // Quiet // Chop $\endgroup$ – Mariusz Iwaniuk May 17 '18 at 18:44
  • $\begingroup$ Will this be constructive: Residue[(I/Pi)/(-x + I/(2 t)), {x, I/(2 t)}], which returns -I/\[Pi]? $\endgroup$ – Αλέξανδρος Ζεγγ May 18 '18 at 5:17

This integral exist only in the principal value sense:

Integrate[(I/Pi)/(-x + I/(2 t)), {x, -Infinity, Infinity},
 Assumptions -> t > 0, PrincipalValue -> True]


That means, the integrand is not in $L^1(\mathbb{R};\mathbb{R})$, but still

$$\lim_{R \to \infty} \int_{-R}^R \frac{\operatorname{i}}{\pi \left(-x+\frac{\operatorname{i}}{2 t}\right)} \, \operatorname{d} x = 1$$

exists. The important point is that

$$ \int_{-S}^R \frac{\operatorname{i}}{\pi \left(-x+\frac{\operatorname{i}}{2 t}\right)} \, \operatorname{d} x = 1$$

only converges if the lower and upper integral boundary $S$ and $R$ converge to $\infty$ with (essentially) same rate. The reason is that the imaginary part of the integrand is

Im[(I/Pi)/(-x + I/(2 t))] // ComplexExpand // TeXForm

$$-\frac{x}{\pi \left(\frac{1}{4 t^2}+x^2\right)},$$

so it is an odd function in $x$ and goes assymptotically like $\frac{1}{x}$ which is not (absolutely) integrable.

| improve this answer | |
  • $\begingroup$ Ok it's clear now, thank you very much $\endgroup$ – Alex May 19 '18 at 10:28
  • $\begingroup$ You're welcome! $\endgroup$ – Henrik Schumacher May 19 '18 at 10:28

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