About the complex error function

  • Context: Graduate 
  • Thread starter Thread starter luisgml_2000
  • Start date Start date
  • Tags Tags
    Complex Error Function
Click For Summary
SUMMARY

The discussion centers on the complex error function, specifically the Faddeyeva function w(z), as described in Abramowitz's reference. The key identity presented is $$\frac{i}{\pi} \int_0^{\infty} \frac{e^{-t^2}}{z-t}\, dt=e^{-z^2}\operatorname{erfc}(-iz)$$, which relates the integral to the complementary error function, defined as $$\operatorname{erfc}(z)=\frac{2}{\sqrt{\pi}}\int_z^\infty e^{-t^2} \, dt$$. The original poster seeks proof for this identity and invites suggestions for further exploration.

PREREQUISITES
  • Understanding of complex analysis, particularly contour integration.
  • Familiarity with special functions, specifically the complementary error function (erfc).
  • Knowledge of integral transforms and their applications in complex functions.
  • Experience with mathematical proofs and derivations in advanced calculus.
NEXT STEPS
  • Study the properties and applications of the Faddeyeva function in complex analysis.
  • Learn about contour integration techniques relevant to proving complex identities.
  • Explore the derivation and applications of the complementary error function (erfc).
  • Review the original paper referenced in the discussion for deeper insights into the identity.
USEFUL FOR

Mathematicians, physicists, and students studying complex analysis or special functions, particularly those interested in advanced mathematical proofs and applications of the complex error function.

luisgml_2000
Messages
42
Reaction score
0
Hello!

I'm studying on my own the complex error function [tex]w(z)[/tex], also known as Faddeyeva function. On page 297 from Abramowitz it is stated that
$$
\frac{i}{\pi} \int_0^{\infty} \frac{e^{-t^2}}{z-t}\, dt=e^{-z^2}\operatorname{erfc}(-iz)
$$
where

[tex] \operatorname{erfc}(z)=\frac{2}{\sqrt{\pi}}\int_z^\infty e^{-t^2} \, dt[/tex]

The former identity is puzzling me and therefore I can't come up with a proof for it. Welcome any suggestions!

Thanks in advance for your attention.
 
Last edited by a moderator:
Physics news on Phys.org

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K