Jul 26, 2018 · I know you are saying that i i is negligible compared to erfi (x x), but how can erfi (x x) have an imaginary part for real x x?

Jan 20, 2019 · Henrick: After reading your comments, we went back and looked at the original problem statement and, indeed, you were right: we needed to find the integral of "x * E^x^2". It …

Aug 11, 2016 · erfi(x) = 2 π−−√ ex2D(x) e r f i (x) = 2 π e x 2 D (x) Where D D is the Dawson Function. This is entirely real valued! Also the scipy implementation gives me a real value too. …

Nov 24, 2021 · I have some analytical results from a physics problem, where the Mathematica gives the results in terms of complex error function. I would like to explore another ...

Mar 4, 2022 · Here, the hypergeometric function converges nowhere and should be understood as a formal description of a divergent asymptotic series. I am looking for a rigorous derivation …

Aug 17, 2017 · $$ \text {erfi}^ {-1} (x) = -i \, \text {erf}^ {-1} (i x) $$ You can show this by considering the definition of the imaginary inverse function $$ \text {erfi} (x) = -i\,\text {erf} (i x) …

Sep 11, 2020 · FresnelC[z] Erfi[z] Erfi[z]} Such identities can be found exploiting MathematicalFunctionData and MathematicalFunction (the latter new in version 12), …

Feb 13, 2022 · An complicated integral involving Erfi function Ask Question Asked 3 years, 7 months ago Modified 3 years, 7 months ago

Showing $\int e^ {x^2} dx = \frac12 \sqrt\pi \operatorname {erfi} (x) + C$ I have seen this identity, but how is it reached? What parts of it make sense and what part is just made into erf?

Mar 13, 2021 · Expansion of Confluent Hypergeometric Function in terms of $\operatorname {erfi} (x)$ Ask Question Asked 4 years, 8 months ago Modified 4 years, 8 months ago