Timeline for Closed form of :$\int_{-1}^1 x^{2k} (\operatorname{erf}(x))^k \,dx $ for $ k$ is even integer and :$\int _{0}^{t}\exp(-x^2 \operatorname{erf}(x))dx$
Current License: CC BY-SA 4.0
12 events
when toggle format | what | by | license | comment | |
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Jun 16, 2018 at 13:56 | vote | accept | zeraoulia rafik | ||
Jun 1, 2018 at 23:16 | comment | added | zeraoulia rafik | @LiviuNicolaescu, have you tried out to see the behavior of that function comparing it with error function ? | |
Jun 1, 2018 at 19:59 | comment | added | user64494 | @Liviu Nicolaescu: I prefer arguments over unbased words. Sorry, I have nothing to discuss with you in such manner. | |
Jun 1, 2018 at 18:48 | comment | added | Liviu Nicolaescu | @user64494 I think it is a legitimate question. "Art for art's sake" would suggest this is a useless question. I don't think it is. | |
Jun 1, 2018 at 17:52 | comment | added | zeraoulia rafik | @user64494, Probably Liviu Nicolaescu think that is easy for evaluation | |
Jun 1, 2018 at 17:50 | comment | added | user64494 | @Liviu Nikolaescu: Please, base your statement. | |
Jun 1, 2018 at 13:15 | comment | added | Liviu Nicolaescu | No, it is not art for art's sake. | |
Jun 1, 2018 at 9:03 | comment | added | user64494 | What for? Isn't it art for art's sake? | |
Jun 1, 2018 at 8:23 | answer | added | Carlo Beenakker | timeline score: 4 | |
May 31, 2018 at 23:57 | comment | added | Gerald Edgar | As a general rule of thumb: since $1$ and $-1$ are not spacial points for the integrand, his definite integral is likely just as difficult as the indefinite integral. So: $\int x^4 \text{erf}(x)^2 dx$ has a known closed form, plug in and get your definite integral. But for $\int x^8 \text{erf}(x)^4 dx$ you are probably out of luck. | |
May 31, 2018 at 23:54 | history | edited | zeraoulia rafik | CC BY-SA 4.0 |
added 2 characters in body; edited title
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May 31, 2018 at 23:30 | history | asked | zeraoulia rafik | CC BY-SA 4.0 |