Talk:Gamma function: Difference between revisions
Content added Content deleted
(int dx f(x) = int f(x)dx) |
(Hmm) |
||
| Line 1: | Line 1: | ||
==Integrals== |
==Integrals== |
||
I prefer the form <math>\int dx\; f(x)</math> (operator-like) instead of the form <math>\int f(x)dx</math>; both are correct; no need to change it the next time:D --[[User:ShinTakezou|ShinTakezou]] 20:01, 5 March 2009 (UTC) |
I prefer the form <math>\int dx\; f(x)</math> (operator-like) instead of the form <math>\int f(x)dx</math>; both are correct; no need to change it the next time:D --[[User:ShinTakezou|ShinTakezou]] 20:01, 5 March 2009 (UTC) |
||
: Hmm, I never saw this form in mathematical literature. Technically, there is no any multiplication of ''f(x)'' by ''dx''. You cannot commute them, if you meant that. And integral operator is not equal to definite integral. The definite integral using the integral operator would be sort of: <math>{\int f}\mid_a^b</math>. --[[User:Dmitry-kazakov|Dmitry-kazakov]] 20:55, 5 March 2009 (UTC) |
|||
Revision as of 20:55, 5 March 2009
Integrals
I prefer the form (operator-like) instead of the form ; both are correct; no need to change it the next time:D --ShinTakezou 20:01, 5 March 2009 (UTC)
- Hmm, I never saw this form in mathematical literature. Technically, there is no any multiplication of f(x) by dx. You cannot commute them, if you meant that. And integral operator is not equal to definite integral. The definite integral using the integral operator would be sort of: . --Dmitry-kazakov 20:55, 5 March 2009 (UTC)
