Le mercredi 15 décembre 2021 à 20:43:07 UTC+1, [email protected] a
écrit :
> See this example:
>
> f(x)=(x+sin(3*x))*exp(-3*x*I)
> g(x)=f(x).expand()
> integral(f(x)-g(x),(x,0,2*pi))
>
> The answer is I*pi, but it should be 0.
>
Huh ?
f has no poles ; therefore, the value of the integrate beteen two given
points is independent of the integration path. In particular,
f(x).integrate(x,a,b)==real(f(x)).integrate(x,a,b)+I*imag(x).integrate(x,a,b)).
You can therefor check your affirmation by executing :
sage: numerical_integral(lambda
u:f(u).real(),0,2*pi)[0]+I*numerical_integral(lambda u:f(u).imag(),0,2*pi)[0]
-3.2262783606579845e-15 - 1.0471975511966007*I
which raises some doubts about your affirmation….
>
> Many other examples (related to Fourier coefficients) give similar errors.
> For instance:
>
> f(x)=(x+cos(x))*exp(-x*I)
> g(x)=f(x).expand()
> integral(f(x)-g(x),(x,0,2*pi))
>
> The answer is -pi, and it should be 0.
>
The previous reasoning also applies here :
```
sage: g(x)=(x+cos(x))*exp(-x*I)
sage: numerical_integral(lambda
u:g(u).real(),0,2*pi)[0]+I*numerical_integral(lambda
u:g(u).imag(),0,2*pi)[0]
3.1415926535897913 + 6.283185307179586*I
```
However, the answers given by the various available algorithms may still be
questioned :
sage: table([[alg,f(x).integrate(x,0,2*pi, algorithm=alg)] for alg in
("maxima", "giac", "sympy", "fricas", "mathematica_free")])
maxima 2/3*I*pi
giac -1/3*I*pi
sympy -1/3*I*pi
fricas -1/3*I*pi
mathematica_free -1.04720000000000*I
sage: table([[alg,g(x).integrate(x,0,2*pi, algorithm=alg)] for alg in
("maxima", "giac", "sympy", "fricas", "mathematica_free")])
maxima 2*I*pi
giac (2*I + 1)*pi
sympy (2*I + 1)*pi
fricas (2*I + 1)*pi
mathematica_free 3.14159000000000 + 6.28319000000000*I
HTH,
It can be easily with sagemath 9.4 in https://sagecell.sagemath.org
>
> Thanks in advance,
>
> Yours,
>
> Juan Luis Varona
>
>
>
>
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