Verify analytical expression: derivative of an integral with undefined function
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Emilio Aguilera Valdes
el 17 de Ag. de 2022
Comentada: Emilio Aguilera Valdes
el 19 de Ag. de 2022
I’m trying to verify an analytical expression i did by hand. The expression is:
What i think is that the result is just the integral argument evaluated in T, but i’m really not sure. I tried to use diff and int, but i don’t know how to write the function Phi(t).
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Torsten
el 17 de Ag. de 2022
As written, the result is 0 since you differentiate with respect to t, not T.
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John D'Errico
el 17 de Ag. de 2022
Editada: John D'Errico
el 17 de Ag. de 2022
You define a function of a variable as I do here:
syms n omega phi(t) T
diff(int(phi*exp(-i*n*omega*t),[0,T]),T)
Looks like you were pretty close after all, but I think you knew that. :)
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Walter Roberson
el 19 de Ag. de 2022
Editada: Walter Roberson
el 19 de Ag. de 2022
The integral with respect to t will not have any remaining t in it. When you differentiate that with respect to t you will get 0. You would need to have t as a bound of the integral for the derivative to be non-zero
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