Verify analytical expression: derivative of an integral with undefined function

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).

[Torsten]

As written, the result is 0 since you differentiate with respect to t, not T.

Answer 1

[John D'Errico]

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)

ans = 

Looks like you were pretty close after all, but I think you knew that. :)

[Emilio Aguilera Valdes]

I notice you differentiate with respect to T, i corrected my picture to enphasize that the integral is with respect to t, this changes de syms part? Something like:

syms n omega phi(t) t
diff(int(phi*exp(-i*n*omega*t),[0,T]),T)

[Torsten]

The integral gives a function of T. Differentiating the function with respect to a variable that does not appear in the function, namely t, gives 0 as I already stated above as a comment.

syms n omega phi(t) T
diff(int(phi*exp(-i*n*omega*t),[0,T]),t)

ans = 

0

[Emilio Aguilera Valdes]

I did not specify that T is not a variable, is a constant.

[Torsten]

Imagine you have a function

f(t) = t^2

Now you integrate this function in the limits [0,T]:

integral_{t=0}^{t=T} f(t) dt = 1/3*T^3.

Now you differentiate this expression with respect to t:

d/dt ( 1/3*T^3 )  = 0

since the value of the integral depends on T, not t.

[Walter Roberson]

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

[Emilio Aguilera Valdes]

Perfect! Thank you very much.