Conversion of summation to integration

Question

Generally when we write $ f=\int ydx$ we are calculating the area of the curve represented by the function y . So how can we calculate the sum of a set of functions such as $$f(z)=\sum_i {A(k_i)e^{(k_i z-wt)} } $$ by an integral like $$ f(z)=\int_{-\infty}^{\infty} {A(k)e^{(kz-wt) }dk} $$ Does this not change the physical dimensions of the function ??

Answer 1

This is a good question. The dimensions must indeed match.

If you think of $A()$ in the summation as giving the height of a rectangle (units length) then there is an implicit length $\Delta (i) = 1$ to get units of area for the answer. (The exponential factor is a unitless weight.)

In the integral $A(x)$ and $dx$ each have units length.

(Perhaps a physicist or applied mathematician will weigh in with a discussion of Fourier series compared to Fourier integrals.)

Answer 2

Now that I see it better, I think there is a typo and the first equation should be

$$ f(z)=\sum_i {A(k_i)e^{(k_i z-wt)} } $$

is that right?

if that is right, then the whole question is just a matter of using $k$ as a discrete of continuous variable. That depends on the interpretation of $k$ on the particular problem you are facing.