# Thread: Integral and Integrand Notation

1. ## Integral and Integrand Notation

Hi All,

I have a query regarding the attached jpg. I am not sure I understand how the LHS = to RHS.

The book states that the 'integral' is a function of t only so a total derivative is used on LHS, but the 'integrand' is a function of x and t, therefore a partial derivative is used on the RHS.

I have marked in red what I understand to be the integral and integrand.

1) How is the integral a function of t, isnt it a function of x because of the dx in the integral?

Thanks

2. Originally Posted by bugatti79
Hi All,

I have a query regarding the attached jpg. I am not sure I understand how the LHS = to RHS.

The book states that the 'integral' is a function of t only so a total derivative is used on LHS, but the 'integrand' is a function of x and t, therefore a partial derivative is used on the RHS.

I have marked in red what I understand to be the integral and integrand.

1) How is the integral a function of t, isnt it a function of x because of the dx in the integral?
You have indicated the integral and the integrand correctly. The integral is not a function of x because it is a definite integral, which means that the x is a dummy variable that does not feature in the result of the integral. To take a simple example, $\int_0^1x\,dx = \frac12$. The integral there is not a function of x because it is just a constant, 1/2. When you integrate a function of two variables, such as $\int_{-\infty}^{\infty}|\psi(x,t)|^2\,dx$, the integral is a function of t, but because it is a definite integral, it is not a function of x.

There is a theorem which says that you can push the operation of differentiation with respect to t past the integral sign. But when you do so, it changes from an ordinary differentiation to a partial differentiation.

3. Originally Posted by Opalg
You have indicated the integral and the integrand correctly. The integral is not a function of x because it is a definite integral, which means that the x is a dummy variable that does not feature in the result of the integral. To take a simple example, $\int_0^1x\,dx = \frac12$. The integral there is not a function of x because it is just a constant, 1/2. When you integrate a function of two variables, such as $\int_{-\infty}^{\infty}|\psi(x,t)|^2\,dx$, the integral is a function of t, but because it is a definite integral, it is not a function of x.

There is a theorem which says that you can push the operation of differentiation with respect to t past the integral sign. But when you do so, it changes from an ordinary differentiation to a partial differentiation.
I see, that makes perfect sense to me now. Perhaps if it was qouted that the 'resulting' integral on the LHS is a function of t I might have spotted it. :-)
So when you have a definite integral for a function of 2 variables x and t to be integrated wrt to t, the resulting integral is a function of x only.

Thanks opalg