Integrating Over Two Variables: Deriving an Expression for dP/dt

[WWCY]

Homework Statement

Hi everyone, I'd appreciate it if someone could help look through my working and check if it makes sense!

I have the following integral:
$$P(t) = \int_{-\infty}^{a} \int_{-\infty}^{-\infty} f_p(p) \ f_{x} (x - pt/m) \ dp \ dx$$
I want to find an expression for $\frac{dP(t)}{dt}$, so I first "bring in" the integral over $x$.
$$P(t) = \int_{-\infty}^{-\infty} f_p(p) \Big [ \int_{-\infty}^{a} f_x (x - pt/m) \ dx \Big ] \ dp$$
Making the substitution of $u = x - pt/m$
$$\frac{d}{dt} \int_{-\infty}^{a} f_x (x - pt/m) \ dx = \frac{d}{dt}\int_{-\infty}^{a - pt/m} f_x (u) \ du = -\frac{p}{m} f_x(a - pt/m)$$
and finally,
$$\frac{dP}{dt} = -\frac{1}{m} \int_{-\infty}^{\infty} f_p (p) . p . f_x (a - pt/m) \ dp$$
Is this right? Many thanks in advance.

Homework Equations

The Attempt at a Solution

 

[jambaugh]

You can also differentiate directly. In general:
$$\frac{d}{dt} \int^{b(t)}_{a(t)} f(x,t)dx = f(b(t),t) b'(t) - f(a(t),t) a'(t) + \int^{a}_b \frac{\partial}{\partial t} f(x,t)dt$$
In your case there is (originally) no $t$ dependency in the limits so:

Let's see...
$$\frac{d}{dt}P(t) = \frac{d}{dt} \int_{-\infty}^{a} \int_{-\infty}^{\infty} f_p(p) \ f_{x} (x - pt/m) \ dp \ dx =$$
$$= \int_{-\infty}^{a} \int_{-\infty}^{\infty} f_p(p) \ f'_{x} (x - pt/m)\frac{-p}{m} \ dp \ dx$$
changing the order of integration:
$$= \int_{-\infty}^{\infty}\frac{-p}{m}f_p(p) \int_{-\infty}^{a}\ f'_{x} (x - pt/m) \ dx \ dp$$
and applying the FTC in its other form (using u substitution for the indefinite integral $du = dx$ and returning to $x$ for the evaluation):
$$= \int_{-\infty}^{\infty}\frac{-p}{m}f_p(p) \ f_{x} (a - pt/m) dp$$

so we agree.

I had some concern with your change of variable which has a $t$ dependency before executing the derivative which is why I worked it out in this alternative fashion. That's something I should meditate on for a bit to be sure if and why it is okay.

 

[WWCY]

Thanks a lot for your time!