1
$\begingroup$

Let's say that $F$ is a nice well-behaved function. How would I compute the following derivative?

$$\frac{\partial}{\partial x} \left\{ \int\limits_0^t \int\limits_{x - t + \eta}^{x + t - \eta} F(\xi,\eta) \,d\xi \,d\eta \right\}$$

This is what I have so far:

$$= \int_0^t \frac{\partial}{\partial x} \left\{ \int\limits_{x - t + \eta}^{x + t - \eta} F(\xi,\eta) \, d\xi \right\} \, d\eta $$

$$= \int_0^t \left\{ F(x - t + \eta, \eta) + F(x + t - \eta, \eta) \right\} \, d\eta $$

$$= \int_0^t F(x - t + \eta, \eta) \,d\eta + \int_0^t F(x + t - \eta, \eta) \, d\eta $$

Is this correct? Or am I missing a minus sign in there?

$\endgroup$
1
$\begingroup$

Since $$\frac{d}{dx}\int_a^x f(x) \, dx=f(x)$$ and $$\frac{d}{dx}\int_x^a f(x) \, dx=-\frac{d}{dx}\int_a^x f(x)=-f(x)$$ and similarly $$\frac{d}{dx}\int_a^{g(x)} f(x) \, dx=f(x)g'(x)$$

You should have, $$\frac{\partial}{\partial x} \left\{ \int\limits_{x - t + \eta}^{x + t - \eta} F(\xi,\eta) d\xi \right\}= \frac{\partial}{\partial x} \left\{ \int\limits_{x - t + \eta}^{0} F(\xi,\eta) d\xi + \int\limits_{0}^{x + t - \eta} F(\xi,\eta) d\xi \right\}\\ =-F(x - t + \eta,\eta) + F(x + t - \eta,\eta)= F(x + t - \eta,\eta)-F(x - t + \eta,\eta) $$ and therefore $$\frac{\partial}{\partial x} \left\{ \int\limits_{0}^{t} \int\limits_{x - t + \eta}^{x + t - \eta} F(\xi,\eta) d\xi d\eta \right\} =\int_0^t F(x + t - \eta,\eta)-F(x - t + \eta,\eta)d\eta$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.