Why isn't the minus sign cancelled in this change of variables? This integral comes from fluid dynamics, specifically it's the fluid velocity when said fluid is subject to small perturbations. Regardless, my question is mathematical, and probably trivial, but I'm kind of scratching my head at it.
Consider:
$$\vec{v}(\vec{x},t)=-\hat{n}\frac{1}{\rho_0}\int\frac{dP'(\xi)}{d\xi}dt \ ; \ \xi=\hat{n}\cdot\vec{x}-ut$$
My notes state that, therefore:
$$\vec{v}(\vec{x},t)= -\hat{n}\frac{1}{\rho_0u}\int\frac{dP'(\xi)}{d\xi}d\xi=-\hat{n}\frac{1}{\rho_0u}P'(\hat{n}\cdot \vec{x}-ut)$$
But it seems to me that the change of variables should be $d\xi=-udt\rightarrow dt=-d\xi/u$ and that therefore the solution should be the same, but without the minus sign. Why isn't this the case?
 A: The source of the minus sign becomes more clear if you leave the limits of integration explicit. You are correct that the sign should go away after the initial change of variables:
$$
-\hat{n}\frac{1}{\rho_0}\int_0^t\frac{dP'(\xi)}{d\xi}dt' = \hat{n}\frac{1}{\rho_0u}\int_x^{x-ut}\frac{dP'(\xi)}{d\xi}d\xi.
$$
However, if we are considering $u > 0$, the upper limit is smaller than the lower limit. This may be undesirable, and can be fixed by swapping the limits of integration, which introduces a minus sign:
$$
-\hat{n}\frac{1}{\rho_0}\int_0^t\frac{dP'(\xi)}{d\xi}dt' = \hat{n}\frac{1}{\rho_0u}\int_x^{x-ut}\frac{dP'(\xi)}{d\xi}d\xi= - \hat{n}\frac{1}{\rho_0u}\int_{x-ut}^x\frac{dP'(\xi)}{d\xi}d\xi.
$$
Note that having or not having the minus sign can be correct, provided that you use the correct limits of integration. Leaving integrals as indefinite leads to confusion if you're not careful.
Also, in general, the sign in the change of variables doesn't matter for definite integrals. If it's negative, you end up having to swap the limits of integration, which cancels the minus sign. This is also essentially why in multivariate integrals, the absolute value of the Jacobian determinant is used for the volume element rather than the signed value.
