# Integration of a scalar function with respect to a vector

I have a scalar function that takes $n$ arguments, $f(x_1, x_2,x_n) = f(\mathbf{x})$, and two vectors also with $n$ elements, $\mathbf{z} = (z_1, z_2\cdots,, z_n)$, and $\Delta\mathbf{z} = (\Delta z_1, \Delta z_2,\cdots ,\Delta z_n)$. Basically, I want to write the integral:

$$\int_{\mathbf{z+\Delta z}}^{\mathbf{z}}f(\mathbf{x})d\mathbf{x} = \int_{z_1+ \Delta z_1}^{z_1}\int_{z_2+ \Delta z_2}^{z_2}f(x_1, x_2,\cdots,x_n)\, dx_1 \, dx_2\,...dx_n$$

but for an arbitrary number of elements/arguments. I believe this is a standard multiple integral, though I've never seen one written using vectors for the limits and differentials; surface and volume integrals are usually expressed by writing a single integral sign for each dimension. Note: $\mathbf{z}$ and $\Delta \mathbf{z}$ are always greater than or equal to zero in my application.

I would write something like $\int_{\prod_{i=1}^{n}[z_i+\Delta z_i,z_i]}f(x) dx$ or define $Q := \prod_{i=1}^{n}[z_i+\Delta z_i,z_i]$ and write $\int_{Q}f(x) dx$.

Judging from the RHS side of your integral I see believe you are trying to notate a standard multiple integral. In general it is fine to write $f(x)dx$ when $x$ is an element of $\mathbb{R}^n$. However, then care must be taken with how you write the "limits" of integration.

First, you don't want to specify a lower and upper limit of integration if you are integrating over a volume. It suggests some form of a path integral (which is even not so well defined unless the integral happens to be path independent). Instead, you want to say something like $$\int_V f(x) dx$$ where $V$ is the rectangle with lower end-point $z + \Delta z$ and upper end point $z$.

However, there is still an issue with this. Based on whether $\Delta z$ in the various dimensions are positive or negative, the sign of the integral may change. So it is probably best to state: $$g(\Delta z)\int_V f(x) dx$$ where $V$ is the rectangle with lower end-point $z$ and upper end point $z + \Delta z$ and $g(\Delta z)$ specifies the correct sign of the integral based on the signs of the $\Delta z$.

Given that $\Delta z$ is positive:

In that case, the impact on the sign of the of the integral is clear. Let $n$ be the dimension of the space. Then you would write: $$(-1)^n\int_V f(x) dx$$ where $V$ is the box ranging from $z$ to $z + \Delta z$.

• BTW, I edited my question to specify that $\Delta \mathbf{z}$ is always positive (or zero). I may be using more than three dimensions ($n>3$), in which case using $V$ to denote my limits of integration might be confusing. – Carlton Aug 28 '15 at 16:19
• @Carlton - I updated the answer given the simplification. It's standard to use a letter like $V$ to denote a region in space when performing a volume integral. You can use a different letter if you like. – muaddib Aug 28 '15 at 16:52