Fourier transform equation question

[Fosheimdet]

In my QFT homework I was asked to prove that $$\int d^3x \int \frac{d^3k}{(2\pi)^3} e^{i\mathbf{k} \cdot (\mathbf{x} - \mathbf{y})} k_j f(\mathbf{x}) = i \frac{df}{dx_j}(\mathbf{y}) $$

Using $\frac{\partial e^{i\mathbf{k} \cdot (\mathbf{x} - \mathbf{y})}}{\partial x^j} = i k_j e^{i\mathbf{k} \cdot (\mathbf{x} - \mathbf{y})}$, the solution is as follows

$$\int d^3x \int \frac{d^3k}{(2\pi)^3} e^{i\mathbf{k} \cdot (\mathbf{x} - \mathbf{y})} k_j f(\mathbf{x}) = -i \int d^3x \int \frac{d^3k}{(2\pi)^3} \frac{\partial e^{i\mathbf{k} \cdot (\mathbf{x} - \mathbf{y})}}{\partial x^j} f(\mathbf{x}) $$

$$ = -i \int d^3x \int \frac{d^3k}{(2\pi)^3} \left( \frac{\partial}{\partial x^j} (e^{i\mathbf{k} \cdot (\mathbf{x} - \mathbf{y})}f(\mathbf{x}) ) - \frac{\partial f(\mathbf x)}{\partial x^j} e^{i\mathbf k \cdot (\mathbf x - \mathbf y)}\right) $$

$$ = i \int d^3x \frac{\partial f(\mathbf x) }{\partial x^j} \int \frac{d^3k}{(2\pi)^3} e^{i\mathbf k \cdot (\mathbf x - \mathbf y)} $$

$$= i \int d^3x \frac{\partial f(\mathbf x)}{\partial x^j} \delta^{(3)}(\mathbf x - \mathbf y) = i \frac{\partial f(\mathbf x)}{\partial x^j} \bigg|_{\mathbf x = \mathbf y}$$

I don't understand what happened to the first term on the RHS of the second equation. Why is $-i\int d^3x \int \frac{d^3k}{(2\pi)^3} \frac{\partial}{\partial x^j} (e^{i\mathbf{k} \cdot (\mathbf{x} - \mathbf{y})}f(\mathbf{x}))=0?$

 

[Orodruin]

It is a total derivative and the function is assumed to go to zero sufficiently fast at infinity.

 

[Fosheimdet]

It is a total derivative and the function is assumed to go to zero sufficiently fast at infinity.

How is it a total derivative though? Isn't $\frac {\partial}{\partial x^j}$ a partial derivative? Also, why can you make the assumption that the function goes to zero at infinity?

 

[Orodruin]

Being a total derivative does not prevent being a partial derivative. It just means that the integrand is a derivative of some function.

An assumption is an assumption. We typically choose to work with such functions. It is related to the domain of the Fourier transform.