This is something that every text book or notes skips to explain in the derivation of the LSZ reduction formula
Suppose we have $$ a_{1}^{\dagger} \equiv \int d^{3} k f_{1}(\mathbf{k}) a^{\dagger}(\mathbf{k})\tag{5.6} $$ where $$ f_{1}(\mathbf{k}) \propto \exp \left[-\left(\mathbf{k}-\mathbf{k}_{1}\right)^{2} / 4 \sigma^{2}\right]\tag{5.7} $$ is an appropriate wave packet, and $\sigma$ is its width in momentum space.
In the derivation of LSZ reduction formula at certain point in this notes Quantum Field Theory by Mark Srednicki at page 50 they have $$\begin{aligned} &-i \int d^{3} k f_{1}(\mathbf{k}) \int d^{4} x e^{i k x}\left(\partial_{0}^{2}-\overleftarrow{\nabla}^{2}+m^{2}\right) \phi(x) \\ &=-i \int d^{3} k f_{1}(\mathbf{k}) \int d^{4} x e^{i k x}\left(\partial_{0}^{2}-\overrightarrow{\nabla}^{2}+m^{2}\right) \phi(x) \end{aligned}\tag{5.10}$$ He claims that here the wave packet is needed to avoid a surface term but I am not seeing how. I am using this to proof it \begin{align*} \int f(y)g''(y)dy &= f(y)g'(y)| - \int g'(y)f'(y)dy\\ \int g(y)f''(y)dy &= g(y)f'(y)| - \int f'(y)g'(y)dy\quad\text{so}\\ \int f(y)g''(y)dy - \int g(y)f''(y)dy &= f(y)g'(y)| - g(y)f'(y)|.\\ \end{align*}
If we take the $x$ part of the Laplacian we would have \begin{align*} \int^{+\infty}_{-\infty}{dr^3 \left(\partial_x^2 e^{-ikr}\right)\phi}=\int^{+\infty}_{-\infty}{dr^3 \ e^{-ikr} \partial_x^2\phi}+\left[e^{-ik_xx}\right]^{\infty}_{-\infty}\int^{+\infty}_{-\infty}{dydze^{-ik_yy-ik_zz}\partial_x\phi}+\int^{+\infty}_{-\infty}{dydzik_xe^{-ikr}\left[\phi\right]^{x=\infty}_{x=-\infty}}. \end{align*}
Now if we assume that $\lim_{r\rightarrow \infty}{\phi}=\lim_{r\rightarrow -\infty}{\phi}=0$ we have that $$\int^{+\infty}_{-\infty}{dydzik_xe^{-ikr}\left[\phi\right]^{x=\infty}_{x=-\infty}}=0.$$
But what about the term
$$\left[e^{-ik_xx}\right]^{\infty}_{-\infty}\int^{+\infty}_{-\infty}{dydze^{-ik_yy-ik_zz}\partial_x\phi}.$$
why it is zero?
