Loosely speaking, if such a function G can be found for the operator L, then, if we multiply the equation 1 for the Green's function by f (s), and then integrate with respect to s, we obtain, ∫ ∫ L G ( x , s ) f ( s ) d s = ∫ ∫ δ δ ( x − − s ) f ( s ) d s = f ( x ) . Because the operator L = L ( x ) is linear and acts only on the variable x (and not on the variable of integration s), one may take the operator L outside of the integration, yielding L ( ∫ ∫ G ( x , s ) f ( s ) d s ) = f ( x ) . This means that