I'm trying to find Green's function for $ nabla ^ 2 phi = S (x) $ for $ x in mathbb {R ^ 3} $ and express the general solution to the Laplace equation in $ mathbb {R ^ 3} $.

To find the function of green, I considered

begin {align}

nabla ^ 2G & = delta left ( underline {x} – underline {x & # 39;} right) \

nabla ^ 2G & = 0 left ( text {when} underline {x} neq underline {x & # 39;} right) \

nabla ^ 2G & = frac {1} {r ^ 2} frac {d} {dr} left (r ^ 2 frac {DG} {dr} right) \

G (r) & = – frac {A} {r} + B, A, B in mathbb {R}.

end {align}

Then consider a ball centered on $ underline {x & # 39;} $ with radio $ a $denoted $ B_a $. So,

begin {align}

iiint_ {B_a} nabla ^ 2G dV & = iiint_ {B_a} delta left ( underline {x} – underline {x & # 39;} right) dV \

iint _ { delta B_a} frac {dG} {dr} bigg | _ {r = a} ds & = 1 text {(divergence theorem)} \

4 pi A & = 1 \

implies A & = frac {1} {4 pi}

end {align}

Therefore (leaving $ B = 0 $ without loss of generality), $$ G (r) = – frac {1} {4 pi r}. $$ Now, $ r = | underline {x} – underline {x & # 39;} | $, meaning $$ G ( underline {x}, underline {x & # 39;}) = – frac {1} {4 pi | underline {x} – underline {x & # 39;} |}. $$

Why can the general solution be expressed as $$ phi ( underline (x)) = – frac {1} {4 pi} iiint _ { mathbb {R ^ 3}} frac {1} {| underline {x} – underline {x & # 39;} |} S ( underline {x & # 39;}) dV & # 39 ;? $$