Second time derivative and laplacian of solution of Cauchy Problem for wave equation

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I am trying to verify that Kirchoff formula solves the Cauchy Problem for wave equation in three dimensions.

However I am having some trouble with the derivatives. The Kirchoff formula is given by $$ u(x,t) = \frac{1}{4 \pi}\int_{S_1(0)} f(x+ty)d\sigma(y) + \frac{t}{4\pi}\int_{S_1(0)}g(x+ty)d\sigma(y)+\frac{t}{4\pi}\int_{S_1(0)}\vec{\nabla}f(x+ty). yd\sigma(y) $$

When I calculate the second time derivative and the laplacian it doesn't fit the wave equation $$\frac{\partial^2 u}{\partial t^2}-\Delta_x u = 0$$ I appreciate any help.

My calculations are bellow

$$ \frac{\partial u}{\partial t} = \frac{1}{4\pi}\int_{S_1(0)}\vec{\nabla} f(x+ty).y d\sigma(y) + \frac{1}{4\pi}\int_{S_1(0)}g(x+ty) d\sigma (y) + \frac{t}{4\pi}\int_{S_1(0)}\vec{\nabla}g(x+ty).y d\sigma (y) + \frac{1}{4\pi}\int_{S_1(0)}\vec{\nabla}f(x+ty).y d\sigma (y) + \frac{t}{4\pi}\int_{S_1(0)}\Delta f(x+ty).y d\sigma (y) $$

$$ \frac{\partial^2 u}{\partial t^2} = \frac{1}{4\pi}\int_{S_1(0)}\Delta f(x+ty) d\sigma (y) + \frac{1}{4\pi}\int_{S_1(0)}\vec{\nabla}g(x+ty).y d\sigma (y) + \frac{1}{4\pi}\int_{S_1(0)}\vec{\nabla}g(x+ty).y d\sigma (y) + \frac{t}{4\pi}\int_{S_1(0)}\Delta g(x+ty) d\sigma (y) + \frac{1}{4\pi}\int_{S_1(0)}\Delta f(x+ty) d\sigma (y) + \frac{1}{4\pi}\int_{S_1(0)}\Delta f(x+ty) d\sigma (y) $$

$$ \Delta_x u = \frac{1}{4\pi}\int_{S_1(0)}\Delta f(x+ty) d\sigma(y) + \frac{t}{4\pi}\int_{S_1(0)}\Delta g(x+ty) d\sigma(y) $$