Suppose $f(x,y)$ has double Fourier series, find Fourier series of $\Delta f$

Question

Suppose $f(x,y)$ has double Fourier series $\sum a_{n1n2} e^{in_1 x} e^{in_2 y}$. Then I have

$$\Delta f(x,y) = \frac{\partial}{\partial x^2} f + \frac{\partial}{\partial y^2}f$$

$$=\frac{\partial}{\partial x^2}\sum a_{n1n2} e^{in_1 x} e^{in_2 y} + \frac{\partial}{\partial y^2}\sum a_{n1n2} e^{in_1 x} e^{in_2 y} $$

$$= \frac{\partial}{\partial x} \sum e^{in_2 y} in_1a_{n1n2} e^{in_1x} + \frac{\partial}{\partial y} \sum e^{in_1 x} in_2 a_{n1 n2} e^{in_2 y}$$

$$=\sum -n_1^2 a_{n1 n2} e^{in_1 x} e^{i n_2 y} + \sum - n_2^2 a_{n1 n2} e^{in_1 x}e^{in_2 y}$$

But the answer should be $\sum (n_1^2+n_2^2) a_{n1 n2} e^{in_1 x} e^{in_2 y}$. I am not sure where I messed up? I feel like I did mess up because the hint says "integration by parts is the thing here" yet I don't even see integral here.

Answer 1

Since no one answers my question, I will post my own answer here. The reason why we might want to use integration by parts is that we need to justify the step where we simply pull down the $-in_1$ and $-in_2$. In other words, we want to show that the Fourier coefficient of derivative of $f$ is just the derivative of Fourier coefficient of $f$.