I'm dealing with the following integral:
$$f(x,y,z) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} [a(k_1,k_2)b(x,k_1,k_2) e^{-i k_2 y}] e^{i k_1 (cz-x)} dk_1 dk_2 $$
Here, $c$ is a constant, while $a$ and $b$ are functions.
My approach has been using integration by parts where term1 is all the terms within square bracket '[]', while term2 is $e^{i k_1 (cz-x)}$.
Thus we can rewrite the expression above as :
$$ \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} [a(k_1,k_2)b(x,k_1,k_2) e^{-i k_2 y}] e^{i k_1 (cz-x)} dk_1 dk_2 \\ = \int_{-\infty}^{+\infty} [a(k_1,k_2)b(x,k_1,k_2) e^{-i k_2 y}] (\int_{-\infty}^{+\infty} e^{i k_1 (cz-x)} dk_1) dk_2 \\ - \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} (\int_{-\infty}^{+\infty} e^{i k_1 (cz-x)} dk_1) \frac{\partial [a(k_1,k_2)b(x,k_1,k_2) e^{-i k_2 y}]}{\partial k_1} dk_1 dk_2 $$
Knowing: $\delta(x-\alpha) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} e^{ip(x-\alpha)} dp$, I write the integral above as:
$$ \int_{-\infty}^{+\infty} [a(k_1,k_2)b(x,k_1,k_2) e^{-i k_2 y}] (\int_{-\infty}^{+\infty} e^{i k_1 (cz-x)} dk_1) dk_2 \\ - \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} (\int_{-\infty}^{+\infty} e^{i k_1 (cz-x)} dk_1) \frac{\partial [a(k_1,k_2)b(x,k_1,k_2) e^{-i k_2 y}]}{\partial k_1} dk_1 dk_2 \\ = \int_{-\infty}^{+\infty} [a(k_1,k_2)b(x,k_1,k_2) e^{-i k_2 y}] (2\pi \delta(cz-x) ) dk_2 \\ - \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} (2\pi \delta(cz-x) ) \frac{\partial [a(k_1,k_2)b(x,k_1,k_2) e^{-i k_2 y}]}{\partial k_1} dk_1 dk_2 $$
Pretty sure there is a fundamental mistake here ... the first term of the expression above now has variable $k_1$ outside the integral ??
