How to calculate the following double integral I would appreciate if you could help me to find the following integral, thank you.
$$f(u)= \int_{-\infty }^{\infty} \int_{-\infty }^{\infty} \frac{e^{ -(x-a)^2/2b^2} }{{b\sqrt {2\pi}}} \frac{e^{ -(y-c)^2/2d^2} }{{d\sqrt {2\pi}}} \delta (xy-u)  dx dy$$
where $\delta()$ is delta function and a, b, c , d are real numbers.
According to MathWorld when a and c are zero the answer would be
$$\frac {K_0 (\frac{|u| }{bd})}{ \pi bd}$$
where $K_0()$ is modified Bessel function.
Now what if a and c are not zero?
I simplified it as follows (not sure if it is correct)
$$f(u)=\frac1{2\pi bd} \int_{-\infty }^{\infty} \frac 1{|y|} e^{ -(\frac uy-a)^2/2b^2} e^{ -(y-c)^2/2d^2}     dy$$
 A: In this answer, it is shown that when composing the dirac delta with $g(x)$, we get
$$
\int_{\mathbb{R}^n} f(x)\,\delta(g(x))\,\mathrm{d}x=\int_{\mathcal{S}}\frac{f(x)}{|\nabla g(x)|}\,\mathrm{d}\sigma(x)
$$
where $\mathcal{S}$ is the surface on which $g(x)=0$ and $\mathrm{d}\sigma(x)$ is standard surface measure on $\mathcal{S}$.
In the given integral, $\mathrm{d}\sigma(x)$ is arclength and
$$
|\nabla g(x)|=|(y,x)|=\sqrt{x^2+y^2}
$$
If we parameterize the curve $xy=u$ by $y=u/x$, we get that
$$
\begin{align}
\mathrm{d}\sigma(x)
&=\sqrt{1+y'^2}\,\mathrm{d}x\\
&=\sqrt{1+\frac{u^2}{x^4}}\,\mathrm{d}x\\
\end{align}
$$
Therefore, the integral becomes
$$
\begin{align}
&\int_{-\infty}^\infty\frac{e^{ -(x-a)^2/2b^2} }{{b\sqrt {2\pi}}} \frac{e^{ -(u/x-c)^2/2d^2} }{{d\sqrt {2\pi}}}\sqrt{1+u^2/x^4}\frac{\mathrm{d}x}{\sqrt{x^2+u^2/x^2}}\\
&=\int_{-\infty}^\infty\frac{e^{ -(x-a)^2/2b^2} }{{b\sqrt {2\pi}}} \frac{e^{ -(u/x-c)^2/2d^2} }{{d\sqrt {2\pi}}}\frac{\mathrm{d}x}{|x|}
\end{align}
$$
which is what you got.
