I was wondering if this has a name (I was taking the Gamma distribution and tried to create a 2D with "wiggly-tail" version of it). And also if there is a way to solve it? I used numerical methods to estimate that the integral (i.e. normalizing-constant) should be around $20, 21$. Wolfram alpha puts it at $5\sqrt 2 \pi\approx 22.2144$.
$$\int_0^\infty \int _{-\infty}^\infty \frac{x e^{-x}}{0.2(\sin(1.5x)+y)^2+0.1}dy dx$$
In the integral over $y$ you can treat $x$ as a parameter and introduce a new integration variable $u=y + \sin(1.5x)$, $du=dy$ to get an integral $$ \int_0^\infty \Big(\int_{-\infty}^\infty \frac{xe^{-x}}{0.2 u^2+0.1} du\Big) dx = \Big( \int_0^\infty xe^{-x} dx\Big)\Big(\int_{-\infty}^\infty \frac{1}{0.2 u^2+0.1} du\Big)$$ Do you know how to calculate this?
$$\int\frac{e^{-x} x}{\frac{1}{5} \left(\sin \left(\frac{3 x}{2}\right)+y\right)^2+\frac{1}{10}}\,dy=5 \sqrt{2} e^{-x} x \tan ^{-1}\left(\sqrt{2} \left(\sin \left(\frac{3 x}{2}\right)+y\right)\right)$$ Assuming $x \geq 0$, then $$\int_{-\infty}^{+\infty}\frac{e^{-x} x}{\frac{1}{5} \left(\sin \left(\frac{3 x}{2}\right)+y\right)^2+\frac{1}{10}}\,dy=5 \pi\sqrt{2}\,x\, e^{-x} $$ One integration by parts to get $$5 \pi\sqrt{2}\,\int_0^p x\, e^{-x}\,dx=5 \pi\sqrt{2} \left(1- (p+1)e^{-p}\right) \to 5\pi \sqrt{2} $$
I'll just add that for the more general case:
$$ \frac{x^ae^{-bx}}{c (\sin(dx)+ey)^2 + f} $$ The integral (normalizing constant) turns to be: $$\frac{\Gamma(a+1)}{b^{a+1}}\frac{\pi}{e\sqrt{cf}} $$
Hence the PDF of this function could be written as:
$$f(x,y) = \frac{b^{a+1}}{\Gamma(a+1)}\frac{e\sqrt{cf}}{\pi}\frac{x^ae^{-bx}}{c (\sin(dx)+ey)^2 + f} $$