joint densities and transformation of random variables

Given the joint density of $x,y$ as $f(x,y) = \frac{2}{x^3y^2}$ where $x\geq 1$ and $y\geq 1$. I am interested in $u=\frac{x}{y}$

Define: $V= Y $

Such that we have $$ Y=g_2(U,V)=V \,,\, X=g_1(U,V)= UV$$

Thus: $$ J = \begin{vmatrix} \frac{d g_1}{dU} & \frac{d g_1}{dV} \\[6pt] \frac{d g_2}{dU} & \frac{dg_2}{dV} \end{vmatrix} = \begin{vmatrix} \frac{d (UV)}{dU} & \frac{d (UV)}{dV} \\[6pt] \frac{d (V)}{dU} & \frac{d (V)}{dV} \end{vmatrix} = \begin{vmatrix} V & U \\[6pt] 0 & 1 \end{vmatrix} = V $$

$$ h(u,v) = f(g_1(u,v), g_2(u,v )) \cdot \mid{v }\mid =f(g_1(u,v), g_2(u,v )) \cdot v = \frac{2v}{ {u^3}{v^3}\cdot v^2 }=\frac{2}{ {u^3}{v^4} } $$

Densities:

$$h(u)= \mathbf{1}_{0<u<1} \int_{\frac{1}{u}}^{\infty} h(u,v)\, dv + \mathbf{1}_{u\geq 1} \int_{u}^{\infty} h(u,v)\, dv $$

for $\mathbf{1}_{0<u<1}$:

$$ \int_{\frac{1}{u}}^{\infty} \frac{2}{ {u^3}{v^4}} \,dv= \frac{2}{3}$$

for $\mathbf{1}_{u\geq 1}$:

$$ \int_{u}^{\infty} \frac{2}{{u^3}{v^4}} \,dv = \frac{2}{3u^6}$$

$$ h(u)= \begin{cases} \frac{2}{3} ,& 0<u<1 \\ \frac{2}{3u^{6}}, & u \geq 1 \end{cases} $$

I observe that if I integrate $h(u)$ over the regions as above, I am not getting one. Does it mean my approach is wrong? I suppose integrating over the entire domain (v then u, in this case) should give 1.

$1\leqslant X, 1\leqslant Y$, $U=X/Y, V=Y$

Then $1\leq V , 1/V\leqslant U$

So the transformed joint density is indeed: $$\begin{align}h(u,v) &=\lvert v\rvert~f(uv,v) \\[2ex]&=\dfrac{2}{u^3v^4}\mathbf 1_{1\leqslant v~,~v^{-1}\leqslant u}\\&=\dfrac{2}{u^3v^4}\mathbf 1_{(0\leq u\lt 1~,~u^{-1}\leqslant v)~\lor~(1\leqslant u~,~1\leqslant v)}\end{align}$$

And the marginals are therefore:$$\begin{align}h_U(u) &=\dfrac{2}{u^3}\mathbf 1_{0\lt u\lt 1}\int_{1/u}^\infty \dfrac{1}{v^4}\,\mathrm d v+\dfrac{2}{u^3}\mathbf 1_{1\leqslant u}\int_{1}^\infty\dfrac{1}{v^4}\,\mathrm dv\\[1ex]&=\dfrac 23\,\mathbf 1_{0\lt u\lt 1}+\dfrac{2}{3u^3}\,\mathbf 1_{1\leqslant u}\\[3ex]h_V(v)&=\dfrac{2}{v^4}\mathbf 1_{1\leqslant v}\int_{1/v}^\infty\dfrac{1}{u^3}\,\mathrm du\\[1ex]&=\dfrac{1}{v^2}\,\mathbf 1_{1\leqslant v}\end{align}$$

And in both cases $$\begin{align}\int_0^1\dfrac 23\,\mathrm d u+\int_1^\infty\dfrac 2{3u^3}\,\mathrm d u &=1\\[2ex]\int_1^\infty\dfrac 1{v^2}\,\mathrm d v &= 1\end{align}$$