How to compute the marginal CDF of a joint density? I am trying to compute the cumulative distribution function of a random  variable $u$ that has the following density:
$$f(u) = \int_{1}^\infty \frac{e^{-4uv}}{v^5}dv$$
for $u \gt  0$.
What I've tried
I've tried computing this integral, giving a function $f(u)$, and then calculating $\int_{-\infty}^x f(u) du$, obtaining the CDF of the density. But I always get some weird results from this integral. I know that there must be a better way to solve this using the fact that this is a joint density, but I don't know how to do it. Can someone please give me any hint or advice?

Warning:  this answer/approach lacks mathematical rigor and relies on Mathematica to do the heavy lifting.
From looking at what is given for the marginal density for $U$
$$f(u)=\int_1^\infty \frac{e^{-4 u v}}{v^5} dv$$
one can surmise that the joint pdf for $U$ and $V$ is proportional to $\frac{e^{-4 u v}}{v^5}$.  I say "proportional" because the integral over the values of $U$ and $V$ integrates to 1/20 and not to 1.  Therefore a multiplier of 20 is needed.
Integrate[Exp[-4 u v]/v^5, {v, 1, \[Infinity]}, {u, 0, \[Infinity]}]
(* 1/20 *)

So the joint pdf is $\frac{20 e^{-4 u v}}{v^5}$.  Integrating over $v$ gets the pdf for $U$:
Integrate[20 Exp[-4 u v]/v^5, {v, 1, \[Infinity]}, Assumptions -> u > 0]

$$\frac{5}{3} \left(e^{-4 u} \left(3-4 u \left(8 u^2-2 u+1\right)\right)-128 u^4 \text{Ei}(-4 u)\right)$$
where $\text{Ei}$ is the exponential integral function.  The cdf is then found with
Integrate[5/3 (E^(-4 u) (3 - 4 u (1 - 2 u + 8 u^2)) - 128 u^4 ExpIntegralEi[-4 u]), {u, 0, z}, 
  Assumptions -> z > 0] /. z -> u

$$\frac{1}{3} \left(-128 u^5 \text{Ei}(-4 u)+e^{-4 u} \left(u \left(3-4 u \left(8 u^2-2 u+1\right)\right)-3\right)+3\right)$$
Here are plots of the pdf and cdf:


Update
I followed @whuber 's suggestion below to treat the pdf symbolically using the general form in his answer to see if Mathematica would then produce a simpler form for the pdf and cdf.  And it did.
pdf = \[Theta] p Integrate[v^(-p) Exp[-\[Theta] u v], {v, 1, \[Infinity]}, 
  Assumptions -> u > 0 && \[Theta] > 0 && p > 0] // FunctionExpand

$$\theta  p (\theta  u)^{p-1} \Gamma (1-p,u \theta )$$
cdf = Simplify[Integrate[pdf, {u, 0, w}, 
 Assumptions -> \[Theta] > 0 && p > 0 && w > 0] /. w -> u // FunctionExpand, 
 Assumptions -> u > 0 && \[Theta] > 0 && p > 0]

$$(\theta  u)^p \Gamma (1-p,u \theta )-e^{\theta  (-u)}+1$$
where $\Gamma (1-p,u \theta )$ is the incomplete gamma function.
