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?
 A: More generally, consider the distribution whose density is
$$f_p(u, \theta) = \theta C_p\int_1^\infty v^{-p} e^{-\theta u v}\,\mathrm{d}v = \theta C_pE_p(\theta u)$$
for $u \gt 0$ and parameters $p\gt 0,$ $\theta \gt 0.$ $E_p$ is the Exponential Integral function and $C_p$ is a normalizing constant (which we will find at the end).  Notice, for future reference, that
$$E_{p+1}(0) = \int_1^\infty v^{-(p+1)}\,\mathrm{d}v = \frac{1}{p}.$$
Moreover, since for any $z\gt 0$
$$
0\le E_p(z) = \int_1^\infty v^{-p} e^{-zv}\,\mathrm{d}v\le  \int_1^\infty e^{-zv}\,\mathrm{d}v = \frac{1}{z},$$
it follows from the Squeeze Theorem (for limits) that
$$\lim_{z\to\infty} E_p(z) = 0.$$
The distribution function (cdf) is, by definition,
$$F_p(u,\theta) = \int_0^u f_p(z)\,\mathrm{d}z =  C_p \int_0^u  E_p(\theta z)\,\theta\mathrm{d}z =  C_p \int_0^{\theta u}  E_p(z)\,\mathrm{d}z.$$
The integrand defining $E_p$ is smooth and decreasing so quickly at its upper limit that we may interchange the operations of differentiating and integrating, showing
$$\begin{aligned}
\frac{\mathrm{d}}{\mathrm{d}z}E_{p+1}(z) &= \frac{\mathrm{d}}{\mathrm{d}z}\int_1^\infty v^{-(p+1)} e^{-zv}\,\mathrm{d}v\\&=\int_1^\infty  \frac{\mathrm{d}}{\mathrm{d}z}v^{-(p+1)} e^{-zv}\,\mathrm{d}v\\&=-\int_1^\infty v^{-p} e^{-zv}\,\mathrm{d}v\\&=-E_p(z).
\end{aligned}$$
It is immediate (by the Fundamental Theorem of Calculus) that
$$F_p(u,\theta) = C_p \left[E_{p+1}(0) - E_{p+1}(\theta u)\right] = C_p \left[\frac{1}{p} - E_{p+1}(\theta u)\right].$$
Finally, since $\lim_{u\to\infty}F_p(u,\theta)=1$ (by the Law of Total Probability), we find
$$1 = \lim_{u\to\infty}C_p \left[\frac{1}{p} - E_{p+1}(\theta u)\right] = \frac{C_p}{p},$$
showing $C_p = p$ and giving

$$F_p(u,\theta) = 1 - pE_{p+1}(\theta u).$$

Set $p=5$ and $\theta=4$ for the answer to the question.  Here are graphs of the density and its cdf:

These were drawn in gray using the analytical solution.  As a check, over them are plotted, in red, the values obtained from numerically computing the single integral for $f$ and the double integral for $F.$  The R code that generated these graphs follows.
#
# Solution.
#
library(expint)
f <- function(u, p, theta) theta * p * expint_En(theta * u, p)
F <- function(u, p, theta) 1 - p * expint_En(theta * u, p+1)
#
# Brute force verification using numerical integration.
#
f. <- Vectorize(function(u, p, theta, ...) {
  theta * p * integrate(function(v) exp(-theta * u * v) / v^p, 1, Inf, ...)$value
}, "u")
F. <- Vectorize(function(u, p, theta, ...) {
  integrate(function(z) f.(z, p, theta), 0, u, ...)$value
}, "u")
#
# Example.
#
p <- 5
theta <- 4
par(mfrow=c(1,2))
curve(f(x, p, theta), 0, 1.5, n=501, lwd=2, col=gray(0.25),
      xlab="u", ylab="Density", 
      main=bquote(f[.(p)](u, .(theta))))
curve(f.(x, p, theta), add = TRUE, col="Red", lwd=2, lty=2)

curve(F(x, p, theta), 0, 1.5, n=501, lwd=2, col=gray(0.25),
      yaxp=c(0,1,1),
      xlab="u", ylab="Probability", 
      main=bquote(F[.(p)](u, .(theta))))
curve(F.(x, p, theta), add = TRUE, col="Red", lwd=2, lty=2)
par(mfrow=c(1,1))

A: 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.
