Prove that $f(x)=exp(-x-e^{-x})$ for $x\in \mathbb{R}$ is a p.d.f and find the c.d.f. Prove that $f(x)=exp(-x-e^{-x})$ for $x\in \mathbb{R}$ is a probability density function and find the cumulative density function.

I think that by proving that $f(x)$ is a pdf, it should be fairly straightforward to find the cdf (if this is not true, would like a hint). I want to focus on the first part.
I am using the following definition.

A continuous random variable X is a RV whose c.d.f. satisfied
$F_X(x)=\mathbb{P}(X\leq x)=\int_{-\infty}^x f_X(u) du$
where $f_X:\mathbb{R} \rightarrow \mathbb{R}$ is a fct s.t. (a)$f_X(u) \geq 0$, $ \forall u \in \mathbb{R}$ and (b) $\int_{-\infty}^{\infty}f_X(u)du=1$
$f_X$ is called the p.d.f. of X

So my approach is to basicaly prove that $f(x)$ satisfies conditions (a) and (b).
Condition (a) seems fairly easy.
$exp(-x-e^{-x})=e^{-x-e^{-x}}=(\frac{1}{e})^{x+\frac{1}{e^x}}=(\frac{1}{e})^{x}(\frac{1}{e})^{\frac{1}{e^x}}$
Since $e$ is a positive constant, the function is also positive for all $x\in \mathbb{R}$. My question is how to prove the second condition, i.e. prove that
$\int_{-\infty}^{\infty}(\frac{1}{e})^{u}(\frac{1}{e})^{\frac{1}{e^u}}du=1$
My attempt so far:
Let $x=e^{-u}$
$dx=-e^{-u} \rightarrow -dx=e^{-u}$
$\int_{-\infty}^{\infty}e^{-u-e^{-u}}du=\int_{-\infty}^{\infty}e^{-u}e^{-e^{-u}}du=\int_{-\infty}^{\infty}-e^{-x}dx=e^{-x}|^{\infty}_{-\infty}=e^{-e^{-u}}|^{\infty}_{-\infty}$
What should I do from here?
 A: Hint:
$$
\frac{d}{dx} exp\left(-\mathrm{e}^{-x}\right) = exp\left(-x -\mathrm{e}^{-x}\right)
$$
A: a) Here it suffices to note that $e^{\text{something}}$ is always positive (for real something).
b) I don't think it is any idea to change back the variables (also, the limits in your $x$-integral are not correct), but since you did so, here is how to continue.
You should insert the limits, or rather
$$
\lim_{R\to+\infty}e^{-e^{-R}}-\lim_{r\to-\infty}e^{-e^{-r}}.
$$
It holds that
$$
\lim_{R\to+\infty}e^{-e^{-R}}=e^0=1,
$$
since $\lim_{R\to+\infty}e^{-R}=0$ and the exponential function is continuous at zero.
It also holds that
$$
\lim_{r\to-\infty}e^{-e^{-r}}=0,
$$
since $-e^{-r}$ tends to $-\infty$ and the exponential function of something going to minus infinity is going to zero.
To conclude, the value becomes
$$
\lim_{R\to+\infty}e^{-e^{-R}}-\lim_{r\to-\infty}e^{-e^{-r}}=1-0=1.
$$
A: The thing to do is to look at a variable transformation:  let $$Y = e^{-X}$$ where $$f_X(x) = e^{-x-e^{-x}}.$$  Then $$f_Y(y) = \frac{1}{y} f_X(-\log y) = \frac{1}{y} e^{\log y - e^{\log y}} = \frac{1}{y} e^{\log y} e^{-y} = e^{-y}, \quad y > 0.$$  Thus $Y \sim \operatorname{Exponential}(1)$.  We trivially see that $$\int_{y=0}^\infty f_Y(y) \, dy = 1,$$ hence $$\int_{x=-\infty}^\infty f_X(x) \, dx = 1$$ because this is a one-to-one transformation of random variables.  Indeed, we have $$F_X(x) = \Pr[X \le x] = \Pr[e^{-X} \ge e^{-x}] = \Pr[Y \ge e^{-x}] = e^{-e^{-x}}.$$
