Integral solution of separable differential equation On page 524 of Tenenbaum's Introduction to Analytic and Probabilistic Number Theory (3rd edition) it is essentially stated that the solution to the first-order differential equation
$$y' = e^{-x}y/x \tag{1}$$
is $$y = Ce^{-J(x)}\tag{2}$$ 
(for a suitable constant $C$) 
where $$J(x) = \int_0^\infty \frac{e^{-x-t}}{x+t}\,dt. \tag{3}$$
Now, I am able to verify that $$J'(x) = -e^{-x}/x \tag{4}$$ 
by differentiating under the integral sign and integrating by parts, which shows that $Ce^{-J(x)}$ really is a solution to (1), but how does one come up with (2) + (3)? Separation of variables does not get me the right answer.
(Equation (1) arises in the study of the Laplace transform $\hat\varrho(s)$ of Dickman's function; specifically, we have $(s\hat\varrho(s))' = e^{-s}\hat\varrho(s)$.)
 A: $$y' = e^{-x}\frac{y}{x} $$
$$\frac{y'}{y} = \frac{e^{-x}}{x} $$
$$\ln|y|=\int \frac{e^{-x}}{x}dx+\text{constant} $$
$$\ln|y|=\int_a^b \frac{e^{-\tau}}{\tau}d\tau +\ln(C)$$
(any constant $a$ , $b$ and $C$)
With change of variable  $\tau=x+t$ :
$$\ln|y|=-\int_{b-x}^{a-x} \frac{e^{-x-t}}{x+t}dt +\ln(C)$$
Since $a$ and $b$ are any constant (relatively to $t$) until no boundary condition is specified, it is possible to set $a=\infty$ and $b=x$ so that :
$$\ln|y|=-\int_{0}^{\infty} \frac{e^{-x-t}}{x+t}dt +\ln(C)$$
which leads to the expected formula :
$$y(x)=C\,e^{-\int_{0}^{\infty} \frac{e^{-x-t}}{x+t}dt}$$
This might appear as a trick. But it is not possible to be more explicit until the boundary conditions for the ODE are missing in the wording of the question.
A: The derivation is actually very simple, as pointed out by a friend of mine. If $y$ is a solution, then separation of variables gives, for all $0 < r < s$, 
$$\log \frac{y(s)}{y(r)} = \int_r^s\!\frac{e^{-t}}{t} \,dt$$
hence
$$y(r) = y(s) \exp\Big(\!\!-\!\!\int_r^s\!\frac{e^{-t}}{t}\,dt\Big).\tag{*}$$
Observe that $C = \lim_{s\to\infty} y(s) = y(1)\exp{\int_1^\infty e^{-t}/t\,dt}$ exists because the integral is convergent. So, taking the limit as $s \to \infty$ in (*) yields 
$$y(r) = C \exp\Big(\!\!-\!\!\int_r^\infty\!\frac{e^{-t}}{t}\,dt\Big)$$
and the desired form follows on making the substitution $t \mapsto t- r$ in the integral ($\infty \mapsto \infty$ and $r \mapsto 0$).
