Integral solution of separable differential equation

Question

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 $(s\hat\varrho(s))' = e^{-s}\hat\varrho(s)$.)

Answer 1

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 $r \mapsto 0$).

Answer 2

$$y' = e^{-x}\frac{y}{x} $$ $$\frac{y'}{y} = \frac{e^{-x}}{x} $$ $$\ln|y|=\int \frac{e^{-x}}{x}dx+\text{constant} $$ $$\ln|y|=\int_a^b \frac{e^{-\tau}}{\tau}d\tau +\ln(C)$$ (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.