On the asymptotics of $\int_0^\infty \exp\big(\!-\!\lambda e^x/x\big)\,dx$

Question

The question is about the asymptotics of $$I(\lambda):=\int_0^\infty \exp\big(\!-\!\lambda e^x/x\big)\,dx$$ as $\lambda\to\infty$. I ask this question here because this is a non-standard form of an integrand that can be treated nevertheless by the saddle point approximation. I also believe the results to be quite elegant. Lastly, this is an integral that Mathematica struggles to numerically evaluate, so I thought that it would be useful to be able to know how to analytically approximate it.

Answer 1

We treat this integral using the saddle-point approximation. Let $f(x)=-\frac{\exp(x)}{x}$. Then we have $f\le 0$, $f(0)=0$, $f(\infty)=0$ and $f$ has a maximum at $x=1$ with $f(1)=-e, f''(1)=-e$. Hence, extending the integration range to the full real line, we have that $$I(\lambda)\sim\int_{-\infty}^\infty \exp\big(\!-\!\lambda\big(e+\frac{e}{2}(x-1)^2\big)\big)\,dx.$$ Doing the Gaussian integral, we find that $$I(\lambda)\sim \frac{\exp\left(-\lambda e\right)}{\sqrt{\lambda}}\sqrt{\frac{2\pi}{e}}.$$

Answer 2

We split the integral at $x=1$ into two parts: $$ I(\lambda ) = \int_0^1 {\exp ( - \lambda {\rm e}^x /x)\,{\rm d}x} + \int_1^{ + \infty } {\exp ( - \lambda {\rm e}^x /x)\,{\rm d}x} . $$ Using the two real branches of the Lambert $W$-function, we perform the substitutions $x = - W_0 (t)$ and $x = - W_{ - 1} (t)$, respectively, to get $$ I(\lambda ) = \int_0^{ - 1/{\rm e}} {\exp (\lambda /t)\,{\rm d}( - W_0 (t))} + \int_{ - 1/{\rm e}}^0 {\exp (\lambda /t)\,{\rm d}( - W_{ - 1} (t))} . $$ Now making the change of variables $t = - 1/s$ yields \begin{align*} I(\lambda ) & = \int_{ + \infty }^{\rm e} {{\rm e}^{ - \lambda s} \,{\rm d}( - W_0 ( - 1/s))} + \int_{\rm e}^{ + \infty } {{\rm e}^{ - \lambda s}\, {\rm d}( - W_{ - 1} ( - 1/s))} \\ & = \int_{\rm e}^{ + \infty } {{\rm e}^{ - \lambda s} \left[ {\frac{{{\rm d}W_0 ( - 1/s)}}{{{\rm d}s}} - \frac{{{\rm d}W_{ - 1} ( - 1/s)}}{{{\rm d}s}}} \right]{\rm d}s} . \end{align*} Finally, with $s = {\rm e} + x$, we deduce $$ I(\lambda ) = {\rm e}^{ - {\rm e}\lambda } \int_0^{ + \infty } {{\rm e}^{ - \lambda x} \left[ {\frac{{{\rm d}W_0 ( - 1/({\rm e} + x))}}{{{\rm d}x}} - \frac{{{\rm d}W_{ - 1} ( - 1/({\rm e} + x))}}{{{\rm d}x}}} \right]{\rm d}x} . $$ Near the branchpoint $-1/{\rm e}$, we have $$ W_{0,1}( - 1/({\rm e} + x)) = - \sum\limits_{n = 0}^\infty {( \mp 1)^n c_n (\sqrt {2\log (1 + x/{\rm e})} )^n } $$ (cf. $(4.13.6)$). Consequently, \begin{align*} W_0 ( - 1/({\rm e} + x)) - W_{ - 1} ( - 1/({\rm e} + x)) & = \sum\limits_{n = 0}^\infty {2c_{2n + 1} (2\log (1 + x/{\rm e}))^{n + 1/2} } \\ & = 2\sqrt {\frac{2}{{\rm e}}} x^{1/2} - \frac{7}{{{\rm 18e}}}\sqrt {\frac{2}{{\rm e}}} x^{3/2} + \ldots \end{align*} as $x\to 0^+$. Therefore, $$ \frac{{{\rm d}W_0 ( - 1/({\rm e} + x))}}{{{\rm d}x}} - \frac{{{\rm d}W_{ - 1} ( - 1/({\rm e} + x))}}{{{\rm d}x}} = \sqrt {\frac{2}{{\rm e}}} x^{ - 1/2} - \frac{7}{{{\rm 12e}}}\sqrt {\frac{2}{{\rm e}}} x^{1/2} + \ldots $$ as $x\to 0^+$. Hence, by Watson's lemma, $$ I(\lambda ) \sim {\rm e}^{ - {\rm e}\lambda } \sqrt {\frac{{2\pi }}{{{\rm e}\lambda }}} \left( {1 - \frac{7}{{24{\rm e}\lambda }} + \ldots } \right) $$ as $\lambda \to +\infty$. It is seen that the natural variable is ${\rm e}\lambda$.