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I would like to compute the integral $$ I(a) = \int_{1}^{\infty} [\exp( \exp(-a y)/y) -1] dy $$ for $a > 0$.

Note that the integrand is decaying very quickly, even more quickly than doubly-exponentially.

Some numerical simulations suggest that this about $I(a) = O(\log a)$.

One can show that $I(a) = O(a)$, because $I(a) \leq J(a)$ where $$ J(a) = \int_{1}^{\infty} [\exp( \exp(-a y)) -1] dy $$ and $J(a)$ has a closed-form solution $J(a) = O(a)$.

What are the tight bounds on $I(a)$ in terms of $a$? Can one get $I(a) = O(\log a)$ or better ?

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Are you interested in $a\to 0$? (I frankly don't understand the results you state in your question) – Fabian Apr 2 '12 at 14:23
It can't be $a \to 0$, since his $J(a) \sim C/a$ as $a \to 0$ is not $O(a)$. So perhaps he wants $a \to +\infty$. – GEdgar Apr 2 '12 at 16:21
$J(a)$ is also not $O(a)$ when $a\to + \infty$... – Fabian Apr 3 '12 at 21:19
up vote 3 down vote accepted

For every $t$ in $(0,1)$, $t\leqslant \mathrm e^t-1\leqslant t+t^2$ hence $J(a)\leqslant I(a)\leqslant J(a)+K(a)$ with $$ J(a)=\int_1^{+\infty}\mathrm e^{-ay}\frac{\mathrm dy}y,\quad K(a)=\int_1^{+\infty}\mathrm e^{-2ay}\frac{\mathrm dy}{y^2}. $$ The change of variables $y=1+(x/a)$ yields $a\mathrm e^{a}J(a)=L(a)$ and $a\mathrm e^{2a}K(a)=M(a)$ with $$ L(a)=\int_0^{+\infty}\frac{\mathrm e^{-x}\mathrm dx}{1+x/a},\quad M(a)=\int_0^{+\infty}\frac{\mathrm e^{-2x}\mathrm dx}{(1+x/a)^2} $$ Using the bounds $1-t\leqslant1/(1+t)\leqslant1$ and $1/(1+t)^2\leqslant1$ for every nonnegative $t$, one sees that $$ 1-a^{-1}\leqslant L(a)\leqslant1,\quad M(a)\leqslant\tfrac12. $$ Thus, $$ a^{-1}\mathrm e^{-a}(1-a^{-1})\leqslant I(a)\leqslant a^{-1}\mathrm e^{-a}(1+\tfrac12\mathrm e^{-a}). $$ In particular, $I(a)=a^{-1}\mathrm e^{-a}+O(a^{-2}\mathrm e^{-a})$ when $a\to+\infty$, that is, $$ \color{red}{\lim\limits_{a\to+\infty}a\mathrm e^aI(a)=1}. $$

As regards the limit $a\to0$, an integration by parts yields $$ L(a)=a\int_0^{+\infty}\mathrm e^{-x}\log(1+x/a)\mathrm dx=aN(a)-a\log(a), $$ with $$ N(a)=\int_0^{+\infty}\mathrm e^{-x}\log(a+x)\mathrm dx. $$ Another integration by parts yields $M(a)=a-aL(2a)$. Since $N(a)=N(0)+o(1)$ and $N(0)$ is finite, one gets $$ a^{-1}L(a)=-\log(a)+O(1),\quad a^{-1}M(a)=1+O(1). $$ Since $\mathrm e^{-a}=1+O(a)$ and $\mathrm e^{-2a}=O(1)$, one sees that $J(a)$ and $J(a)+K(a)$ are both $-\log(a)+O(1)$. Finally, when $a\to0$, $I(a)=-\log(a)+O(1)$, and in particular, $$ \color{red}{\lim\limits_{a\to0}\frac{I(a)}{\log a}=-1}. $$

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Let's change variables: $y = 1 - \frac{1}{a} \log(u)$, where $ 0<u<1$. This gives: $$ I(a) = \int_0^1 \frac{1}{a} \cdot \left( \exp\left(\frac{a \cdot \mathrm{e}^{-a} \cdot u}{a - \log(u)}\right) - 1\right) \cdot \frac{\mathrm{d} u}{u} $$ The integrand is an increasing function, with zero right limit at $u=0$, and increasing to $$ \lim_{u \uparrow 1} \frac{1}{a \cdot u} \cdot \left( \exp\left(\frac{a \cdot \mathrm{e}^{-a} \cdot u}{a - \log(u)}\right) - 1\right) = \frac{1}{a} \left( \exp\left( \mathrm{e}^{-a}\right) -1 \right) $$ enter image description here

From here, it is intuitively clear, that large $a$ asymptotic behavior of $I(a)$ is $$ I(a) = \mathcal{O}\left( \frac{1}{a} \left( \exp\left( \mathrm{e}^{-a}\right) -1 \right) \right) $$

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