Integral of an exponential of an exponential

Question

I have been trying to study this function of $\alpha$:

$$f(\alpha) = \int_{-\infty}^{\infty}1-e^{-\alpha e^{-\frac{1}{2}x^2}}\mathrm{d}x$$

but all I can obtain is an oscillating series, by expanding $f(\alpha)$ at $\alpha=0$,

$$f(\alpha) = \int_{-\infty}^{\infty} \left[a e^{-\frac{x^2}{2}} - \frac{1}{2} a^2 e^{-x^2} + \frac{1}{6} a^3 e^{-\frac{3x^2}{2}} - \frac{1}{24} a^4 e^{-2x^2} + \dots \right]\mathrm{d}x$$

then integrating term by term, giving,

$$f(\alpha) = a\sqrt{2 \pi } - \frac{1}{2}a^2\sqrt{\pi } + \frac{1}{3} a^3\sqrt{\frac{\pi }{6}} - \dots$$

Also, the function $f(\alpha)$ looks almost linear when I plot it. Is there anyway to do this integral, without leaving the answer as an oscillating series?

Answer 1

As you have (correctly) computed, $f(\alpha)=\sqrt{2\pi}g_{1/2}(\alpha)$, where $$g_\lambda(x)=\sum_{n=1}^\infty(-1)^{n-1}\frac{x^n}{n!\,n^\lambda}$$ doesn't seem to have a closed form (if $\lambda$ is not an integer).

An $x\to+\infty$ asymptotics of $g_\lambda(x)$ for $\Re\lambda>-1$ (cf. this question) can be obtained using$$g_\lambda(x)=\sum_{n=0}^\infty\frac{(-1)^n x^{n+1}}{n!(n+1)^{1+\lambda}}=\frac{x}{\Gamma(1+\lambda)}\sum_{n=0}^\infty\frac{(-x)^n}{n!}\int_0^1 y^n(-\log y)^\lambda\,dy\\=\frac{x}{\Gamma(1+\lambda)}\int_0^1 e^{-xy}(-\log y)^\lambda\,dy=\frac{1}{\Gamma(1+\lambda)}\int_0^x e^{-z}(\log x-\log z)^\lambda\,dz.$$ Now, informally, the idea is to use the binomial series for $(1-t)^\lambda$ at $t=\log z/\log x$. For convergence to hold, we have to split $\int_0^x=\int_0^{1/x}+\int_{1/x}^x$ then. But the series we get (in the limit) is in powers of $1/\log x$. Hence the $\int_0^{1/x}$ term is small compared to these powers, and can be neglected; further, $\int_{1/x}^{x}e^{-z}(\log z)^n\,dz$, appearing as the "coefficient" of $(1/\log x)^n$, can be (asymptotically!) replaced with $\int_0^\infty e^{-z}(\log z)^n\,dz$ [equal to $\Gamma^{(n)}(1)$] for the same reason. All these arguments can be made rigorous. Finally, we get $$g_\lambda(x)\asymp\frac{(\log x)^\lambda}{\Gamma(1+\lambda)}\sum_{n=0}^{(\infty)}\binom{\lambda}{n}\Gamma^{(n)}(1)\left(\frac{-1}{\log x}\right)^n\qquad(x\to+\infty)$$ Observe that this series diverges (as it often happens). And $x$ must be really large for this to be used as a practical approximation. The following table illustrates this for our value of $\lambda=1/2$: $$\begin{array}{c|c|c|} & x=10^3 & x=10^6 \\ \hline n=0 & 2.96567483 & 4.19409756 \\ \hline 0\leqslant n\leqslant 1 & 3.08958150 & 4.28171281 \\ \hline 0\leqslant n\leqslant 2 & 3.07421374 & 4.27627949 \\ \hline 0\leqslant n\leqslant 3 & 3.07727557 & 4.27682075 \\ \hline \text{exact} & 3.07645422 & 4.27673523 \\ \hline \end{array}$$