# Estimate $\displaystyle\int_0^\infty\frac{t^n}{n!}e^{-e^t}dt$ accurately.

How can I obtain good asymptotics for $$\gamma_n=\displaystyle\int_0^\infty\frac{t^n}{n!}e^{-e^t}dt\text{ ? }$$

[This has been already done] In particular, I would like to obtain asymptotics that show $$\sum_{n\geqslant 0}\gamma_nz^n$$

converges for every $z\in\Bbb C$.

N.B.: The above are the coefficients when expanding $$\Gamma \left( z \right) = \sum\limits_{n \geqslant 0} {\frac{{{{\left( { - 1} \right)}^n}}}{{n!}}\frac{1}{{n + z}}} + \sum\limits_{n \geqslant 0} {{\gamma _n}{z^n}}$$

ADD Write $${c_n} = \int\limits_0^\infty {{t^n}{e^{ - {e^t}}}dt} = \int\limits_0^\infty {{e^{n\log t - {e^t}}}dt}$$

We can use something similar to Laplace's method with the expansion $${p_n}\left( x \right) = g\left( {{\rm W}\left( n \right)} \right) + g''\left( {{\rm W}\left( n \right)} \right)\frac{{{{\left( {x - {\rm W}\left( n \right)} \right)}^2}}}{2}$$

where $g(t)=n\log t-e^t$. That is, let $$\begin{cases} w_n={\rm W}(n)\\ {\alpha _n} = n\log {w_n} - {e^{{w_n}}} \\ {\beta _n} = \frac{n}{{w_n^2}} + {e^{{w_n}}} \end{cases}$$

Then we're looking at something asymptotically equal to $${C_n} = \exp {\alpha _n}\int\limits_0^\infty {\exp \left( { - {\beta _n}\frac{{{{\left( {t - {w_n}} \right)}^2}}}{2}} \right)dt}$$

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if you write $g(n,t):=t^ne^{-e^{t}}/n!$, then you can swap sum and integral by Fubini-Tonelli. And you know that $\sum g(n,t)z^n$=e^{tz}$. – Alex R. Aug 9 '13 at 1:51 If your only goal is to show that the series converges, I think you can circumvent the estimation issue by just noting that an entire function has a power series that converges everywhere. And certainly the Gamma function with its poles removed is entire. – Potato Aug 9 '13 at 2:27 @Potato You're assuming what I want to get to! =) – Pedro Tamaroff Aug 9 '13 at 2:30 @PeterTamaroff You say you want to show that series converges in the question! – Potato Aug 9 '13 at 2:37 @Potato The point is the following. Put $$\Gamma(x)=\int_0^\infty t^{x-1}e^{-t}dt$$ Splitting at$1$, we expand$e^{-t}$around zero in the first integral to obtain the poles, and use$t^x=e^{x\log t}$and expand in power series to obtain a series expansion with the coefficients I get. With this, one aims to show that we may extend this to$\Bbb C$. – Pedro Tamaroff Aug 9 '13 at 2:41 ## 3 Answers To get an asymptotic estimate we can use a trick similar to one used to find the asymptotics of$n!$. You wrote $$c_n = \int_0^\infty t^n e^{-e^t}\,dt = \int_0^\infty \exp\left(n\log t - e^t\right)\,dt = \int_0^\infty \exp f_n(t)\,dt$$ and saw that$f_n(t)$has a maximum at$t = W(n). We thus scale the integration variable by $$t = W(n)(1+s),$$ which yields \begin{align} c_n &= W(n)^{n+1} \int_{-1}^\infty \exp\left\{n\left[\log(1+s) - W(n)^{-1} e^{W(n)s}\right]\right\}\,ds \\ &= W(n)^{n+1} \int_{-1}^\infty \exp\left\{n g_n(s)\right\}\,ds \end{align} The largest contribution to the integral now comes from a neighborhood ofs=0, and there we have $$g_n(s) = -\frac{1}{W(n)} - \frac{1+W(n)}{2}\,s^2 + \cdots.$$ The details of the Laplace method can be worked out as usual to arrive at the conclusion that \begin{align} c_n &\sim W(n)^{n+1} \int_{-\infty}^{\infty} \exp\left\{n\left[-\frac{1}{W(n)} - \frac{1+W(n)}{2}\,s^2\right]\right\}\,ds \\ &= W(n)^{n+1} e^{-n/W(n)} \sqrt{\frac{2\pi}{n(1+W(n))}}. \end{align} Below is a plot ofc_n$in blue and this asymptotic in purple for$1 < n < 10$. Below that is a plot of$W(n)^{-n-1} e^{n/W(n)} c_n$in blue and$\sqrt{\frac{2\pi}{n(1+W(n))}}$in purple for$1 < n < 50$. We can then use the estimate $$\log n - \log\log n < W(n) < \log n$$ from this answer, true for$n > e$, to find that $$c_n < (\log n)^{n+1} e^{-n/\log n} \sqrt{\frac{2\pi}{n(1+\log n - \log\log n)}}$$ for$n$large enough. Since$n! \geq (n/e)^n \sqrt{2\pi n}we obtain the asymptotic bound \begin{align} \gamma_n &= \frac{c_n}{n!} \\ &< \left(\frac{e \log n}{n}\right)^{n+1} e^{-1-n/\log n} (1+\log n-\log\log n)^{-1/2} \end{align} which clearly decreases faster than exponentially. From this it follows that\sum \gamma_n z^n$converges for all$z$. - (+1) This is very similar to the method I showed in chat yesterday. I didn't feel like justifying all the steps of the Laplace method, so I came up with an alternate answer. – robjohn Aug 9 '13 at 4:30 (+1) Wasn't expecting any less ;) – Pedro Tamaroff Aug 9 '13 at 4:44 For$t>0$, it is true that$e^t>t^2$. Using this bound on the integral gives $$\int_0^\infty\frac{t^n}{n!}e^{-e^t}dt < \int_0^\infty\frac{t^n}{n!}e^{-t^2}dt.$$ The last integral is just$\frac{1}{2}$of $$\frac{\Gamma\left( \frac{n+1}{2} \right)}{n!}.$$ This decays quickly enough to give convergence in the entire complex plane. - The bounding integral is equal to$\frac1{2n!}\Gamma\left(\frac{n+1}{2}\right)$, but that is not$\frac1{(2n-1)!!}$. It is monotonic, and for odd$n$, it is$\lower{4pt}{\frac{\left(\frac{n-1}{2}\right)!}{n!}}\le\raise{4pt}{\frac1{\left(‌​\frac{n+1}{2}\right)!}}. This, also, is enough to get convergence for the whole complex plane. – robjohn Aug 9 '13 at 3:57 @robjohn I didn't add that part, but thanks for letting me know. – Potato Aug 9 '13 at 3:58 \begin{align} \sum_{n=1}^\infty\gamma_nz^n &=\sum_{n=1}^\infty\int_0^\infty\frac{t^nz^n}{n!}e^{-e^t}\,\mathrm{d}t\\ &=\int_0^\infty e^{tz}e^{-e^t}\,\mathrm{d}t\\ &=\int_0^\infty e^{t(z-1)}e^{-e^t}\,\mathrm{d}e^t\\ &=\int_{1}^\infty u^{z-1}e^{-u}\,\mathrm{d}u\\ &=\Gamma(z,1) \end{align} The Upper Incomplete Gamma Function is an entire function. According to this answer, the power series for an entire function has an infinite radius of convergence.\color{#C0C0C0}{\text{idea mentioned in chat}}\begin{align} \int_0^\infty\frac{x^{n-1}}{(n-1)!}e^{-e^x}\,\mathrm{d}x &=\frac1{(n-1)!}\int_1^\infty\log(t)^{n-1}e^{-t}\frac{\mathrm{d}t}{t}\\ &=\frac1{n!}\int_1^\infty\log(t)^ne^{-t}\,\mathrm{d}t\\ &=\frac1{n!}\int_1^\infty e^{-t+n\log(\log(t))}\,\mathrm{d}t\\ \end{align} Looking at the function\phi(t)=-t+n\log(\log(t))$, we see that it reaches its maximum when$t\log(t)=n$; i.e.$t_0=e^{\mathrm{W}(n)}=\frac{n}{\mathrm{W}(n)}$. Using the estimate $$\mathrm{W}(n)\approx\log(n)-\frac{\log(n)\log(\log(n))}{\log(n)+1}$$ from this answer, at$t_0, \begin{align} \phi(t_0) &=-n\left(\mathrm{W}(n)+\frac1{\mathrm{W}(n)}-\log(n)\right)\\ &\approx n\log(\log(n)) \end{align} and \begin{align} \phi''(t_0) &=-\frac{\mathrm{W}(n)+1}{n}\\ &\approx-\frac{\log(n)}{n} \end{align} According to the Laplace Method, the integral would be asymptotic to \begin{align} \frac1{n!}\sqrt{\frac{-2\pi}{\phi''(t_0)}}e^{\phi(t_0)} &\approx\frac1{n!}\sqrt{\frac{2\pi n}{\log(n)}}\log(n)^n\\ &\approx\frac1{\sqrt{2\pi n}}\frac{e^n}{n^n}\sqrt{\frac{2\pi n}{\log(n)}}\log(n)^n\\ &=\frac1{\sqrt{\log(n)}}\left(\frac{e\log(n)}{n}\right)^n \end{align} which dies away faster thanr^{-n}$for any$r$. Analysis of the Approximation to Lambert W For$x\ge e$, the approximation $$\mathrm{W}(x)\approx\log(x)\left(1-\frac{\log(\log(x))}{\log(x)+1}\right)$$ attains a maximum error of about$0.0353865$at$x$around$67.9411$. At least that same precision is maintained for$x\ge\frac53$. For all$x\gt1$, this approximation is an underestimate. - Rob, this is from a course on real analysis, so complex analysis is really out of the question as I told Potato. Also, notice I want to obtain accurate asymptotics. Could you continue with what I wrote using Lambert's$\rm W$? – Pedro Tamaroff Aug 9 '13 at 4:24 @PeterTamaroff: okay. I was fooled by all the mention of$\mathbb{C}\$ in the question. I will append the stuff I mentioned yesterday in chat, some of which is similar to what Antonio Vargas has posted. – robjohn Aug 9 '13 at 4:34
No problem, Rob. If I recall correctly you got estimates in terms of the logarithn and the exponential which make things quite simpler! – Pedro Tamaroff Aug 9 '13 at 4:42