Asymptotic behavior of a sequence of integrals of non-analytic functions

Question

I am interested in the asymptotic behavior of the sequence $(I_n)_{n=1}^\infty$ of integrals $$I_n=\int_0^1(1-x)^ne^{-1/x}\,dx.$$ Motivation: It is straightforward to show that, for any $f\in C[0,1]$ with $f(0)\neq0$, $$\int_0^1(1-x)^nf(x)\,dx\sim\frac{f(0)}{n}.$$ (One method is to approximate $f$ by a $C^1$ function, then integrate by parts.) In fact, using essentially the same argument, one can show that for any $f\in C^\infty[0,1]$, if there is a $k\in\mathbb{N}$ such that $f^{(k)}(0)\neq0$ and $f(0),f'(0),\ldots,f^{(k-1)}(0)=0$, then $$\int_0^1(1-x)^nf(x)\,dx\sim\frac{f^{(k)}(0)}{n^{k+1}}.$$ So what about smooth functions which have every derivative at $0$ equal to zero? Such functions are either identically zero (boring) or non-analytic, so appeals to Taylor series expansions will not help to analyze them. The choice $f(x)=e^{-1/x}$ for $x\in(0,1]$ and $f(0)=0$ is one of the classical examples of a smooth but non-analytic function.

The integration by parts argument doesn't seem as helpful here - we can use the formula $$\frac{d^k}{dx^k}e^{-1/x}=e^{-1/x}\sum_{\ell=1}^k(-1)^{k+\ell}\binom{k}{\ell}\binom{k-1}{\ell-1}x^{-(k+\ell)}$$ (proved here) to rewrite the integral as $$I_n=\frac{k!}{(n+k)!}\sum_{\ell=1}^k(-1)^{k+\ell}\binom{k}{\ell}\binom{k-1}{\ell-1}\int_0^1(1-x)^{n+k}\frac{e^{-1/x}}{x^{k+\ell}}\,dx,$$ but this doesn't seem to make things any easier, except to see that the sequence $(I_n)_{n=1}^\infty$ must converge to zero faster than $n^k$ for any $k\in\mathbb{N}$ (as expected).

Any ideas for how to attack this?

Answer 1

I shall derive the leading asymptotics of $I_n$. The change of variables from $x$ to $s$ via $s = 1/(x\sqrt n )$ gives \begin{align*} I_n & = \int_0^1 {(1 - x)^n {\rm e}^{ - 1/x} {\rm d}x} = \frac{1}{{\sqrt n }}\int_{1/\sqrt n }^{ + \infty } {\left( {1 - \frac{1}{{\sqrt n s}}} \right)^n {\rm e}^{ - \sqrt n s} \frac{{{\rm d}s}}{{s^2 }}} \\& = \frac{1}{{\sqrt n }}\int_{1/\sqrt n }^{ + \infty } {\exp \left( {n\left[ {\log \left( {1 - \frac{1}{{\sqrt n s}}} \right) + \frac{1}{{\sqrt n s}}} \right]} \right){\rm e}^{ - \sqrt n (s + 1/s)} \frac{{{\rm d}s}}{{s^2 }}} . \end{align*} We assume that $n\ge 5$ and split the range of integration into two parts at $s=1/2$. For the finite interval, we note that \begin{align*} & \frac{1}{{\sqrt n }}\int_{1/\sqrt n }^{1/2} {\exp \left( {n\left[ {\log \left( {1 - \frac{1}{{\sqrt n s}}} \right) + \frac{1}{{\sqrt n s}}} \right]} \right){\rm e}^{ - \sqrt n (s + 1/s)} \frac{{{\rm d}s}}{{s^2 }}} \\ & \le \frac{1}{{\sqrt n }}{\rm e}^{ - (5/2)\sqrt n } \int_{1/\sqrt n }^{1/2} {\exp \left( {n\left[ {\log \left( {1 - \frac{1}{{\sqrt n s}}} \right) + \frac{1}{{\sqrt n s}}} \right]} \right)\frac{{{\rm d}s}}{{s^2 }}} \\ & \le \frac{1}{{\sqrt n }}{\rm e}^{ - (5/2)\sqrt n } \int_{1/\sqrt n }^{1/2} {\frac{{{\rm d}s}}{{s^2 }}} \le \frac{{\sqrt n }}{2}{\rm e}^{ - (5/2)\sqrt n } = o\!\left( {\frac{1}{{n^{5/4} }}{\rm e}^{ - 2\sqrt n } } \right), \end{align*} since $\log(1-w)-w<0$ for all $w<1$. Now for $s>1/2$ and $n\ge 5$, we have $$ n\left[ {\log \left( {1 - \frac{1}{{\sqrt n s}}} \right) + \frac{1}{{\sqrt n s}}} \right] = - \frac{1}{{2s^2 }} + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right) $$ uniformly in $s$. Whence, on the infinite part of the integration rage \begin{align*} & \frac{1}{{\sqrt n }}\int_{1/2}^{ + \infty } {\exp \left( {n\left[ {\log \left( {1 - \frac{1}{{\sqrt n s}}} \right) + \frac{1}{{\sqrt n s}}} \right]} \right){\rm e}^{ - \sqrt n (s + 1/s)} \frac{{{\rm d}s}}{{s^2 }}} \\ &=\frac{1}{{\sqrt n }}\int_{1/2}^{ + \infty } {\exp \left( { - \frac{1}{{2s^2 }}} \right){\rm e}^{ - \sqrt n (s + 1/s)} \frac{{{\rm d}s}}{{s^2 }}} \left( {1 + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right)} \right). \end{align*} By Laplace's method applied to the sole saddle point at $s=1$, we find $$ \frac{1}{{\sqrt n }}\int_{1/2}^{ + \infty } {\exp \left( { - \frac{1}{{2s^2 }}} \right){\rm e}^{ - \sqrt n (s + 1/s)} \frac{{{\rm d}s}}{{s^2 }}} = \sqrt {\frac{\pi }{{\rm e}}} \frac{1}{{n^{3/4} }}{\rm e}^{ - 2\sqrt n } \left( {1 + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right)} \right) $$ as $n\to+\infty$. Therefore, collecting all the partial results, $$ I_n = \sqrt {\frac{\pi }{{\rm e}}} \frac{1}{{n^{3/4} }}{\rm e}^{ - 2\sqrt n } \left( {1 + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right)} \right) $$ as $n\to+\infty$. This is in agreement with Svyatoslav's comment above.

Addendum. I provide the complete asymptotic expansion without proof: \begin{align*} I_n & \sim \sqrt {\frac{\pi }{{\rm e}}} \frac{1}{{n^{3/4} }}{\rm e}^{ - 2\sqrt n } \sum\limits_{k = 0}^\infty {\frac{{a_k }}{{n^{k/2} }}} \\ & = \sqrt {\frac{\pi }{{\rm e}}} \frac{1}{{n^{3/4} }}{\rm e}^{ - 2\sqrt n } \!\left( {1 - \frac{{43}}{{48}}\frac{1}{{n^{1/2} }} - \frac{{2111}}{{4608}}\frac{1}{n} + \frac{{2865793}}{{3317760}}\frac{1}{{n^{3/2} }} + \frac{{218305349}}{{{\rm }637009920}}\frac{1}{{n^2 }} - \ldots } \right) \end{align*} as $n\to+\infty$, with $${\small a_k = \frac{1}{{k!}}\sum\limits_{m = 0}^k {\frac{1}{{4^m }}\binom{k}{m}\left[ {\frac{{{\rm d}^{2m} }}{{{\rm d}t^{2m} }}\left( {\frac{{{\rm e}^{(1 - 1/t^2 )/2} }}{{t^{k - m + 2} }}\left( {\frac{{(t - 1)^2 }}{{t + 1/t - 2}}} \right)^{m + 1/2} \left[ {\frac{{{\rm d}^{k - m} }}{{{\rm d}s^{k - m} }}\exp \left( {\sum\limits_{r = 1}^{k - m} {\frac{{ - t^{-2}}}{{r + 2 }}s^r } } \right)} \right]_{s = 0} } \right)} \right]_{t = 1} } } $$ for any $k\ge 0$.

Answer 2

The solution by @Gary provides a rigorous proof of the first asymptotic term. It might make sense, though, to get next asymptotic term $\Big(O\big(1/\sqrt n \,\big)\Big)$. For example, it might be useful for practical calculations, given the fact that convergence is slow.

The evaluation is similar to the solution by @Gary above. $$I_n=\int_0^1(1-x)^ne^{-1/x}dx\overset{t=\frac1x}{=}\int_1^\infty e^{-t+n\ln(1-\frac1t)}\frac{dt}{t^2}$$ Decomposing logarithm $$=\int_1^\infty e^{-t-\frac nt-\frac n{2t^2}-\frac n{3t^3}-...}\,\,\frac{dt}{t^2}$$ $$\overset{t=\sqrt n\,x}{=}\frac1{\sqrt n}\int_{\frac1{\sqrt n}}^\infty e^{-\sqrt n(x+\frac1x)-\frac1{2x^2}-\frac1{3x^3\sqrt n}-...}\,\,\frac{dx}{x^2}$$ Applying the Laplace' method, we find that the minimum of the function $f(x)=x+\frac1x$ is at $x=1$. Decomposing $f(x)$ near this point $$f(x)=2+(x-1)^2-(x-1)^3+(x-1)^4+...$$ For our purpose (finding second asymptotic term) it is enough to keep only these four terms of the decomposition. Making the substitution $\,x-1=t$ $$I_n=\frac{e^{-2\sqrt n}}{\sqrt n}\int_{-1+\frac1{\sqrt n}}^\infty e^{-\sqrt n(t^2-t^3+t^4-...)-\frac1{2(t+1)^2}-\frac1{3(t+1)^3\sqrt n}-...}\frac{dt}{(1+t)^2}$$ Making the substitution $\,t=\frac x{\sqrt[4]{n}}$, expanding integration to $-\infty$ and keeping only contributing terms $\Big(\sim1+O\big(\frac1{\sqrt n}\big)\Big)$ $$I_n\sim\frac{e^{-2\sqrt n}}{n^\frac34}\int_{-\infty}^\infty e^{-x^2}\Big(1+\frac{x^3}{\sqrt[4]{n}}-\frac{x^4}{\sqrt n}-\frac1{2\big(1+\frac x{\sqrt[4]{n}}\big)^2}+\frac12\big(\frac{x^3}{\sqrt[4]{n}}-\frac1{2\big(1+\frac x{\sqrt[4]{n}}\big)^2}\big)^2\Big)\,\frac{\Big(1-\frac1{2\sqrt n}\Big)}{\big(1+\frac x{\sqrt[4]{n}}\big)^2}\,dx$$ Decomposing the integrand up to $\sim\frac1{\sqrt n}$ terms (for example, $\big(1+\frac x{\sqrt[4]{n}}\big)^{-2}=1-\frac{2x}{\sqrt[4]{n}}+\frac{3x^2}{\sqrt n}+...$), we finally get a simple expression: $$I_n\sim\frac{e^{-2\sqrt n}}{n^\frac34\sqrt e}\int_{-\infty}^\infty e^{-x^2}\Big(1-\frac1{3\sqrt n}+\frac{x^6}{2\sqrt n}-\frac{2x^4}{\sqrt n}\Big)dx$$ Integrating, $$\boxed{\,\,I_n=\sqrt\frac\pi e\,\frac{e^{-2\sqrt n}}{n^\frac34}\Big(1-\frac{43}{48}\frac1{\sqrt n}+O\big(\frac1n\big)\Big)\,\,}$$

---

Quick numeric check shows that the approximation works already for small $n$:

$n=9\quad I_n=0.0003503\quad appr\approx 0.0003596$ $n=25\quad I_n=3.5346\cdot10^{-6}\quad appr\approx 3.5832\cdot10^{-5}$ $n=100\quad I_n=6.3535\cdot10^{-11}\quad appr\approx 6.3793\cdot10^{-11}$ $n=10000\quad I_n=1.4743\cdot10^{-90}\quad appr\approx 1.4744\cdot10^{-90}$