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I am trying to find the leading order term in the asymptotic expansion for $$\int_x^1 e^{-1/t} \, dt$$ as $x\to 0^+$.

I attempted to do a $u$-substitution by taking $u=\frac{1}{t}$ so $du=-t^{-2} dt$ but then the integral will turn into $\int_\infty^1 -\frac{e^{-u}}{u^2} du$ which has no $x$ dependence so it would not be possible to find an asymptotic expansion in terms of $x$. I also cannot expand the integrand of the integral with Taylor series and integration by parts will not work, so I am not really sure how to find an expansion for this. Any help would be greatly appreciated.

leonbloy
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mathim1881
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  • The lower bound of the integral after $u$-substitution is $\frac{1}{x}$. Anyway, you can just differentiate this expression using the fundamental theorem of calculus. – Qiaochu Yuan Sep 20 '22 at 20:33
  • @QiaochuYuan Thanks, I didnt realize this. I tried that and am left with the integral being equal to $1/e-xe^{-1/x}+\int_{1/x}^1 \frac{e^{-y}}{y}dy$. I feel like this might be related to exponential integral but I am not very sure how to relate it? – mathim1881 Sep 20 '22 at 20:58
  • Let's back up a second here. If $f(x)$ is a continuously differentiable function at a point $x = 0$, then the first two terms in the asymptotic expansion as $x \to 0$ are $f(0) + x f'(0)$. Here $f(0)$ is some definite integral; in any case it's a constant. $f'(0)$ can be evaluated by differentiating the original indefinite integral and then substituting $x = 0$. What do you get when you do that? – Qiaochu Yuan Sep 20 '22 at 21:01
  • @QiaochuYuan When I do that I get $\frac{d}{dx} \int_x^1 e^{-1/t} dt=-e^{-1/x}$. Putting in $x=0$ gives $0$. – mathim1881 Sep 20 '22 at 21:28
  • @QiaochuYuan I do not understand?? The limit you asked for is 0, is it not? – mathim1881 Sep 21 '22 at 16:58
  • @QiaochuYuan Do you mean that it is actually approaching $0$ from the negative? – mathim1881 Sep 21 '22 at 17:58
  • Ah, wow, never mind, I've made a silly error, sorry for misleading you! $0$ is the correct answer; this is the linear term of the asymptotic expansion. I don't know if "leading order term" means the linear term or the first term past the constant term which is nonzero, though. – Qiaochu Yuan Sep 21 '22 at 18:00
  • @QiaochuYuan Haha no problem, and sorry I did mean the first non constant term. I ended up getting that this integral is equal to $\frac{1}{e}-xe^{-1/x}+Ei(-1)-Ei(-1/x)$ where $Ei(x)$ is exponential integral of $x$ Now, I found an asymptotic expansion of the exponential integral, but I am wondering about the $xe^{-1/x}$ Should I do Taylor expansion on this and combine with asymptotic expansion of exponential integral to find the nonconstant leading order term? – mathim1881 Sep 21 '22 at 18:12
  • @QiaochuYuan Well actually, I cannot expand $xe^{-1/x}$ as $x\to 0$, so would i just treat this term as $0$ in the asymptotic expansion since it goes to $0$ exponentially fast? – mathim1881 Sep 21 '22 at 18:15
  • I would just take the derivative again to compute the quadratic term and not worry too much about the exact value of the integral. But yes, that term goes to zero exponentially fast so it doesn't contribute anything at any finite order. – Qiaochu Yuan Sep 21 '22 at 18:16
  • @QiaochuYuan Ok, thanks that makes sense. Sorry to ask so much, but I just realized that my asymptotic expansion for the exponential integral will blow up if I take asymptotic expansion of $Ei(-1/x)$ as $x\to 0$ since the asymptotic expansion found on wikipedia would blow up if we put in $-1/x$. Any idea how I can deal with this? – mathim1881 Sep 21 '22 at 18:25

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With the notation of the DLMF, your integral is $$ \mathrm{e}^{-1} - x\mathrm{e}^{ - 1/x} + \operatorname{Ei}( - 1) + E_1 (1/x) \sim \mathrm{e}^{-1} + \operatorname{Ei}( - 1) +x \mathrm{e}^{ - 1/x} \sum\limits_{n = 1}^\infty {( - 1)^n n!x^n } $$ as $x\to 0^+$ (cf. $(6.12.1)$). Numerically $\mathrm{e}^{-1} + \operatorname{Ei}( - 1) = 0.14849550677592204792\ldots$.

Gary
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