How to get the asymptotic expansion for the integral $$\int_{0}^{1}\exp(-x/t)dt$$ in the limit $x\rightarrow 0$ ? I took $x/t=u$ and did integration by parts (IP) but if I keep doing IP, I get a series in $\exp(-x)$ times increasing powers of $x^{-1}$; this does not look like a valid asymptotic expansion. Another way I tried to use is to write $1/t$ as a geometric series in $1-t$ but it does not help either. Can I just take the integrand to be $\exp(-x)$ because the maximum contribution to the integral is from $t=1$ and then just expand $\exp(-x)\approx 1-x$ in the limit $x\rightarrow 0$ and call that as the asymptotic expansion ?

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    $\begingroup$ Substituting $u = x/t$ gets you something you can work with. $\endgroup$ Oct 29, 2014 at 15:54
  • $\begingroup$ Your integral equals $e^{-x}-\Gamma(0,x)$ where $\Gamma(a,z)$ is the incomplete gamma function. Mathematica gives the expansion $1+x\log x - (1-\gamma)x-x^2/2+O(x^3)$, where $\gamma$ is the Euler–Mascheroni constant. $\endgroup$ Jan 5, 2018 at 11:11

1 Answer 1


If you suppose $x>0$ and put $v=x/t$, you get $$F(x)=x\int_x^{+\infty}\frac{\exp(-v)}{v^2}dv=x\int_1^{+\infty}\frac{\exp(-v)}{v^2}dv+x\int_x^1\frac{\exp(-v)}{v^2}dv$$ Hence $$F(x)=x\int_1^{+\infty}\frac{\exp(-v)}{v^2}dv+x\int_x^1\frac{\exp(-v)-1+v}{v^2}dv+x\int_x^1\frac{1-v}{v^2}dv$$


$$F(x)=x\int_1^{+\infty}\frac{\exp(-v)}{v^2}dv+x\int_x^1\frac{\exp(-v)-1+v}{v^2}dv-x+1+x\log x$$

Now $$\frac{\exp(-v)-1+v}{v^2}=\sum_{k\geq 2}\frac{(-1)^k}{k!}v^{k-2}$$ and $$ \int_x^1\frac{\exp(-v)-1+v}{v^2}dv=\sum_{k\geq 2}\frac{(-1)^k}{k!}(\frac{1-x^{k-1}}{k-1})$$

Finally: $$F(x)=1+x\log x+cx+\sum_{k\geq 2}\frac{(-1)^{k-1}}{k!(k-1)}x^k$$ with $$c=\sum_{k\geq 2}\frac{(-1)^k}{k!(k-1)}+\int_1^{+\infty}\frac{\exp(-v)}{v^2}dv-1$$


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