I am trying analyze an integral of the form

$$I(\varepsilon)=\int_0^\infty f(t,\varepsilon) \,dt$$

where $\varepsilon$ is a small real parameter. The function $f(t,\varepsilon)$ is very complicated, so the integral cannot be evaluated analytically, but it is enough for me to understand its behavior around $\varepsilon=0$. The function $f(t,\varepsilon)$ has a Taylor expansion $f(t,\varepsilon) =\sum_{n=0}^\infty f_n(t) \varepsilon^n$, so the naive answer would be $I(\varepsilon)=\sum_{n=0}^\infty \varepsilon^n\int_0^\infty f_n(t) \, dt$.

However, some of the integrals $\int_0^\infty f_n(t) \, dt$ in the series diverge (to be more specific, the integrands behave as $e^{-t}t^{-m}$, so the integrals diverge around 0). Is there any way to compute the expansion in such case? I have tried various substitutions, division of the integral into several parts, etc., but I always get some kind of divergent or ambiguous result.

As a simple example, consider the integral

$$I(\varepsilon)=\int_0^\infty \frac{\varepsilon\, e^{-t}}{t^2+\varepsilon^2} \, dt$$

This integral can be evaluated explicitly using special functions and its expansion is

$$I(\varepsilon)=\frac{\pi }{2}+\varepsilon (\log(\varepsilon ) + \gamma -1)+\cdots$$

The expansion of the integrand is $$\frac{\varepsilon\, e^{-t}}{t^2+\varepsilon^2}=\frac{e^{-t} \varepsilon}{t^2} -\frac{e^{-t} \varepsilon^3}{t^4}+\frac{e^{-t} \varepsilon ^5}{t^6}+\cdots,$$ which means that integrals of all terms in the series diverge. I would like to know how to get the expansion of $I(\varepsilon)$ without computing $I(\varepsilon)$ explicitly.

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    $\begingroup$ do the change of variable $t=\epsilon s$. $\endgroup$ Commented Jan 11 at 15:05
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    $\begingroup$ One thing is for certain: you cannot obtain the $\varepsilon \log \varepsilon$ term using solely a power series expansion. $\endgroup$
    – Diffusion
    Commented Jan 11 at 19:52
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    $\begingroup$ This expansion of the integrand is pointless since the main contribution to the integral comes from $t\lesssim\epsilon$, but then nothing gets small in the series. $\endgroup$ Commented Jan 11 at 20:01
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    $\begingroup$ I don't think there can be a general method for the general problem. In the example, one way saying it is that you are trying to expand the harmonic (on $\mathbb C^+$) function $u(x,y)=P*e^{-|t|}$ about $(0,0)$, along vertical directions $(0,\epsilon)$. However, since $e^{-|t|}$ is not smooth at $t=0$, power series expansions are ruled out. $\endgroup$ Commented Jan 11 at 20:19
  • $\begingroup$ This question is too general. But it seems like the notion of asymptotic expansion might help. $\endgroup$
    – Nemo
    Commented Jan 12 at 7:44

2 Answers 2


Write $$I(\epsilon)=\epsilon\int_0^1 \frac{e^{-s}}{\epsilon^2+s^2}, ds +\epsilon\int_1^\infty \frac{e^{-s}}{\epsilon^2+s^2}, ds =: J(\epsilon)+K(\epsilon).$$ Then $$J(\epsilon)=\epsilon \int_0^1 \sum_{n=0}^\infty \frac{(-1)^n s^n}{n!(\epsilon^2+s^2)}\, ds= \arctan \epsilon^{-1}-\frac {\epsilon}{2}\log (1+\epsilon^{-2})+\epsilon \sum_{n \geq 2}\frac{(-1)^n}{n!}\int_0^1 \frac{s^n}{\epsilon^2+s^2}\, ds $$ so that $J(\epsilon) \approx \frac \pi 2-\epsilon+\epsilon \log \epsilon +\epsilon \sum_{n \geq 2}\frac{(-1)^n}{n!(n-1)}+O(\epsilon)^2$. Also $$ K(\epsilon)=\epsilon \int_1^\infty \frac{e^{-s}}{s^2(1+\epsilon^2 s^{-2})}\, ds=\sum_{n=0}^\infty(-1)^n \epsilon^{2n+1}\frac{e^{-s}}{s^{2n+2}}\, ds=\epsilon \int_1^\infty \frac{e^{-s}}{s^2}\, ds+O(\epsilon^2). $$ Summing all together $$I(\epsilon)=\frac \pi 2+\epsilon \log \epsilon +\epsilon \left (\sum_{n \geq 2}\frac{(-1)^n}{n!(n-1)}+\int_1^\infty \frac{e^{-s}}{s^2}\, ds -1 \right )+O(\epsilon^2).$$

  • $\begingroup$ Curiously, the value of the series and the value of the integral , according to Maple, have no simple expression; yet their sum does, namely just the Euler-Mascheroni constant. $\endgroup$ Commented Jan 12 at 19:46
  • $\begingroup$ Yes, but I do not now a proof... $\endgroup$ Commented Jan 12 at 19:54
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    $\begingroup$ The sum is $\int_0^1 \frac{e^{-x}+x-1}{x^2}dx$ (integrate term-wise), and the integral of $\frac{e^{-s}}{s^2}$ can be transformed, via integration by parts, to the integral of $\frac{e^{-s}}{s}$. And now the expression in the brackets can be computed from the known asymptotics for the exponential integral ($\gamma$ comes from $Ei(x) = - \gamma - \log(x) + O(x)$ for small $x$). $\endgroup$ Commented Jan 12 at 20:25
  • $\begingroup$ @AlekseiKulikov Thank you, this closes the gap. $\endgroup$ Commented Jan 12 at 20:33
  • $\begingroup$ Thank you, I will try whether this approach helps me with the actual function I am trying to expand. $\endgroup$ Commented Jan 15 at 15:51

Some partial progress: following the suggestion in the comments, make the substitution $t\mapsto \varepsilon t$ to arrive at the integral $$I(\varepsilon) = \int_0^\infty \frac{e^{-\varepsilon t}}{1+t^2}\,dt$$ Now, consider first the portion of the integral over $[0,1/\varepsilon]$, \begin{aligned} g(\varepsilon)= \int_0^{1/\varepsilon} \frac{e^{-\varepsilon t}}{1+t^2}\,dt. \end{aligned} You can show that the remaining part over $[1/\varepsilon, \infty)$ is $O(\varepsilon^2)$. Using limits as necessary, we have that $$g(\varepsilon)-g(0)= \int_0^{1/\varepsilon} \frac{e^{-\varepsilon t}-1}{1+t^2}\,dt - \varepsilon + O(\varepsilon^2). $$ Then, note that $$\int_0^{1/\varepsilon} \frac{e^{-\varepsilon t}-1}{1+t^2}\,dt=\int_0^{1/\varepsilon}\frac{-\varepsilon t}{1+t^2}dt +O(\varepsilon)=\varepsilon \log(\varepsilon)+ O(\varepsilon).$$ At least, this calculation shows how to get the $\varepsilon \log(\varepsilon)$ in the expansion. What is left is to show that $$\int_0^{1/\varepsilon}\frac{e^{-\varepsilon t}-1}{1+t^2}\,dt = -\varepsilon \log(\varepsilon) + \gamma \varepsilon+ O(\varepsilon^2).$$


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