5
$\begingroup$

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.

$\endgroup$
7
  • 2
    $\begingroup$ do the change of variable $t=\epsilon s$. $\endgroup$ Commented Jan 11 at 15:05
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

1
$\begingroup$

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 $$I(\epsilon)=\frac \pi 2+\epsilon \log \epsilon +\epsilon \left (\sum_{n \geq 2}\frac{(-1)^n}{n!(n-1)}+\int_1^\infty \frac{e^{-s}}{s^2}\, ds -1 \right )+O(\epsilon^2).$$

$\endgroup$
7
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
0
$\begingroup$

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 $$\int_0^{1/\varepsilon}\frac{e^{-\varepsilon t}-1}{1+t^2}\,dt = -\varepsilon \log(\varepsilon) + \gamma \varepsilon+ O(\varepsilon^2).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.