"Formal" asymptotic approximation of an integral

Question

I wanted to asymptotically evaluate the following integral: $$ \mathcal{I}(\epsilon)\equiv\int_{-\infty}^{\infty}dx\int_{-|x|}^{|x|}dy f(y)\,\exp{\left(-\frac{|x|}{\epsilon}\right)}\,, $$ for $\epsilon\sim0$ ($\epsilon>0$), where $f:\mathbb{R}\mapsto\mathbb{R}^+$, $f(y)\in C^{\infty}$, with an absolut maximum on $x_0=0$. Moreover $f(y)$ is an even function of $y$ and it normalized to one: $\int_{-\infty}^{\infty}f(y)dy=1$. I wanted to know how to formally treat this integral other than naively expand $f(y)$ inside the integral: $$ \mathcal{I}(\epsilon)\stackrel{?}{=}\int_{-\infty}^{\infty}dx\int_{-|x|}^{|x|}dy \left(f(0)+\frac{1}{2}f''(0)\,y^2\right)\,\exp{\left(-\frac{|x|}{\epsilon}\right)}+O\left(\epsilon^6\right)=4f(0)\epsilon^2+4f''(0)\epsilon^4+O\left(\epsilon^6\right)\,. $$ I don't think this naiv expansion is formal and I wanted to know how to do it properly. Moreover it would be useful to know some good references where to start to study asymptotics of integrals. Thanks!

Answer 1

Besides regular asymptotic methods, one can use symmetry and swap the order of integration to obtain $$I = 4\int_0^\infty dy \int_y^\infty f(y)\exp\left(-\frac{x}{\epsilon}\right)$$ $$ = 4\epsilon^2\int_0^\infty dt \: f(\epsilon t)\exp(-t) = 4f(0)\epsilon^2 +O(\epsilon^4)$$

We used the fact that $f$ is even to deduce that the next order correction would be $\epsilon^4$ rather than $\epsilon^3$.

Your "naive" expansion can be more rigorously justified in the final integral in terms of $t$. Swapping the integral with the summation for the Taylor series is allowed because the series is uniformly convergent on the interval of integration.