I am baffled with a seemingly a straightforward problem. Suppose we are given the following integral:
\begin{equation} f(a)\,=\,\int_{0}^{\infty} \frac{x^4}{x^4+a^4} e^{-x}, \end{equation} and we want to determine the dependence of $f(a)$ on $a$ when $a\ll 1$. Apparently this integral can be solved using Mathematica. Taylor expanding the result, which is a Meijer G-function, it turns out that $f(a)$ is analytic in $a$.
In the specific case of this integral, it's possible to use a trick so that one can directly Taylor expand the integrand (Taylor expanding the integrand of $f(a)-f(0)$ after $x\to x'=a x$). But I'm not interested in this particular integral and am mentioning this as a simple example.
Now here is what I find paradoxical: Let's try to do this in a more pedestrian way by breaking up the integration range and Taylor expanding the exponential when x is small and the rest of the integrand when x is large. Interchanging the integration and summation is justified by Fubini's theorem (if I'm not mistaken, $\int \sum |c_n(x)| <\infty$ or $\sum\int |c_n(x)|<\infty$).
Now, breaking up the integral can be done in two ways. Either,
\begin{equation} f(a)=\int_{0}^{1} \frac{x^4}{x^4+a^4} e^{-x} + \int_{1}^{\infty} \frac{x^4}{x^4+a^4} e^{-x}\,, \end{equation} or
\begin{equation} f(a)=\int_{0}^{2a} \frac{x^4}{x^4+a^4} e^{-x} + \int_{2a}^{\infty} \frac{x^4}{x^4+a^4} e^{-x}\,. \end{equation} $\frac{x^4}{x^4+a^4}$ can be Taylor expanded and the integration ranges are within the convergence radius in both cases. The Taylor expansion in both cases results in a series that's uniformly convergent and therefore one should be able to interchange integration and summation.
The former case, where the integration range is broken up at $1$, gives an analytic result in $a$. Curiously, the latter (breaking up the integral at $2a$) gives non-analytic terms (see below) and I cannot figure out how to reconcile this with the exact result. The lower integration ranges in both cases give analytic expressions in $a$.
\begin{equation} \int_{2a}^{\infty} \frac{x^4}{x^4+a^4} e^{-x}\,=\, \sum_{n=0}^{\infty} \int_{2a}^{\infty} \frac{(-1)^n a^{4n}}{x^{4n}}e^{-x} \,=\, \sum_{n=0}^{\infty}(-1)^{n}a^{4n}\Gamma(1-4n,2a). \end{equation} Using the series expansion of the upper incomplete $\Gamma$-function, there will be terms of the form $\frac{-(-1)^n}{(4n-1)!} a^{4n} \ln(a)$.
I would like to know whether the Taylor expansion is not justified (if so, why precisely), or, although hard to imagine, is it that somehow these non-analytic terms sum up to an analytic result. Thanks.
I figured it out. There is no contradiction. Both integrals do exactly the same result, which coincides with what one finds from doing the integral using Mathematica and series expanding the quoted Meijer G-function.
The $a^{4n}\ln(a)$ terms do appear in the case where the integration range is broken up at $1$. They show up in the lower-range integral ($\int_{0}^{1}\cdots$). I should have been more careful. Sorry for the confusion.
The direct Taylor expansion that I alluded to was incorrect and does fail.