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.

This post imported from StackExchange Mathematics at 2014-06-16 11:25 (UCT), posted by SE-user S.G.