# Commutativity of integration and Taylor expansion of the integrand in an integral

I am baffled with a seemingly a straightforward problem. Suppose we are given the following integral:

$$f(a)\,=\,\int_{0}^{\infty} \frac{x^4}{x^4+a^4} e^{-x},$$ 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,

$$f(a)=\int_{0}^{1} \frac{x^4}{x^4+a^4} e^{-x} + \int_{1}^{\infty} \frac{x^4}{x^4+a^4} e^{-x}\,,$$ or

$$f(a)=\int_{0}^{2a} \frac{x^4}{x^4+a^4} e^{-x} + \int_{2a}^{\infty} \frac{x^4}{x^4+a^4} e^{-x}\,.$$ $\frac{x^4}{x^4+a^4}$ 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 $a$.

$$\int_{2a}^{\infty} \frac{x^4}{x^4+a^4} e^{-x}\,=\, \sum_{n=0}^{\infty} \int_{2a}^{\infty} \frac{(-1)^n a^{4n}}{x^{4n}}e^{-x} \,=\, \sum_{n=0}^{\infty}(-1)^{n}a^{4n}\Gamma(1-4n,2a).$$ 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 think your computation of the Taylor expansion of the latter case is wrong in the sense that it is not a Taylor expansion. Your integral boundaries depend on $a$ which has to be taken into account. – Dominik Jan 21 '14 at 16:37
• Thanks Dominik. But that's exactly what I would like to know. Does that mean I can have a Taylor expansion for a function (here $f(a)$), and at the same time a series expansion that involve non-analytic terms of the above form? – S.G. Jan 21 '14 at 17:00
• What series expansion did you use for $\Gamma(1-4n,2a)$? I don't see where the terms with $\ln(a)$ come from. – Pulsar Jan 21 '14 at 18:54
• Thanks Pulsar. You can expand $\Gamma(m,a)$ in Mathematica. One way to see the log terms is the following: $\Gamma(m,a)=\int_{a}^{\infty} \text{d}s s^{m-1}e^{-s}$. For $a\ll 1$, this can be written as $\int_{a}^{1}\text{d}s \,s^{m-1}e^{-s}+\Gamma(m,1)$. Now for $m\leq 0$, the remaining integral can be dealt with by Taylor expanding the exponential. For any $m$ there will be a $s^m$ term from the Taylor expansion of the integral that, together with the $s^{m-1}$ term, will leave you $\frac{1}{s}$. Then, doing the $s$ integral gives the log term. See the expansion of $\Gamma(0,a)$ on Wikipedia. – S.G. Jan 21 '14 at 19:23
• Incidentally, this question is more suited for math SE. I'll ask for a migration. – Pulsar Jan 21 '14 at 20:05

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.