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

+ 4 like - 0 dislike
30 views

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.
asked Jan 21, 2014 in Mathematics by S.G. (30 points) [ no revision ]
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.

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

This post imported from StackExchange Mathematics at 2014-06-16 11:25 (UCT), posted by SE-user S.G.
What series expansion did you use for $\Gamma(1-4n,2a)$? I don't see where the terms with $\ln(a)$ come from.

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

This post imported from StackExchange Mathematics at 2014-06-16 11:25 (UCT), posted by SE-user S.G.
Hmm, my approach doesn't work; the denominators become zero. I'm deleting my answer again.

This post imported from StackExchange Mathematics at 2014-06-16 11:25 (UCT), posted by SE-user Pulsar
Incidentally, this question is more suited for math SE. I'll ask for a migration.

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

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

1 Answer

+ 1 like - 0 dislike

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.

This post imported from StackExchange Mathematics at 2014-06-16 11:25 (UCT), posted by SE-user S.G.
answered Jan 22, 2014 by S.G. (30 points) [ no revision ]

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