Suppose we have a bounded function $f:\mathbb{R} \to \mathbb{C}$. I want to compute $$ \int_\mathbb{R} e^{-x^2} f(x) dx $$

Of course this integral exists. I know that $f$ has a Taylor expansion which is valid for all of $\mathbb{R}$, say $$ f(x)=\sum_{r=0}^\infty a_r \frac{x^r}{r!} $$

Is it generally true that

$$ \int_\mathbb{R} e^{-x^2} f(x) dx = \sum_{r=0}^\infty \frac{a_r}{r!} \int_\mathbb{R} e^{-x^2} x^r dx$$

Of course all the integrals on the right hand side also exist. However, I cannot use the theorem of uniform convergence, since the Taylor expansion does not converge uniformly on $\mathbb{R}$ but of course on any compact subset. So my guess is that a this equality should hold anyways, in particular since the sum is not some wierdly constructed counterexample but a regular Taylor series.

How could I prove such a result. In fact I do not necessarily need to know how to prove it but just know it. Is there a reference?

  • $\begingroup$ Of course validity will depend on the coefficients $a_r$. You could try the basic limit theorems of Lebesgue theory, and see if they work in your case: monotone convergence, dominated convergence. $\endgroup$
    – GEdgar
    Jun 6, 2011 at 12:37
  • $\begingroup$ ok, I would like it to hold in a pretty general form. Of course you are right and this is not true. But we may impose the condition that the right hand side converges. Since the integral on the right hand side is roughly $\Gamma(r/2)$ it should be enough to require $a_r$ being at most polynomial in $r$. Will this suffice? $\endgroup$
    – wood
    Jun 6, 2011 at 12:45
  • $\begingroup$ It does hold in "pretty general" form. In fact, I suspect you would have to work pretty hard to find an example of $\int_\mathbb{R} e^{-x^2} \sum_{r} a_r x^r dx \ne \sum_{r} a_r \int_\mathbb{R} e^{-x^2} x^r dx$ where both sides exist. $\endgroup$
    – GEdgar
    Jun 6, 2011 at 12:49
  • $\begingroup$ the problem is that in my case the $a_r$ are pretty complicated expressions. I might be able to find bounds for them but I want to start doing that only after having a sufficient condition. unfortunately monotone convergence will not work, since I work with complex valued functions and I doubt that the real resp imaginary parts will behave monotonously. And dominated convergence is also not really applicable because I do not know a lot about the partial sums of the Taylor expansion. $\endgroup$
    – wood
    Jun 6, 2011 at 12:58

1 Answer 1


Summation over $\mathbb{N}$ is actually integration with respect to a measure with discrete "atoms", that is a measure defined on $\mathbb{N}$ that assignes a real or complex number to every natural number.

In our case we can define a measure $\mu$ on $\mathbb{N}$ via $$ \mu (r) := \frac{a_r}{r!} $$ If we next define a function on the product measure space $(\mathbb{R}, d x) \times (\mathbb{N}, \mu)$ via $$ g(x, r) := \exp(- x^2) x^r $$ then your question becomes a candidate for an application of Fubini's theorem.

Edit: to avoid making $\mu$ a signed measure, we can redefine $$ \mu (r) := \frac{1}{r!} $$ and $$ g(x, r) := a_r \; \exp(- x^2) \; x^r $$

The question is of course: are the coefficients $a_r$ such that Fubini's theorem is applicable? That depends on the coefficients, of course...

  • 4
    $\begingroup$ I don't buy this argument. First of all, your measure $\mu$ will be a signed measure. How do you know that $\sum |a_r|/r!$ converges so that the several rearrangements of the sums that seem implicitly involved here can be justified? $\endgroup$
    – t.b.
    Jun 6, 2011 at 13:12
  • $\begingroup$ @Theo, Tim You write "first of all". If I could indeed prove that in my particular case $\sum |a_r|/r!$, do you have other objections against this answer. Honestly I do not have a lot of knowledge about measure theory, so I cannot say anything about the answer myself $\endgroup$
    – wood
    Jun 6, 2011 at 14:37
  • $\begingroup$ Hm, of course we don't know that everything converges, which is why I did not say "Fubini's theorem says that sum and intergral are always interchangeable", but "the question is a candiate for an application of Fubini's theorem". Then I don't think that it is really a problem that the $a_r$ may be negative, turning $\mu$ into a signed measure, because we can of course shift them from the measure to the function $g(x, r)$, or am I missing something? $\endgroup$ Jun 6, 2011 at 14:39
  • $\begingroup$ @claudi85: No further objections and having $\sum |a_r|/r! \lt \infty$ is certainly enough. I read the first version of the answer and didn't see that it was changed while I was thinking about it (the crucial "candidate for" was added then). @Tim: No you're not missing anything, but in order to do this, you seem to need that the function $g$ be integrable wrt the product measure and it seems to me that this is essentially equivalent to $\sum |a_n|/r! \lt \infty$, but I may be wrong here, as I haven't thought very deeply about this problem. $\endgroup$
    – t.b.
    Jun 6, 2011 at 15:20
  • 3
    $\begingroup$ Well, since we are told that the series $\sum a_r x^r/r!$ has infinite radius of convergence, then of course $\sum |a_r|/r! < \infty$. $\endgroup$
    – GEdgar
    Jun 6, 2011 at 18:32

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