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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?

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    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.2011-06-06

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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...

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    @GEdgar: Well, that's rather embarrassing...2011-06-06