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I've been going through this proof.

And I'm wondering what allows me to change the order of the integral and the infinite sum.

$$\int_{-\infty}^{\infty} \left( \sum_{n \ge 0} \frac{2^n t^n x^n}{n!} \right) e^{-x^2} dx = \sum_{n \ge 0} \frac{2^n t^n}{n!} \int_{-\infty}^{\infty} x^n e^{-x^2} \, dx.$$

I know the result that for a function series that converges uniformly on $[a, b]$ the following holds:

$$ \int^{b}_{a} \sum_{n=0}^{\infty} f_n(x) \ dx = \sum_{n=0}^{\infty} \int_a^b f_n(x) \ dx$$

But here the limits of integration are $-\infty$ and $+\infty$, and I'm not sure what to make of it.

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The series can be written as $$ \sum_{n=0}^\infty \frac{(2tx)^n}{n!}e^{-x^2}\leq \sum_{n=0}^\infty \frac{(2|t||x|)^n}{n!}e^{-x^2}=e^{-(x^2-2|t||x|)}=e^{-[(|x|-|t|)^2-|t|^2]}=e^{t^2}e^{-(|x|-|t|)^2} $$ Obviously, the right hand term is integrable (it is essentially a Gaussian distribution centered at $|t|$). Call it $f(x;t)$. Now let the partial sum be denoted as $g_N(x;t)=\sum_{n=0}^N \frac{(2tx)^n}{n!}e^{-x^2}$. Clearly, $|g_N(x;t)|\leq \sum_{n=0}^N \frac{(2|t||x|)^n}{n!}e^{-x^2}\leq f(x;t)$ (since all terms in sum are positive). By dominated convergence theorem, $$ \lim_{N\to\infty}\int_{-\infty}^\infty g_N(x;t)\,dx=\int_{-\infty}^\infty\lim_{N\to\infty}g_N(x;t)\,dx. $$ For the left hand side, partial sums can always be pulled out of integrals. The conclusion then holds.

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    $\begingroup$ Thanks! But I'm left wondering if the same can be achieved without resorting to Lebesgue integration - with respect to the Riemann integral? $\endgroup$ – Darks Mar 26 at 20:48
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Let $\mu$ be a positive measure on $X.$ If $S_N(x)\to S(x)$ pointwise $\mu$ a.e. on $X,$ and there exists $f\in L^1(X,\mu)$ such that $|S_N(x)|\le f(x)$ on $X$ for each $N,$ then

$$\int_X S_N \,d\mu \to \int_X S \,d\mu.$$

This is just straight up DCT. All that's left to do is think of our problem, where $t$ is fixed, identify $S_N$ as

$$\sum_{n=0}^{N}\frac{2^nt^nx^n}{n!}e^{-x^2}$$

and identify $f(x)= \exp (2|t||x|-x^2)\dots$

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