# Interchange of integral and sum

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.

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.

• Thanks! But I'm left wondering if the same can be achieved without resorting to Lebesgue integration - with respect to the Riemann integral?
– Dark
Mar 26 '20 at 20:48

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