# I need to prove $\int\sum a_k=\sum\int a_k$ for a certain $\{a_n\}$

I've been solving a problem from my integral calculus class and I've found I need to prove that: $$\int^\infty_{-\infty}\sum^\infty_{k=0}((-1)^k\frac{(ax)^{2k}}{(2k)!}e^{-x^2})dx=\sum^\infty_{k=0}((-1)^k\frac{a^{2k}}{(2k)!}\int^\infty_{-\infty}x^{2k}e^{-x^2}dx)$$ To be said, I need to prove that I can swap the integral and the sum. I think it has something to do with the DCT (Dominated Convergence Theorem), does that mean that I just need to find a measurable and integrable function $$g(x)$$ so that $$|f(x)|\leq g(x),\forall x\in\mathbb{R}?$$ I know to solve the rest of the problem using that property, I just need to prove that I can swap the integral and the sum.

• Perhaps you could bound the summation by $e^{ax}$? – copper.hat Jun 1 at 18:24
• But is bounding it enough to be able to swap the integral and the sum? That's my question – Alejandro Bergasa Alonso Jun 1 at 18:26
• @copper.hat Also how do I prove that $e^{ax}$ bounds my expression? – Alejandro Bergasa Alonso Jun 1 at 19:02

I'm not sure I understood correctly your last remark.
If you are asking if you can use DCT to change the order of summation and integration, then: Yes.
Denote $$f_n(x) = \sum_{k=0}^n g_k(x),\quad f(x) = \sum_{k=0}^\infty g_k(x)$$

You can prove with DCT that: $$\lim_{n\to\infty} \int_{-\infty}^\infty f_n(x)dx = \int_{-\infty}^\infty f(x)dx$$ but the lefthand side is actually equal to (using linearity of the integral, because it is a finite sum): $$\lim_{n\to\infty} \int_{-\infty}^\infty f_n(x)dx = \lim_{n\to\infty} \int_{-\infty}^\infty \sum_{k=0}^n g_k(x) dx = \lim_{n\to\infty} \sum_{k=0}^n \int_{-\infty}^\infty g_k(x) dx = \sum_{k=0}^\infty \int_{-\infty}^\infty g_k(x) dx$$

So, over all we got: $$\sum_{k=0}^\infty \int_{-\infty}^\infty g_k(x) dx = \int_{-\infty}^\infty \sum_{k=0}^\infty g_k(x)dx$$ Which is just the order of summation and integration changed.
To summarize, if the partial sums of the series are dominated as prescribed in DCT, then you can change order of summation and integration.

If I understood your last remark incorrectly, and you meant that you are having problem showing that the partial sums are dominated, let me know and I'll add it to my answer

EDIT:
Here is how I would have shown the partial sums are dominated.
Firstly, $$\left| \sum_{k=0}^n (-1)^k \frac{(ax)^{2k}}{(2k)!} e^{-x^2} \right| = e^{-x^2} \left| \sum_{k=0}^n (-1)^k \frac{(ax)^{2k}}{(2k)!} \right|$$ Now note that the sum is the Taylor approximation of $$\cos(ax)$$, So: $$\left| \sum_{k=0}^n (-1)^k \frac{(ax)^{2k}}{(2k)!} \right| \xrightarrow{n\to\infty} \left| \cos(ax) \right| \le 1$$ As such, there exists $$N \in \mathbb{N}$$ such that $$\forall n \ge N$$: $$\left| \sum_{k=0}^n (-1)^k \frac{(ax)^{2k}}{(2k)!} \right| \le 2$$ We got that $$\forall n \ge N$$ $$\left| \sum_{k=0}^n (-1)^k \frac{(ax)^{2k}}{(2k)!} e^{-x^2} \right| = e^{-x^2} \left| \sum_{k=0}^n (-1)^k \frac{(ax)^{2k}}{(2k)!} \right| \le 2e^{-x^2} \in L^1$$ and it is enough for DCT that the sequence is eventually dominated, the finite number of elements in the sequence are negligible.

• I also have problems proving that my expression is dominated also, so it would be so helpful if you help me prove it. Thanks for answering! – Alejandro Bergasa Alonso Jun 1 at 19:18

Note that $$|\sum^n_{k=0}((-1)^k\frac{(ax)^{2k}}{(2k)!}e^{-x^2} | \le \sum^\infty_{k=0} {|ax|^k \over k!} e^{-x^2} = e^{-|ax| -x^2}$$ and the latter is integrable.

If we let $$f_n(x) = \sum^n_{k=0}((-1)^k\frac{(ax)^{2k}}{(2k)!}e^{-x^2}$$ we see that the dominated convergence theorem gives $$\lim_n \int f_n \to \int \lim_n f$$ and linearity of the integral gives $$\int f_n = \sum^n_{k=0}((-1)^k\frac{a^{2k}}{(2k)!}) \int_{-\infty}^\infty x^{2k}e^{-x^2}dx$$.