evaluate $\int_\mathbb{R}\cos(yx)\,e^{-x^2/2}\,\lambda(dx)$

Question

(assumption:$(\mathcal{B},\lambda)$ is measure on $\mathbb{R}$, $\forall (a,b]\in\mathcal{B}$ and $\lambda((a,b])=b-a$)

I'm trying to solve the following problem using $\int_\mathbb{R}x^{2k}\,e^{-x^2/2}\,\lambda(dx)=\sqrt{2\pi}\frac{(2k)!}{2^kk!}\,\,(k\in\mathbb{Z}_{\geq0})$.
\begin{align} \int_\mathbb{R}\cos(yx)\,e^{-x^2/2}\,\lambda(dx)=?\qquad(y\in\mathbb{R}) \end{align} I tried to solve it as follows ; \begin{align} \int_\mathbb{R}\cos(yx)\,e^{-x^2/2}\,\lambda(dx)&=\int_\mathbb{R}\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}(yx)^{2n}\,e^{-x^2/2}\,\lambda(dx)\tag{1} \\ &=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}y^{2n}\int_{\mathbb{R}}x^{2n}\,e^{-x^2/2}\,\lambda(dx)\tag{2} \\ &=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}y^{2n} \cdot \sqrt{2\pi}\frac{(2n)!}{2^nn!} \\ &=\sqrt{2\pi}\,e^{-y^2/2} \end{align} However, the transformation from (1) to (2) does not seem correct because $f_n(x)=\frac{(-1)^n}{(2n)!}(yx)^{2n}\,e^{-x^2/2}$ is not always positive.
(I recognize that for the exchange between $\int$ and $\sum$ to be valid, $f_n(x)$ always be positive and $\mathcal{B}$-measurable.)
How should we solve this problem?

Answer 1

Your solution is perfectly fine if we can justify the interchange of the summation and integration. One way to do this would be to use Fubini's theorem, however as you mentioned in the comments that you are unfamiliar with it, we'll do without it and instead rely on the Dominated Convergence Theorem (DCT).

So fix $y\in\mathbb{R}$, and set

$$f_n(x)=\sum_{j=0}^n\frac{(-1)^j}{(2j)!}(xy)^{2j}e^{-\frac{x^2}{2}},\quad f(x)=\lim_{n\to\infty}f_n(x).$$

What we need to show is that

$$\int_\mathbb{R}f(x)~\mathrm{d}x=\lim_{n\to\infty}\int_\mathbb{R}f_n(x)~\mathrm{d}x,$$

as that justifies what you have done already. So observe that

\begin{align*} \lvert f_n(x)\rvert &\leq e^{-\frac{x^2}{2}}\sum_{j=0}^n\frac{(xy)^{2j}}{(2j)!} \\ &\leq e^{-\frac{x^2}{2}}\sum_{j=0}^\infty\frac{(xy)^{2j}}{(2j)!} \\ &=\frac{1}{2}\left(e^{xy}+e^{-xy}\right)e^{-\frac{x^2}{2}}, \end{align*}

and in a similar way we can dominate $f$ by the same function. As $x\mapsto\frac{1}{2}\left(e^{xy}+e^{-xy}\right)e^{-\frac{x^2}{2}}$ is integrable, the result now follows by the DCT.

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An alternative solution to the problem is to turn it into a differential equation. I'll leave some of the details to you. So set

$$I(y)=\int_{\mathbb{R}}\cos(yx)e^{-\frac{x^2}{2}}~\mathrm{d}x,$$

and observe that

\begin{align*} I'(y) &\overset{(*)}{=}\int_\mathbb{R}\frac{\partial}{\partial y}\cos(yx)e^{-\frac{x^2}{2}}~\mathrm{d}x \\ &=-\int_\mathbb{R}x\sin(yx)e^{-\frac{x^2}{2}}~\mathrm{d}x \\ &\overset{(**)}{=}-y\int_\mathbb{R}\cos(yx)e^{-\frac{x^2}{2}}~\mathrm{d}x \\ &=-yI(y), \end{align*}

where $(*)$ can be justified by using the DCT (or a measure theoretic version of the Leibniz integral rule, such as the one on Wikipedia), and $(**)$ comes from integrating by parts. Now the ODE

$$I'(y)+yI(y)=0$$

can be solved either using that it is separable, or using an integrating factor (I'll leave the details to you). Solving yields that

$$I(y)=Ce^{-\frac{y^2}{2}}$$

for some $C\in\mathbb{R}$. Finally, observe that

$$C=I(0)=\int_\mathbb{R}e^{-\frac{x^2}{2}}~\mathrm{d}x=\sqrt{2\pi},$$

yielding that

$$I(y)=\sqrt{2\pi}e^{-\frac{y^2}{2}}$$

as was desired.