Compute $\int_\mathbb{R} \cos(ty) d\delta_x(y)$

Question

Compute $$\int_\mathbb{R} \cos(ty) d\delta_x(y), $$ for some fixed $t,x \in \mathbb{R}$, where $\delta_x$ is the point-density measure at $x$, that is: $$\forall A \in \mathcal{B}(\mathbb{R}): \delta_x(A) = \left\{ \begin{array}{ll} 1, x \in A \\ 0, x \notin A \end{array} \right. $$

I have tried using the change of variables formula and the push-forward measure. We can see that $$\forall b \in \mathbb{R}: \delta_x((-\infty, b]) = \left\{ \begin{array}{ll} 1, b \geq x \\ 0, b < x \end{array} \right. = \chi_{[x, \infty)}(b), $$ the characteristic function of $[x, \infty)$. Therefore, we have that $$\int_\mathbb{R} \cos(ty) d\delta_x(y) = \int_\mathbb{R} \cos(ty) \cdot \chi_{[x,\infty)}(y) d\mathcal{L}^1(y) = \int_{[x, \infty)} \cos(ty) d\mathcal{L}^1(y), $$ but I don't know how to compute the last integral. Are there other ways to compute this?

Remark I know that this integral comes from $$\int_\mathbb{R} e^{-ity} d\delta_x(y). $$ which is the Fourier transform of a delta-function, but I am looking for a more elementary solution.

Answer 1

$\int_{\Bbb{R}}\cos{ty}d\delta_x(y)=\int_{\Bbb{R} \setminus \{x\}}\cos{ty}d\delta_x(y)+\int_{\{x\}}\cos{ty}d\delta_x(y)=0+ (\cos{tx})\delta_x(\{x\})=\cos{tx}$

Answer 2

Re-wording Marios's solution
Prove the more general fact. If $f : \mathbb R \to \mathbb R$ is any function, then $\int_{\mathbb R} f(y)\;\delta_x(y) = f(x)$. Indeed, the function $f$ is equal $\delta_x$-almost everywhere to the constant $f(x)$, so they have the same integral.