Is the equation below true for $ x>0$ ?
$$\lim_{x\rightarrow0}\left(\int_0^\infty \exp(-Rx)\cos(R(y-t)) \, dR\right)=\pi \delta(t-y)$$
Actually i don't understand why the multiplier $\pi$ exists in the RHS of the equation. i must be sure of that.
Thanks.
This a distributional limit equation. To prove it, fix $x$ and $y$, and set $$\psi(t;x,y) = \int_0^\infty \exp(-Rx)\cos(R(y-t))\, dR \qquad (t \in \Bbb R)$$ By integration by parts, $\psi(t;x,y) = \frac{x}{x^2 + (y-t)^2}$. Thus, for any $\phi \in C_c^\infty(\Bbb R)$, $$\lim_{x\to 0^+}\int_{-\infty}^\infty \psi(t;x,y)\phi(t)\, dt = \pi\phi(y)$$
On the other hand, $$\int_{-\infty}^\infty \pi\delta(t - y)\phi(t)\, dt = \pi\phi(y)$$ Hence $\lim_{x\to 0^+} \psi(t;x,y) = \pi\delta(t-y)$ distributionally.