I solved the distributional limit as seen below

$$\lim_{x\rightarrow+0}\left(\int_0^\infty \exp(\frac{-ARx}{2})\cos(R(y-t)) \, dR\right)=\frac{2\pi \delta(t-y)}{A}$$

where $ -1<y<1 $ , $ -1<t<1 $ and $A$ is a positive constant.

I am not sure about the existence of the multiplier $\frac{2}{A}$ on the rigth hand side of the equation. Does it exist? if it does not, what is the correct solution.

Best wishes..

  • $\begingroup$ Did you find this answer? $\endgroup$ – science Feb 8 '15 at 21:03
  • $\begingroup$ i supposed that i found the answer. but i am not sure whether it is true or not. Thats why i am here. $\endgroup$ – Onur Feb 8 '15 at 21:11
  • $\begingroup$ @Omur: How did you find this answer? $\endgroup$ – science Feb 8 '15 at 21:33
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    $\begingroup$ 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. $\endgroup$ – Onur Feb 8 '15 at 21:39
  • $\begingroup$ i am not sure about the solution if the inside of the exponential is multiplied by A/2. $\endgroup$ – Onur Feb 8 '15 at 21:44

Rewrite your integral as $$ I(A,t,y)=\lim_{x\rightarrow0+}\Re\int_{0}^{\infty}e^{-ARx/2} e^{iR(y-t)} =\lim_{x\rightarrow0+}\Re\frac{1}{Ax/2-i(y-t)}=\lim_{x\rightarrow0+}\frac{Ax/2}{(Ax/2)^2+(y-t)^2} $$

It's well known that such expressions are a representation of Dirac's Delta distribution: $\delta(y)=\lim_{x\rightarrow0}\frac{x}{x^2+y^2}$

We obtain $$ I(A,t,y)=\frac{2 \pi}{A}\delta\left(\frac{4}{A^2}(y-t)\right) $$

Using the identiy $\delta(ax)=\frac{1}{|a|}\delta(x)$ this simplifies to $$ I(A,t,y)=\frac{A\pi}{2}\delta\left(y-t\right) $$ So i think your solution is not correct.

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    $\begingroup$ something unclear? $\endgroup$ – tired Feb 10 '15 at 20:19
  • $\begingroup$ I think there is a mistake in last two equation, they should be $$ I(A,t,y)=\frac{2 \pi}{A}\delta\left(\frac{2}{A}(y-t)\right)=\pi\delta\left(y-t\right) $$ Am i wrong? $\endgroup$ – Onur Mar 6 '15 at 20:16

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