2
$\begingroup$

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..

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.