Calculate limits as distributions

[Haorong Wu]

TL;DR Summary

How to calculate the following limits, when viewed as distributions?

Hi, there. I am reading this thesis. On page 146, it reads that

when viewed as distributions, one can show that the following limits holds:
$$\lim_{r\rightarrow \infty}\frac {\sin ((Q-Q')r)}{Q-Q'}=\pi \delta(Q-Q') ,$$
$$\lim_{r\rightarrow \infty}\frac {\cos ((Q+Q')r)}{Q+Q'}=0 .$$

I do not know how to calculate the limits when they are viewed as distributions. I am trying to integrate a test function with the limits. So I try ($Q$ is defined as $Q>0$) $$\lim_ {r\rightarrow \infty} \int_{0}^\infty dQ \cos ((Q-Q')r )\frac {\sin ((Q-Q')r)}{Q-Q'}=\frac \pi 2,$$ while $\int_{-\infty}^\infty dQ \cos ((Q-Q')r ) \delta (Q-Q')=1$. Then I only have $\lim_{r\rightarrow \infty}\frac {\sin ((Q-Q')r)}{Q-Q'}=\pi \delta(Q-Q') /2$. Is this wrong? Thanks.

 

[Philip Koeck]

TL;DR Summary: How to calculate the following limits, when viewed as distributions?

Hi, there. I am reading this thesis. On page 146, it reads that
I do not know how to calculate the limits when they are viewed as distributions. I am trying to integrate a test function with the limits. So I try ($Q$ is defined as $Q>0$) $$\lim_ {r\rightarrow \infty} \int_{0}^\infty dQ \cos ((Q-Q')r )\frac {\sin ((Q-Q')r)}{Q-Q'}=\frac \pi 2,$$

I have no real answer for you, just a comment.
You need to show that the limit on the left of your equation has the sifting property just like a delta distribution has.
So the integral you state should give cos(Q') if you put cos(Q) (rather than cos(Q-Q')) into the integrand.
You ought to show this in general though: f(Q) is turned into f(Q') by the sifting integral.

 

[pasmith]

I would expand $$\sin r(Q - Q') = \frac{e^{ir(Q-Q')} -e^{-ir(Q-Q')}}{2i}$$ and express the integral $$\int_{-\infty}^\infty f(Q) \frac{\sin(r(Q-Q'))}{Q-Q'}\,dQ$$ as a sum of fourier transforms.

 

[Haorong Wu]

Thanks, @Philip Koeck and @pasmith. I will try to demonstrate the first expression.

Suppose $F(\omega)$ is the Fourier transform of $f(Q)$, i.e., $f(Q)=(2\pi)^{-1/2} \int d\omega F(\omega) e^{-i\omega Q}$. Then the integral \begin{align}
&~~\lim_{r\rightarrow \infty} \int_0^\infty dQ f(Q) \frac {\sin ((Q-Q')r)}{Q-Q'} \nonumber \\
&=\lim_{r\rightarrow \infty} \int_0^\infty dQ (2\pi)^{-1/2} \int d\omega F(\omega) e^{-i\omega Q} \frac {\sin ((Q-Q')r)}{Q-Q'} \nonumber \\
&=(2\pi)^{-1/2} \int d\omega F(\omega) e^{-i\omega Q' } \lim_{r\rightarrow \infty} \int_0^\infty dQ e^{-i\omega (Q-Q')} \frac {\sin ((Q-Q')r)}{Q-Q'}.\nonumber
\end{align} Letting $x=Q-Q'$, we have $\lim_{r\rightarrow \infty} \int_0^\infty dQ e^{-i\omega (Q-Q')} \frac {\sin ((Q-Q')r)}{Q-Q'}=\lim_{r\rightarrow \infty}\int_{-Q'}^\infty dx e^{-i\omega x}\frac {\sin (xr)}{x}$. Further, setting $y=xr$. it becomes \begin{align}&~~\lim_{r\rightarrow \infty}\int_{-\infty}^\infty dy e^{-i\omega y/r}\frac {\sin (y)}{y} \nonumber \\
&=\lim_{r\rightarrow \infty}\int_0^\infty dy (e^{-i\omega y/r}\frac {\sin (y)}{y}+e^{i\omega y/r}\frac {\sin (y)}{y})\nonumber \\
&=\lim_{r\rightarrow \infty}\int_0^\infty dy 2\cos(\frac {\omega y}{r})\frac {\sin (y)}{y}\nonumber \\ &=\int_0^\infty dy 2\frac {\sin (y)}{y}=\pi. \nonumber\end{align}
Therefore, $\lim_{r\rightarrow \infty} \int_0^\infty dQ f(Q) \frac {\sin ((Q-Q')r)}{Q-Q'}=\pi (2\pi)^{1/2} \int d\omega F(\omega) e^{-i\omega Q' }=\pi f(Q')$. Hence $\lim_{r\rightarrow \infty} \frac {\sin ((Q-Q')r)}{Q-Q'}=\pi \delta(Q-Q')$.

Is the demonstration correct?

 

[Philip Koeck]

Thanks, @Philip Koeck and @pasmith. I will try to demonstrate the first expression.

Suppose $F(\omega)$ is the Fourier transform of $f(Q)$, i.e., $f(Q)=(2\pi)^{-1/2} \int d\omega F(\omega) e^{-i\omega Q}$. Then the integral \begin{align}
&~~\lim_{r\rightarrow \infty} \int_0^\infty dQ f(Q) \frac {\sin ((Q-Q')r)}{Q-Q'} \nonumber \\
&=\lim_{r\rightarrow \infty} \int_0^\infty dQ (2\pi)^{-1/2} \int d\omega F(\omega) e^{-i\omega Q} \frac {\sin ((Q-Q')r)}{Q-Q'} \nonumber \\
&=(2\pi)^{-1/2} \int d\omega F(\omega) e^{-i\omega Q' } \lim_{r\rightarrow \infty} \int_0^\infty dQ e^{-i\omega (Q-Q')} \frac {\sin ((Q-Q')r)}{Q-Q'}.\nonumber
\end{align} Letting $x=Q-Q'$, we have $\lim_{r\rightarrow \infty} \int_0^\infty dQ e^{-i\omega (Q-Q')} \frac {\sin ((Q-Q')r)}{Q-Q'}=\lim_{r\rightarrow \infty}\int_{-Q'}^\infty dx e^{-i\omega x}\frac {\sin (xr)}{x}$. Further, setting $y=xr$. it becomes \begin{align}&~~\lim_{r\rightarrow \infty}\int_{-\infty}^\infty dy e^{-i\omega y/r}\frac {\sin (y)}{y} \nonumber \\
&=\lim_{r\rightarrow \infty}\int_0^\infty dy (e^{-i\omega y/r}\frac {\sin (y)}{y}+e^{i\omega y/r}\frac {\sin (y)}{y})\nonumber \\
&=\lim_{r\rightarrow \infty}\int_0^\infty dy 2\cos(\frac {\omega y}{r})\frac {\sin (y)}{y}\nonumber \\ &=\int_0^\infty dy 2\frac {\sin (y)}{y}=\pi. \nonumber\end{align}
Therefore, $\lim_{r\rightarrow \infty} \int_0^\infty dQ f(Q) \frac {\sin ((Q-Q')r)}{Q-Q'}=\pi (2\pi)^{1/2} \int d\omega F(\omega) e^{-i\omega Q' }=\pi f(Q')$. Hence $\lim_{r\rightarrow \infty} \frac {\sin ((Q-Q')r)}{Q-Q'}=\pi \delta(Q-Q')$.

Is the demonstration correct?

Looks good to me.