# Calculate limits as distributions

• I
• Haorong Wu
Well done!In summary, the conversation discusses how to calculate limits when they are viewed as distributions. The conversation covers using a test function to calculate the limits and integrating with the sifting property. The process involves using the Fourier transform to show that the limit on the left of the equation has the sifting property, leading to the conclusion that the limit is equal to the product of the test function and the delta distribution. The summary concludes that the demonstration provided in the conversation is correct.
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.

Haorong Wu said:
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.

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.

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?

Philip Koeck
Haorong Wu said:
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.

Haorong Wu

• Calculus
Replies
12
Views
1K
• Calculus
Replies
2
Views
1K
• Calculus
Replies
5
Views
1K
• Calculus
Replies
2
Views
1K
• Calculus
Replies
1
Views
1K
• Calculus
Replies
1
Views
1K
• Calculus
Replies
1
Views
1K
• Calculus
Replies
3
Views
1K
• Calculus
Replies
4
Views
666