Take the limit of $\frac{\sin(2\pi T \tau)}{\pi \tau}$ with respect to $T$ and $\tau$

Question

I am given a sine cardinal:

$$f(\tau) = \dfrac{\sin(2\pi T \tau)}{\pi \tau}$$

where $T$ is a positive real number, and I have been asked to take the limit of $f(\tau)$ with respect to $T$ and $\tau$

Both of these limits seems to be hard.

• For instance, it is not clear what to do for: $$\lim\limits_{T \to \infty}f(\tau) =\lim\limits_{T \to \infty} \dfrac{\sin(2\pi T \tau)}{\pi \tau}$$

• Similarly,

$$\lim\limits_{\tau \to 0}f(\tau) =\lim\limits_{\tau \to 0} \dfrac{\sin(2\pi T \tau)}{\pi \tau}$$

while I am aware of the result: $\lim\limits_{t \to 0} \dfrac{\sin(t)}{t} = 1$ and you cannot use l'Hopital, it seems that $\lim\limits_{\tau \to 0} \dfrac{\sin(2\pi T \tau)}{\pi \tau}$ is quite different compared to the former e.g., more stuff in the argument of $\sin$.

Any idea how I can proceed with either limits?

Answer 1

Let me rewrite them:

$$\lim_{T \to \infty} \frac{\sin(2\pi T\tau)}{\pi\tau}=\frac{1}{\pi \tau}\cdot\color{red}{\left[\lim_{T \to \infty} \sin(2\pi T \tau)\right]}$$

$$\lim_{\tau \to 0} \frac{\sin(2\pi T\tau)}{\pi\tau} =2T\cdot \color{blue}{\left[\lim_{\tau \to 0} \frac{\sin(2\pi T\tau)}{2\pi T\tau }\right]}$$

Can you evaluate the colored limit?

Answer 2

A remark regarding your comments in Siong Thye Goh post:

Note that if $\ \displaystyle I=\int_{-\infty}^{+\infty} f(t)\mathop{dt}$ exists (i.e. if it is finite) then $\displaystyle\lim\limits_{T\to+\infty}\int_{-T}^Tf(t)\mathop{dt}=I$.

But the reciprocal is not true.

For the improper integral to exists you need to proove that $\displaystyle \lim\limits_{a\to\,-\infty\\b\to\,+\infty}\int_a^bf(t)dt$ exists and has the same value whichever path we choose for the couple $(a,b)$ to go to infinity.

In particular here the value $\delta(\tau)$ is not finite for $\tau=0$.

So in the strict sense the improper integral does not exists, and neither does $\lim\limits_{T\to+\infty}\dfrac{\sin(2\pi\tau T)}{\pi\tau}$

You can notice that for $\tau=0$ then $\displaystyle \int_a^b e^0\mathop{df}=b-a$ has no limit for $(a,b)\to(-\infty,+\infty)$.

You need to be extra careful when mixing results in the sense of distributions and classical results with normal functions.