Ideal sampling using sinc funcion Let $ x(t) $ be a bandwidth limited signal such as $ \forall |\omega|&gt;\frac{\pi}{T} : X^F(\omega)=0 $ while $ X^F(\omega)$ is the signal's Fourier Transform.

Let us denote $y[n]=\int_{-\infty}^{\infty}x(\tau)sinc(\frac{\tau-nT}{T}) d\tau$

I noticed that my professor said that because the signal is bandwidth limited and the width of the $sinc$ function is $T$ and it's orthogonal property, and with regard to ideal sampling and reconstruction we can say that $y[n]=T\cdot x[nT]$.

I understand that we can treat $sinc$ as a delta function somehow but I'm can't find the right mathematical explenation which had lead to this result.

I'd happy for some insights.
Your professor is using sloppy notation. When one writes the sample value $x[\cdot]$, it is generally accepted that the argument is an integer (which $nT$ might not be) and we relate this to the continuous-time signal $x(t)$ via $x[n] = x(nT)$. So, what your professor should have said is that $y[n]$ has value $T\cdot x[n] = T\cdot x(nT)$, and not that $y[n] = T\cdot x[nT]$ (even if $T$ happens to be an integer).  An outline of a proof that $y[n] = T\cdot x[n] = T\cdot x(nT)$ is given below.

If $x(t)$ is a continuous-time signal bandlimited to $\left(-\frac{1}{2T}, \frac{1}{2T}\right)$, that is, its Fourier transform $X(f)$ has the property that $X(f) = 0$ for all $|f| \geq \frac{1}{2T}$,then it can be reconstructed from its samples spaced $T$ seconds apart. So, let's consider the signal $x(t)\operatorname{sinc}\left(\frac{t}{T}\right)$ where
$$\operatorname{sinc}(t) = \begin{cases}\frac{\sin(\pi t)}{\pi t},&amp; t \neq 0,\\1, &amp; t = 0.\end{cases}$$
The Fourier transform of $\operatorname{sinc}(t)$ is $\operatorname{rect}(f)$ and so the Fourier transform of $\operatorname{sinc}\left(\frac{t}{T}\right)$ is $T\operatorname{rect}(fT)$, and the Fourier transform of $x(t)\operatorname{sinc}\left(\frac{t}{T}\right)$ is the convolution  $X(f)\circledast T\operatorname{rect}(fT)$ of the Fourier transforms $X(f)$ and $T\operatorname{rect}(fT)$ of $x(t)$ and $\operatorname{sinc}\left(\frac{t}{T}\right)$ respectively.  Now, in principle, the convolution of a function and a rectangular pulse is easy to compute but I refuse to write out the answer because it is not needed: what we need to find is the value of $\int_{-\infty}^\infty x(t)\operatorname{sinc}\left(\frac{t}{T}\right) \mathrm dt$ which, if you think about it a bit, is actually the value of the Fourier transform of $x(t)\operatorname{sinc}\left(\frac{t}{T}\right)$ evaluated at $f=0$. That is,
\begin{align}
\int_{-\infty}^\infty x(t)\operatorname{sinc}\left(\frac{t}{T}\right) \mathrm dt &amp;= X(f)\circledast T\operatorname{rect}(fT)\bigg |_{f=0}\\
&amp;= T\int_{-\infty}^\infty X(\lambda)\operatorname{rect}((0-\lambda) T) \, \mathrm d\lambda\\
&amp;= T\int_{-\infty}^\infty X(\lambda)\operatorname{rect}(\lambda T) \, \mathrm d\lambda &amp; \scriptstyle\text{rect is an even function}\\
&amp;= T\int_{-\frac{1}{2T}}^{\frac{1}{2T}}X(\lambda) \, \mathrm d\lambda &amp;\scriptstyle\text{rect has finite support}\\
&amp;= T\int_{-\infty}^\infty X(\lambda) \, \mathrm d\lambda &amp;\scriptstyle{X(\lambda)=0 ~\text{for}~|\lambda| &gt; \frac{1}{2T}}\\
&amp;= T
\cdot x(0).
\end{align}
Thus, $y[0] \overset{\text{def}}{=} \int_{-\infty}^\infty x(t)\operatorname{sinc}\left(\frac{t}{T}\right) \mathrm dt$ equals $T\cdot x(0)$. I leave it as an exercise for the reader to work out the Fourier transform of $\operatorname{sinc}\left(\frac{t-nT}{T}\right)$ and show that $y[n] \overset{\text{def}}{=} \int_{-\infty}^\infty x(t)\operatorname{sinc}\left(\frac{t-nT}{T}\right) \mathrm dt$ works out to be $T\cdot x(nT)$ and so if we define the samples of $x(t)$ as $x[n] = x(nT)$, then we have that $y[n] = T\cdot x[n]$ for all $n$.