Integral through Fourier Transform and Parseval's Identity $$
\int_{-\infty}^{\infty}{\rm sinc}^{4}\left(\pi t\right)\,{\rm d}t\,.
$$
Can you help me evaluate this integral with the help of Fourier Transform and Parseval Identity. I could not see how it is implemented. Thank you..

There are two correct answers to this question, depending on how you understand sinc. My guess is that your convention is $\operatorname{sinc}x = \frac{\sin x}{x}$. 
I don't know your FT convention, but I will use $\hat f(\xi)=\int f(x)e^{-2\pi i \xi x}\,dx$. Then
$$\hat \chi_{[-a,a]}(\xi)= \int_{-a}^a e^{-2\pi i \xi x}\,dx = 
\frac{e^{2\pi i a\xi}-e^{-2 \pi i a\xi}}{2\pi i \xi} =  2a\operatorname{sinc}(2 \pi a \xi)$$
To square the righthand side,   convolve $\chi_{[-a,a]}$ with itself. This convolution is $f(x) = (2a-|x|)^+$. Thus, $\hat f(\xi) = 4a^2 \operatorname{sinc}^2(2 \pi a \xi)$. Since $$
\int_{\mathbb R} f^2
= 2\int_0^{2a} (2a-x)^2\,dx = \frac{16 a^3}{3}
$$
Parseval's identity implies 
$$
\int_{\mathbb R} \operatorname{sinc}^4(2 \pi a \xi)\,d\xi
 = \frac{1}{16a^4}\cdot \frac{16a^3}{3} = \frac{1}{3a}
$$
