I know how to determine the energy of $\text{sinc}^2(kt)$, but how does this change when I need to find the energy of $(kt)\cdot \text{sinc}^2(kt)$.

If the $(kt)$ in front is squared (in energy integral) does that make the energy of this infinity ? the energy integral in time domain would be $$E_{\infty}=\int_{-\infty}^{\infty}(k^2t^2) \cdot \text{sinc}^4(kt)dt$$

But then Wolfram Alpha returned that the integral does not converge?

If multiplication in the time domain is convolution in freq domain

So, how do I handle the $(k^2 t^2)$ portion out front?

  • $\begingroup$ If the integral diverges to infinity, then there's your answer; the signal in question has infinite energy. $\endgroup$
    – Jason R
    Commented Jul 1, 2016 at 12:56
  • $\begingroup$ This integral is not divergent; the integrand decays as $1/t^2$. $\endgroup$
    – Matt L.
    Commented Jul 1, 2016 at 21:20

1 Answer 1


The energy of the signal $x(t)=kt\,\text{sinc}^2(kt)$ is finite. Note that with $\text{sinc}(kt)=\sin(kt)/(kt)$, $x(t)$ can be written as


The energy of $x(t)$ is


Using Parseval's theorem, $E_x$ can also be written in terms of $X(j\omega)$, the Fourier transform of $x(t)$:


With the Fourier transform relation

$$f(t)g(t)\Longleftrightarrow\frac{1}{2\pi}F(j\omega)\star G(j\omega)$$

where $\star$ denotes convolution, we get from $(1)$


where $\text{rect}(\omega,k)$ is a rectangular function with support $[-k,k]$. From $(3)$ we get


Plugging $(4)$ into $(2)$ finally yields


where I have implicitly assumed that $k>0$ holds. Since $x^2(t)$ is independent of the sign of $k$, $E_x$ must also be independent of the sign of $k$. So if negative values of $k$ are allowed, the result is

$$E_x=\frac{\pi}{2|k|},\quad k\neq 0\tag{6}$$

Even WolframAlpha agrees with me:

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