Energy of $\text{sinc}^2(kt)$ function multiplied by variable
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

$$x(t)=\sin(kt)\frac{\sin(kt)}{kt}\tag{1}$$

The energy of $x(t)$ is

$$E_x=\int_{-\infty}^{\infty}x^2(t)dt=\int_{-\infty}^{\infty}\left(\sin(kt)\frac{\sin(kt)}{kt}\right)^2dt$$

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

$$E_x=\frac{1}{2\pi}\int_{-\infty}^{\infty}|X(j\omega)|^2d\omega\tag{2}$$

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 $(1)$

$$\begin{align}X(j\omega)&amp;=\frac{1}{2\pi}\frac{\pi}{j}\left[\delta(\omega-k)-\delta(\omega+k)\right]\star\frac{\pi}{k}\text{rect}(\omega,k)\\&amp;=\frac{\pi}{2kj}\left[\text{rect}(\omega-k,k)-\text{rect}(\omega+k,k)\right]\tag{3}\end{align}$$

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

$$|X(j\omega)|^2=\frac{\pi^2}{4k^2}\text{rect}(\omega,2k)\tag{4}$$

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

$$E_x=\frac{1}{2\pi}\frac{\pi^2}{4k^2}\int_{-2k}^{2k}d\omega=\frac{\pi}{2k}\tag{5}$$

where I have implicitly assumed that $k&gt;0$ 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: