# Computing the integral $\int\limits_{-\infty}^{\infty}(t^2-1)\delta(t)\:dt$

Have I solved this problem correctly?

\begin{align} \int\limits_{-\infty}^{\infty}\left(t^2-1\right)\delta\left(t\right)\:dt&=\int\limits_{-\infty}^{\infty}t^2\delta\left(t\right)\:dt-\int\limits_{-\infty}^{\infty}\delta\left(t\right)\:dt\tag{1} \\ &=\int\limits_{-\infty}^{\infty}t^2\delta\left(t\right)\:dt-1\tag{2} \\ &=0-1=\boxed{-1.}\tag{3} \end{align}

Such that $\delta\left(t\right)$ is the dirac-delta function.

Thanks!

• I don't really see the point in splitting up the integral, we can just evaluate $t^2-1$ at $t=0$ directly no? – anon Dec 23 '14 at 22:34
• +1 You did fine but you don't need to split the integral as @anon already pointed out. – Felix Marin Dec 23 '14 at 23:06

$$\int_{-\infty}^\infty f(t)\delta(t)\, dt = f(0)$$ for $f(t)=t^2-1$ you have $$\int_{-\infty}^\infty (t^2-1)\delta(t)\, dt = -1.$$