Not sure how to finish this integral $\int_{-\infty}^\infty x^3 \delta(x^2-2)dx$

Question

Dirac delta is a symmetric function defined as $$\int_{-\infty}^\infty f(t)\delta(t-A)dt = f(A)$$ Find the value of $$\int_{-\infty}^\infty x^3 \delta(x^2-2)dx$$

SOLUTION:

Let $t=x^2$ then $dt=2xdx \rightarrow dx = dt/(2x)$ and the integral is $$\int_{-\infty}^\infty tx \delta(t-2)dt/(2x) = \int_{-\infty}^\infty t/2 \delta(t-2)dt = f(A).$$ I am not quite sure what to do now in terms of finishing the problem...

Answer 1

You can use the property

$$ \int_{-\infty}^{+\infty}f(x)\delta(g(x))~{\rm d}x = \sum_i \frac{f(x_i)}{|g'(x_i)|} $$

where $x_i$ are the roots of $g$: $g(x_i) = 0$. In your case you have $g(x) = (x - 2^{1/2})(x + 2^{1/2})$, $x_1 = 2^{1/2}$ and $x_2 = -2^{1/2}$, $g'(x) = 2x$ and $f(x) = x^3$.

Putting everything together

\begin{eqnarray} \int_{-\infty}^{+\infty}x^3\delta(x^2 - 2)~{\rm d}x = \frac{(2^{1/2})^{3}}{|2(2^{1/2})|} + \frac{(-2^{1/2})^{3}}{|2(-2^{1/2})|} = \cdots \end{eqnarray}

can you take it from here?

Answer 2

OP's integral $$\int_{\mathbb{R}}\! \mathrm{d}x~ x^3 \delta(x^2-2)~=~0$$ is zero because it's an odd integrand on a symmetric interval.