Integral of a delta function

Question

How can I calculate the integral of the form, \begin{equation} I = \int_{-\infty}^\infty dx \,\delta\left(b-\sqrt{x^2+a}\right) , \end{equation} assuming positive values for $a$?

Answer 1

Just use the definition of the Delta of a function:

$$\delta[f(x)] = \sum_R \frac{\delta(x - x_R)}{|f'(x_R)|}$$

Where the sum runs all over the $R$ that are, the roots of $f(x)$.

In your case it's simple to see that the roots are

$$x_R = \pm \sqrt{b^2-a}$$

Whence

$$\delta\left(b - \sqrt{x^2+a}\right) = \frac{\delta(x + \sqrt{b^2-a}) + \delta(x - \sqrt{b^2-a})}{\frac{\sqrt{b^2-a}}{\sqrt{b^2}}}$$

If you assume $b >0$ you can simplify the above expression into

$$b \frac{\delta(x - \sqrt{b^2-a}) + \delta(x + \sqrt{b^2-a})}{\sqrt{b^2-a}}$$

The integral now is trivial since you just need to apply the well know rules of Dirac Delta integration.