quantum mechanics - Don't understand the integral over the square of the Dirac delta function

In Griffiths' Intro to QM [1] he gives the eigenfunctions of the Hermitian operator $\hat{x}=x$ as being

$$g_{\lambda}\left(x\right)~=~B_{\lambda}\delta\left(x-\lambda\right)$$

(cf. last formula on p. 101). He then says that these eigenfunctions are not square integrable because

$$\int_{-\infty}^{\infty}g_{\lambda}\left(x\right)^{*}g_{\lambda}\left(x\right)dx ~=~\left|B_{\lambda}\right|^{2}\int_{-\infty}^{\infty}\delta\left(x-\lambda\right)\delta\left(x-\lambda\right)dx ~=~\left|B_{\lambda}\right|^{2}\delta\left(\lambda-\lambda\right) ~\rightarrow~\infty$$

(cf. second formula on p. 102). My question is, how does he arrive at the final term, more specifically, where does the $\delta\left(\lambda-\lambda\right)$ bit come from?

My total knowledge of the Dirac delta function was gleaned earlier on in Griffiths and extends to just about understanding

$$\tag{2.95}\int_{-\infty}^{\infty}f\left(x\right)\delta\left(x-a\right)dx~=~f\left(a\right)$$

(cf. second formula on p. 53).

References:

1. D.J. Griffiths, Introduction to Quantum Mechanics, (1995) p. 101-102.

Answer

You need nothing more than your understanding of

$$\int_{-\infty}^\infty f(x)\delta(x-a)dx=f(a)$$ Just treat one of the delta functions as $f(x)\equiv\delta(x-\lambda)$ in your problem. So it would be something like this: $$\int\delta(x-\lambda)\delta(x-\lambda)dx=\int f(x)\delta(x-\lambda)dx=f(\lambda)=\delta(\lambda-\lambda)$$ So there you go.