Well, the Dirac delta function $\delta(x)$ is a distribution, also known as a generalized function.
One can e.g. represent $\delta(x)$ as a limit of a rectangular peak with unit area, width $\epsilon$, and height $1/\epsilon$; i.e.
$$\tag{1} \delta(x) ~=~ \lim_{\epsilon\to 0^+}\delta_{\epsilon}(x), $$
$$\tag{2} \delta_{\epsilon}(x)~:=~\frac{1}{\epsilon} \theta(\frac{\epsilon}{2}-|x|)
~=~\left\{ \begin{array}{ccc} \frac{1}{\epsilon}&\text{for}& |x|<\frac{\epsilon}{2}, \\
\frac{1}{2\epsilon}&\text{for}& |x|=\frac{\epsilon}{2}, \\
0&\text{for} & |x|>\frac{\epsilon}{2}, \end{array} \right. $$
where $\theta$ denotes the Heaviside step function with $\theta(0)=\frac{1}{2}$.
The product $\delta(x)^2$ of the two Dirac delta distributions does strictly speaking not$^1$ make mathematical sense, but for physical purposes, let us try to evaluate the integral of the square of the regularized delta function
$$\tag{3} \int_{\mathbb{R}}\! dx ~\delta_{\epsilon}(x)^2
~=~\epsilon\cdot\frac{1}{\epsilon}\cdot\frac{1}{\epsilon}
~=~\frac{1}{\epsilon} ~\to~ \infty
\quad \text{for} \quad \epsilon~\to~ 0^+. $$
The limit is infinite as Griffiths claims.
It should be stressed that in the conventional mathematical theory of distributions, the eq. (2.95) is a priori only defined if $f$ is a smooth test-function. In particular, it is not mathematically rigorous to use eq. (2.95) (with $f$ substituted with a distribution) to justify the meaning of the integral of the square of the Dirac delta distribution. Needless to say that if one blindly inserts distributions in formulas for smooth functions, it is easy to arrive at all kinds of contradictions! For instance,
$$ \frac{1}{3}~=~ \left[\frac{\theta(x)^3}{3}\right]^{x=\infty}_{x=-\infty}~=~\int_{\mathbb{R}} \!dx \frac{d}{dx} \frac{\theta(x)^3}{3} $$ $$\tag{4} ~=~\int_{\mathbb{R}} \!dx ~ \theta(x)^2\delta(x)
~\stackrel{(2.95)}=~ \theta(0)^2~=~\frac{1}{4}.\qquad \text{(Wrong!)} $$
--
$^1$ We ignore Colombeau theory. See also this mathoverflow post.
Using the following expression for the Dirac delta function: $$\delta(k-k')=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i(k-k')x}\mathrm{d}x$$
show that if $\Psi(x,t)$ is normalized at time $t=0$, then the corresponding momentum space wave function $\Phi(p_x,t)$ is also normalized at time $t=0$.
Good.
By definition of the momentum space wave function, $$\Phi(p_x,0)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}e^{\frac{i}{\hbar}p_xx}\Psi(x,0)\mathrm{d}x$$ and $$\Phi^*(p_x,0)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}e^{\frac{i}{\hbar}p_xx}\Psi^*(x,0)\mathrm{d}x$$
No. $$\Phi^*(p_x,0)\neq\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}e^{\frac{i}{\hbar}p_xx}\Psi^*(x,0)\mathrm{d}x$$
Just write out the real and imaginary parts of the previous integral and flip the sign of the imaginary one. And the $x,$ $dx$ and $y,$ $dy$ are dummy variables, they are just the name of something that changes and so can be called anything. Call the second one $dy$ instead of $dx$ to make your life easier later. But keep the $p_x$ as $p_x$ since that isn't a dummy. It isn't changing, it's the fixed value on the left hand side over in $\Phi^*(p_x,0).$
To check if $\Phi(p_x,0)$ is normalized, we'd like to check if $\Phi(p_x,0)\Phi^*(x,0)$ integrates to $1$ over all values of $p_x$,
Another mistake, this one might just be a typo.
$$=\frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{2i\frac{p_x}{\hbar}yx}\Psi^*(y,0)\Psi(x,0)\mathrm{d}y\mathrm{d}x$$
Also wrong, and not just because of previous mistakes carrying forward but because of a new mistake. $e^Ae^B=e^{A+B}$ and you can check that $A,$ $B,$ and $A+B$ all have the same units, dimensionless units.
I'm not exactly sure how to make the leap and start making use of some properties of the Dirac delta.
After you get it to look exactly like the delta you can use some change of variables to use the one property of the delta.
Best Answer
This is essentially a notational issue. In Dirac notation, a pure state is represented by a vector (ket) in Hilbert space $|\Psi\rangle$ which contains all the information that one can have about that state. The eigenvectors of momentum are then denoted as $|p\rangle$. This notation is representationally independent - i.e. it does not assume you are in the position representation or the momentum representation. The ket $|p\rangle$ represents the state and says nothing about in which representation it's in. In order to get $|p\rangle$ in the position representation we take the inner product of it with position eigenvector $|x\rangle$. So the position representation of $|p\rangle$ is denoted in Dirac notation by $\langle x |p\rangle$. This position representation of the momentum eigenvector is exactly that Griffiths calculated on page 103.