From my limited knowledge of this subject, I would say that a non-normalizable wave-function wouldn't really make any physical sense.
Remember, the wave-function is a function whose value squared evaluated between two points represents the probability that the particle will be found between those two points. So, the restriction that wave functions be normalized is basically just a nod to reality - the particle must be found SOMEWHERE.
Normally, the restriction is: $$ \int_{-\infty}^ {\infty} |\psi(x)|^{2} dx = 1, $$ i.e. the probability of finding the particle if you looked between $-\infty$ and $\infty$ is 1. Having a probability greater than one of finding the particle between these bounds would not make any physical sense.
Having a wave-function described by the equations you posted above would imply that there is an infinite chance of finding the particle anywhere.
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
The momentum operator $P$ doesn't have a normalizable eigenfunction. One writes the eigenfunction as $$\chi(x)=\frac{1}{\sqrt{2\pi}}e^{ikx}$$ The factor comes in here. If we consider the finite boundary, say $L$. Then
$$\chi_n(x)=\frac{1}{\sqrt{L}}e^{ik_nx}$$ The completeness required: $$\sum_{n=-\infty}^\infty \frac{1}{L}e^{ik_nx}e^{-ik_{n'}x}=\delta (x-x').\ \ \ \ x,x'\in [-L/2,L/2]$$
Now let $L\rightarrow \infty $ then $$\sum_{n=-\infty}^\infty\left\{\cdots \right\}\rightarrow \int dn \{\cdots \}=\int dk\left( \frac{dn}{dk}\right)\{\cdots\}$$ where $$\frac{dn}{dk}=\frac{L}{2\pi}$$ therefore, $$\int \frac{1}{2\pi} e^{ik_nx}e^{-ik_{n'}x} \ dk=\delta (x-x')$$
where we reorganize $$\chi_n(x)=\frac{1}{\sqrt{2\pi}} e^{ik_nx}$$ as desired.