Question 3.2: Find the eigenfunctions and eigenvalues of the momentum oper...

Let f_p(x) be the eigenfunction and p the eigenvalue:

-i\hbar \frac{d}{dx} f_{p}(x)=pf_{p}(x)       (3.30)

The general solution is

f_{p}(x )=Ae^{ipx/\hbar }

This is not square-integrable for any (complex) value of p—the momentum operator has no eigenfunctions in Hilbert space.
And yet, if we restrict ourselves to real eigenvalues, we do recover a kind of ersatz “orthonormality.” Referring to Problems 2.23(a) and 2.26,

\int_{-\infty }^{\infty }{f^{*}_{\acute{p}}(x)}f_{p} (x)dx=\left|A\right|^{2} \int_{-\infty }^{\infty } e^{i(p-\acute{p})x/\hbar }dx=\left|A \right|^{2} 2\pi \hslash \delta (p-\acute{p})                  (3.31)

If we pick A=1/\sqrt{2\pi \hslash }  so that

f_{p}=\frac{1 }{\sqrt{2\pi \hslash}} e^{ipx/\hbar }     (3.32)

then

\left\langle f_{\acute{p} }|f_{p} \right\rangle =\delta (p-\acute{p} )        (3.33)

which is reminiscent of true orthonormality (Equation 3.10)—the indices are now continuous variables, and the Kronecker delta has become a Dirac delta, but otherwise it looks just the same. I’ll call Equation 3.33 Dirac orthonormality.
Most important, the eigenfunctions (with real eigenvalues) are complete, with the sum (in Equation 3.11) replaced by an integral: Any (square-integrable) function f(x) can be written in the form

<f_m|f_n>=\delta_{mn}          (3.10)

f(x)=\sum\limits_{n=1}^{\infty}{c_n f_n(x)}.         (3.11)

f(x)=\int_{-\infty }^{\infty }{c(p)f_{p}(x)dp}= \frac{1}{\sqrt{2\pi \hbar} } \int_{-\infty }^{\infty }{c(p) e^{ipx/ \hbar }}dp     (3.34)

The “coefficients” (now a function,c(p)) are obtained, as always, by Fourier’s trick:

\left\langle f_{\acute{p} }|f_{p}\right\rangle \int_{-\infty }^{\infty }{c(p)\left\langle f_{\acute{p} }|f_{p} \right\rangle }dp= \int_{-\infty }^{\infty }{c(p)}\delta (p-\acute{p} )dp=c(\acute{p})       (3.35)

Alternatively, you can get them from Plancherel’s theorem (Equation 2.103); indeed, the expansion (Equation 3.34) is nothing but a Fourier transform.

f(x)=\frac{1}{\sqrt{2\pi } } \int_{-\infty }^{+\infty }{F(k)e^{ikx}dk}\Longleftrightarrow F(k)= \frac{1}{\sqrt{2\pi } }\int_{-\infty }^{+\infty }{f(x)e^{ikx}dx}                        (2.103)