A seeming contrdiction in deriving wave function for delta function potential

[kof9595995]

First of all, let me copy the standard solution from Griffiths, section 2.5, just for the sake of clarity.
Potential$$V(x) = - \alpha \delta (x)$$

The bound state eigenfunction:
$$\psi (x) = \left\{ \begin{array}{l}<br /> B{e^{\kappa x}}{\rm{ (}}x \le 0{\rm{)}} \\ <br /> B{e^{ - \kappa x}}{\rm{ (}}x \ge 0{\rm{)}} \\ <br /> \end{array} \right.{\rm{ where }}\kappa = \frac{{\sqrt { - 2mE} }}{\hbar }$$
Integrate time independent Schrödinger's equation from -infinitesimal to +infinitesimal to get an imposed condition
$$- \frac{{{\hbar ^2}}}{{2m}}\int_{ - \varepsilon }^{ + \varepsilon } {\frac{{{d^2}\psi }}{{d{x^2}}}} dx + \int_{ - \varepsilon }^{ + \varepsilon } {V(x)} \psi (x)dx = E\int_{ - \varepsilon }^{ + \varepsilon } {\psi (x)dx}$$
For right hand side, since the we're integrating a finite function in a infinitesimal region
$$\int_{ - \varepsilon }^{ + \varepsilon } {\psi (x)dx} = 0$$
So we have
$$- \frac{{{\hbar ^2}}}{{2m}}\mathop {\lim }\limits_{\varepsilon \to 0} ({\left. {\frac{{d\psi }}{{dx}}} \right|_{ + \varepsilon }} - {\left. {\frac{{d\psi }}{{dx}}} \right|_{ - \varepsilon }}) = \alpha \psi (0) \Rightarrow - 2B\kappa = - \frac{{2m\alpha }}{{{\hbar ^2}}}B$$
so $$\kappa = \frac{{m\alpha }}{{{\hbar ^2}}}$$
and consequently$$E = - \frac{{m{\alpha ^2}}}{{2{\hbar ^2}}}$$ and $$B = \frac{{\sqrt {m\alpha } }}{\hbar }(normalization)$$

And here comes my problem: instead of integrating a infinitesimal region, I tried to integrate the whole x-axis just to check the consistency
$$- \frac{{{\hbar ^2}}}{{2m}}\int_{ - \infty }^{ + \infty } {\frac{{{d^2}\psi }}{{d{x^2}}}} dx + \int_{ - \infty }^{ + \infty } {V(x)} \psi (x)dx = E\int_{ - \infty }^{ + \infty } {\psi (x)dx}$$
in which:
$$\int_{ - \infty }^{ + \infty } {\psi (x)dx} = \frac{2}{\kappa }$$
$$- \frac{{{\hbar ^2}}}{{2m}}\int_{ - \infty }^{ + \infty } {\frac{{{d^2}\psi }}{{d{x^2}}}} dx = - \frac{{{\hbar ^2}}}{{2m}}({\left. {\frac{{d\psi }}{{dx}}} \right|_{ + \infty }} - {\left. {\frac{{d\psi }}{{dx}}} \right|_{ - \infty }}) = 0$$
Finally,
$$- \alpha \psi (0) = \frac{{2E}}{\kappa } \Leftrightarrow - \alpha B = \frac{{2E}}{\kappa }$$
Now if we sub in what we've got in the first part, e.g.$$\kappa = \frac{{m\alpha }}{{{\hbar ^2}}}$$, $$E = - \frac{{m{\alpha ^2}}}{{2{\hbar ^2}}}$$, $$B = \frac{{\sqrt {m\alpha } }}{\hbar }$$
we find
$$\frac{{\sqrt {m\alpha } }}{\hbar } = 1$$

This is obviously incorrect, since m and alpha are arbitrary. So where did I get wrong?

 

[xepma]

You forgot the normalization of the wavefunction. The integral you quote should be:
$$\int_{ - \infty }^{ + \infty } {\psi (x)dx} = \frac{2B}{\kappa }$$
If you do the same trick, you'll just get 1=1 instead of what you have.

 

[kof9595995]

Ah,you're right. Thanks