Cannot solve step function problem in Boas mathematical physics. I'm trying to solve a problem in Boas(3ed.), Mathematical physics book.
Although I put my 3 days to solve it, I couldn't get a solution written on the page.
The problem is to show that
$\int_{-\infty}^{\infty} \phi(x)F''(x) dx$ = $\phi(0)+2\phi'(0)$
for F(x) which is defined:
$F(x) = \begin{cases}
x-2 & \mbox{, x>0} \\
0 & \mbox{, x<0}
\end{cases} $
and any test function $\phi(x)$,
And my solution is below:
$\int_{-\infty}^{\infty} \phi(x)F''(x) dx$
= $\lim_{t \to 0-}\int_{-\infty}^{t} \phi(x)F''(x) +\lim_{t \to 0+}\int_{t}^{\infty} \phi(x)F''(x)$
=$\lim_{t \to 0+}\int_{t}^{\infty} \phi(x)F''(x)$
=$\lim_{t \to 0+} [\phi(x)F'(x)|_{t}^{\infty} - \int_{t}^{\infty} \phi'(x)F'(x) dx]$
=$\lim_{t \to 0+} [\phi(x)F'(x)|_{t}^{\infty} - \phi'(x)F(x)|_{t}^{\infty} + \int_{t}^{\infty} \phi'(x)F'(x) dx]$
=$\lim_{t \to 0+} [\phi(x)F'(x)|_{t}^{\infty} - \phi'(x)F(x)|_{t}^{\infty} + \int_{t}^{\infty} x\phi''(x) dx - 2\int_{t}^{\infty} \phi''(x) dx]$
=$\lim_{t \to 0+}[-F'(t)\phi(t)+F(t)\phi'(t)+ x\phi'(x)|_{t}^{\infty} - \int_{t}^{\infty} \phi'(x) dx-2\phi(x)|_{t}^{\infty}]$
=$-\phi(0)-2\phi'(0)+\phi(0)+2\phi'(0)=0$.
Is there any mistake or what I missed?
If there is, Can you help me please?
 A: You're going wrong at the very first step, where you write
$$\int_{-\infty}^{\infty} \phi(x)F''(x)\,dx=\lim_{t \to 0-}\int_{-\infty}^{t} \phi(x)F''(x)\,dx +\lim_{t \to 0+}\int_{t}^{\infty} \phi(x)F''(x)\,dx.$$
By that reasoning, we could also incorrectly write
$$1=\int_{-\infty}^{\infty} \delta(x)\,dx=\lim_{t \to 0-}\int_{-\infty}^{t} \delta(x)\,dx +\lim_{t \to 0+}\int_{t}^{\infty} \delta(x)\,dx=0 + 0.$$
Rather, apply integration by parts directly:
$$\begin{align}\int_{-\infty}^{\infty} \phi F''\,dx
&=[\phi F']_{-\infty}^\infty-\int_{-\infty}^{\infty}\phi' F'\,dx\\[1em]
&=[\phi F']_{-\infty}^\infty-\left\{[\phi'F]_{-\infty}^\infty-\int_{-\infty}^{\infty} \phi'' F\,dx \right\}\\[1em]
&=\int_{-\infty}^{\infty} \phi'' F\,dx\\[1em]
&=\int_{0}^{\infty} \phi''(x)(x-2)\,dx\\[1em]
&=\int_{0}^{\infty}x \phi''(x)\,dx-2\int_{0}^{\infty} \phi''(x)\,dx\\[1em]
&=[x\phi']_0^\infty-\int_{0}^{\infty} \phi'(x)\,dx-2[\phi']_{0}^{\infty}\\[1em]
&=[x\phi']_0^\infty-[\phi]_{0}^{\infty}-2[\phi']_{0}^{\infty}\\[1em]
&=\phi(0)+2\phi'(0)
\end{align}$$

The above result is consistent with the fact that $F''=\delta-2\delta',$ which can be derived directly using the following properties of the Dirac delta (as discussed by Boas):

*

*$\quad(1_{x>0})'=\delta(x),$

*$\quad x\,\delta'(x)=-\delta(x),$

*$\quad\phi(x)\delta'(x)=-\phi'(0).$
A: Integration by parts gives for every $a>0$
\begin{align}
\int_{-a}^{+a}\phi(x)F''(x)\,dx=\phi(x)F'(x)\Big|_{x=-a}^{x=+a}-\int_{-a}^{+a}\phi'(x)F'(x)\,dx\,.
\end{align}
From $F'(x)=1_{(0,+\infty)}(x)-2\delta_0(x)$ we get for the RHS
$$
\phi(a)-\int_0^{+a}\phi'(x)\,dx+2\phi'(0)=\phi(a)-\phi(a)+\phi(0)+2\phi'(0)=\phi(0)+2\phi'(0)\,.
$$
