Cannot solve step function problem in Boas mathematical physics.

Question

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?

Answer 1

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 $$ \phi(a)-\int_0^{+a}\phi'(x)\,dx+2\phi'(0)=\phi(a)-\phi(a)+\phi(0)+2\phi'(0)=\phi(0)+2\phi'(0)\,. $$

Answer 2

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: $$\begin{align}\int_{-\infty}^{\infty} \phi F''\,dx &=[\phi F']_{-\infty}^\infty-\int_{-\infty}^{\infty}\phi' F'\,dx\\[1em] &=[\phi F']_{-\infty}^\infty-\left\{[\phi'F]_{-\infty}^\infty-\int_{-\infty}^{\infty} \phi'' F\,dx \right\}\\[1em] &=\int_{-\infty}^{\infty} \phi'' F\,dx\\[1em] &=\int_{0}^{\infty} \phi''(x)(x-2)\,dx\\[1em] &=\int_{0}^{\infty}x \phi''(x)\,dx-2\int_{0}^{\infty} \phi''(x)\,dx\\[1em] &=[x\phi']_0^\infty-\int_{0}^{\infty} \phi'(x)\,dx-2[\phi']_{0}^{\infty}\\[1em] &=[x\phi']_0^\infty-[\phi]_{0}^{\infty}-2[\phi']_{0}^{\infty}\\[1em] &=\phi(0)+2\phi'(0) \end{align}$$

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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):

1. $\quad(1_{x>0})'=\delta(x),$
2. $\quad x\,\delta'(x)=-\delta(x),$
3. $\quad\phi(x)\delta'(x)=-\phi'(0).$