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If $f(0)=g(0)=0$ and $f''$ and $g''$ are continuous, show that

$$\int _{ 0 }^{ a }{ f(x)g''(x)dx } =f(a)g'(a)-f'(a)g(a)+ \int _{ 0 }^{ a }{ f''(x)g(x)dx } $$

What I did:

1) We know that $$f(x)g(x)=\int { f'(x)g(x)dx } +\int { f(x)g'(x)dx } $$

which further evaluates to: $$\int { f(x)g'(x)dx } =f(x)g(x)-\int { f'(x)g(x)dx } $$

2) We now use integration by parts.

We let $u=f(x)$, $du=f'(x)$, $v=g'(x)$, $dv=g''(x)$

which gives us the following: $$\int _{ 0 }^{ a }{ f(x)g''dx } =f(x)g'(x)-\int _{ 0 }^{ a }{ f'(x)g'(x)dx } $$

3) Now we utilize integration by parts once more to evaluate the integrand on the right hand side, $\int _{ 0 }^{ a }{ f'(x)g'(x)dx } $ as follows:

Let $s=f'(x)$, $ds=f''(x)$, $t=g(x)$, $dt=g'(x)$

so we get that:

$$\int { f'(x)g'(x)dx } =f'(x)g(x)-\int { f''(x)g(x)dx } $$

4) By substituting in $f'(x)g(x)-\int { f''(x)g(x)dx } $ in place of $\int { f'(x)g'(x)dx } $, we get the following:

$$\int _{ 0 }^{ a }{ f(x)g''(x)dx } =f(x)g'(x)-f'(x)g(x)+\int { f''(x)g(x)dx } $$

So, at this point I believe that I have the right idea, but I don't know how I am supposed to end up with those $a$'s in the $f(a)g′(a)−f′(a)g(a)$ part. I would appreciate any help/guidance in this regard.

Cherry_Developer
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    You seem to have forgotten to apply the limits to the middle term which is why your final result don't make sense (the left hand side is a number but the right hand side is a function of $x$). What we do know is that $$\int_0^af''(x)g(x){\rm d}x = [f'(x)g(x)]_0^a - \int_0^a f'(x)g'(x){\rm d}x$$ and $$\int_0^a f'(x)g'(x){\rm d}x = [f(x)g'(x)]_0^a - \int_0^a f(x)g''(x){\rm d}x$$ where $[h(x)]_0^a = h(a) - h(0)$. Notice the way the limits is also taken on the $uv$-term in the integration by parts. – Winther Sep 14 '16 at 02:34
  • @Winther I am having trouble with understanding why it works out the way you mentioned. Why do we just evaluate the middle part and ignore the $- \int_0^a f(x)g''(x){\rm d}x$ part when evaluating from $0$ to $a$. I apologize if this comes across as a stupid question, but my main reason for posting this in the first place was to actually understand what I'm doing, rather than to just copy to correct answer and call it a day. – Cherry_Developer Sep 14 '16 at 03:07
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    It's perhaps best to go back to scratch and see where the integration by parts formula comes from to understand why it has to be so. We have

    $$\frac{d}{dx}[u(x) v(x)] = u'(x)v(x) + u(x)v'(x)$$

    Integrating this over $[0,a]$ we get

    $$\int_0^a\frac{d}{dx}[u(x) v(x)],{\rm d}x = \int_0^a(u'(x)v(x) + u(x)v'(x)),{\rm d}x$$

    The left hand side becomes $[u(x)v(x)]_0^a$ while the right hand side can be split up as (the integral is linear) $\int_0^a u'(x)v(x),{\rm d}x + \int_0^au(x)v'(x)),{\rm d}x$. Rearranging this formula you get the integration by parts formula.

     – Winther Sep 14 '16 at 03:14
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    One can also get to this formula from the indefinate integral formulation of integration by parts. The way we impose limits on $\int uv'{\rm d}x = uv - \int u'v {\rm d}x$ is to do it on all three terms. So when you take $\int \to \int_a^b$ you must also take $uv \to [uv]_a^b$, i.e. $u(x)v(x)$ becomes $u(b)v(b) - u(a)v(a)$. – Winther Sep 14 '16 at 03:19
  • @Winther I am aware of the origin of the integration by parts formula, that is where I got my step 1) from. I believe that my problem is that I am conflating what is supposed to be done when integrating by parts an indefinite integral with what is supposed to be done when integrating by parts a definite integral. From what I now understand, $a$ is a constant, and by treating is as a number, your solution makes more sense to me now. I hope that it at least sounds like I am actually getting this now. – Cherry_Developer Sep 14 '16 at 03:20
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    Good. It's not uncommon to have problems with this. The indefinate integral formulation of integration by parts is the root of alot of confusion, see e.g. this question for a funny little "paradox". – Winther Sep 14 '16 at 03:34

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