Integrating product of 2nd-order partial and same-variable function.

Question

I've been going through my Calculus textbook but can't seem to understand what I'm actually trying to do with this integral I have.

It is the integral of the form:

$$ \int^b_a \left[ \frac {\partial^2f(x)}{\partial x^2}g(x) - \frac {\partial^2g(x)}{\partial x^2}f(x) \right] \, dx $$

So I'm guessing it would be something that could be done with integration by parts, yet it seems incredibly weird since I'm used to integrating actual functions like $ f(x) = x^2 $ and I'm insecure in my answer, and not sure how to integrate a product of a partial.

My guess was that the result would be something of the form $$ \left[ \frac {\partial f(x)}{\partial x}g(x) - \int \frac{\partial f(x)}{\partial x} + \frac {\partial g(x)}{\partial x}f(x) - \int \frac{\partial g(x)}{\partial x} \right]^b_a $$ if done by integration by parts but it really doesn't look right.

Any tips would be appreciated.

Answer 1

Some other ways to write it:

Integrating the first term in each product, $$ \int_a^b (f''(x) g(x) - g''(x) f(x) ) = \left[ f'(x)g(x) - g'(x) f(x) \right]_a^b - \int_a^b ( f'(x)g'(x) - g'(x) f'(x) ) = \left[ f'(x)g(x) - g'(x) f(x) \right]_a^b. $$

Or, with $dh(x) = h(x) dx,$: $$ \int_a^b g d(f') - \int_a^b f d(g')(x) = \left[ f'g - g' f \right]_a^b - \int_a^b ( g'df - f' dg ) = \left[ f'g - g' f \right]_a^b. $$

Answer 2

Let $h(x) = f'(x)$, $k(x)=g'(x)$... Don't worry about changing $g(x)$ and $f(x)$, though. It's just easier to integrate by parts like this. The first summand is \begin{equation}\int_{[a,b]}h'(x)g(x)dx = h(x)g(x)|_{[a,b]} - \int_{[a,b]}h(x)g'(x)dx\end{equation} And the second \begin{equation}-\int_{[a,b]}k'(x)f(x)dx = -k(x)f(x)|_{[a,b]} + \int_{[a,b]}k(x)f'(x)dx\end{equation} Now reverse $h(x) = f'(x)$, $k(x)=g'(x)$ and sum the terms up: \begin{equation}f'(x)g(x)|_{[a,b]} - \int_{[a,b]}f'(x)g'(x)dx -g'(x)f(x)|_{[a,b]} + \int_{[a,b]}g'(x)f'(x)dx = f'(x)g(x)|_{[a,b]} -g'(x)f(x)|_{[a,b]} \end{equation}