## Empty one year ago Fun integral problem

1. Empty

$\Large \int\limits_a^b f''(x)g(x)dx +\int\limits_b^a g''(x)f(x)dx$ Evaluate

2. Empty

Good luck! :)

3. ikram002p

*watching this *

4. Empty

*drum roll*

5. freckles

like to make a conjecture $(f(x))'|_a^b \cdot g(x)-(g(x))'|_a^b f(x)$

6. freckles

oops missing pluggin a and b into those one things

7. anonymous

Sturm-Liouville operators are self-adjoint, meaning $$\langle Lf,g\rangle=\langle f,Lg\rangle$$ over a suitable Hilbert space

8. freckles

$[(f(x))' g(x)]_a^b-[(g(x))' f(x)]_a^b$

9. anonymous

you can verify this by just integrating by parts, although in this case it is not as simple as there are no suitable boundary conditions: \begin{align*}\int_a^b f''(x) g(x)\, dx&=f'(x)g(x)\bigg|_a^b -\int_a^b f'(x) g'(x)\, dx\\&=\bigg[f'(x)g(x)-f(x)g'(x)\bigg]_a^b+\int_a^b f(x) g''(x)\, dx\end{align*}

10. perl

$$\large \int\limits_a^b f''(x)g(x)dx +\int\limits_b^a g''(x)f(x)dx \\= \large \int\limits_a^b f''(x)g(x)dx -\int\limits_a^b g''(x)f(x)dx \\= \large \int\limits_a^b [f''(x)g(x) -g''(x) f(x)]dx$$

11. freckles

oh integration by parts awesome @oldrin.bataku

12. anonymous

anyways, since $$\int_b^a f(x)g''(x) \, dx=-\int_a^b f(x)g''(x)\, dx$$ the remaining integral cancels out and we have $$\bigg[f'(x) g(x)-f(x)g'(x)\bigg]_a^b$$

13. Empty

Without using IBP (Which is admittedly better since it's systematic and doesn't rely on tricks) we can us this cheap trick is if you can recognize: $\frac{d}{dx}[f'(x)g(x)-f(x)g'(x)] = f''(x)g(x)-f(x)g''(x)$ Similarly you can use this sort of trick to integrate: $\int \vec v \times \vec v '' d t = \vec v \times \vec v'$ Fun stuff :D

14. perl

nice problem :)