# Integration by parts

1. Jan 6, 2008

### quasar987

[SOLVED] Integration by parts

1. The problem statement, all variables and given/known data
I've been starring at this for 30 minutes and can't figure out what's wrong. I end up with 1/xi instead of xi. The book says,

$$\int_{-\infty}^{+\infty}te^{-t^2/2}\sin(\xi t)dt=\int_{-\infty}^{+\infty}e^{-t^2/2}\xi \cos(\xi t)dt$$

3. The attempt at a solution

I set $u=te^{-t^2/2}$ and $dv = \sin(\xi t)$. So I get $du=e^{-t^2/2}-t^2e^{-t^2/2}$ and unless I'm completely crazy, $v=-\xi^{-1}\cos(\xi t)[/tex], so that $$\int_{-\infty}^{+\infty}te^{-t^2/2}\sin(\xi t)dt=te^{-t^2/2}\xi^{-1}\cos(\xi t)|_{-\infty}^{+\infty} + \int_{-\infty}^{+\infty}e^{-t^2/2}\xi^{-1}\cos(\xi t)dt - \int_{-\infty}^{+\infty}t^2e^{-t^2/2}\xi^{-1}\cos(\xi t)dt = \int_{-\infty}^{+\infty}e^{-t^2/2}\xi^{-1}\cos(\xi t)dt$$ HELP! 2. Jan 6, 2008 ### Rainbow Child What is [itex] v$?

$$I=-\int_{-\infty}^{+\infty}(e^{-t^2/2})'\sin(\xi t)dt=-te^{-t^2/2}\sin(\xi t)|_{-\infty}^{+\infty} + \int_{-\infty}^{+\infty}e^{-t^2/2}(\sin(\xi t))'dt = \int_{-\infty}^{+\infty}e^{-t^2/2}\xi\cos(\xi t)dt$$

3. Jan 6, 2008

### CompuChip

I can't entirely follow what you're doing with the u and dv. I learned partial integration as
$$\int_a^b f'(x) g(x) dx = - \left. f(x) g(x) \right|_a^b + \int_a^b f(x) g'(x) dx$$
Applying this to
$$f'(t) = t e^{-t^2/2}, \qquad \implies \qquad f(t) = - e^{-t^2/2}$$
and
$$g(t) = \sin(\xi t), \qquad \implies \qquad g'(t) = \xi \cos(\xi t)$$
then gives me
$$\int_{-\infty}^{+\infty}te^{-t^2/2}\sin(\xi t)dt = \left. e^{-t^2/2} \sin(\xi t) \right|_{-\infty}^{+\infty} + \int_{-\infty}^{+\infty} -e^{-t^2/2}\xi \cos(\xi t)dt.$$
The boundary term vanishes so that gives exactly minus the result you'd like (probably a minus error on my side) but definitely a $\xi$ and not $\xi^{-1}$. As it should be (otherwise you'd get strange results for $\xi = 0$).

I should learn to type LaTeX even faster, Rainbow Child beat me by 5 whole minutes [/edit]

Last edited: Jan 6, 2008
4. Jan 6, 2008

### quasar987

I see my mistake: the last term does not vanish. Thx!