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[SOLVED] Integration by parts
I've been staring 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,
[tex]\int_{-\infty}^{+\infty}te^{-t^2/2}\sin(\xi t)dt=\int_{-\infty}^{+\infty}e^{-t^2/2}\xi \cos(\xi t)dt[/tex]
I set [itex]u=te^{-t^2/2}[/itex] and [itex]dv = \sin(\xi t)[/itex]. So I get [itex]du=e^{-t^2/2}-t^2e^{-t^2/2}[/itex] and unless I'm completely crazy, [itex]v=-\xi^{-1}\cos(\xi t)[/tex], so that
[tex]\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[/tex]
HELP!
Homework Statement
I've been staring 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,
[tex]\int_{-\infty}^{+\infty}te^{-t^2/2}\sin(\xi t)dt=\int_{-\infty}^{+\infty}e^{-t^2/2}\xi \cos(\xi t)dt[/tex]
The Attempt at a Solution
I set [itex]u=te^{-t^2/2}[/itex] and [itex]dv = \sin(\xi t)[/itex]. So I get [itex]du=e^{-t^2/2}-t^2e^{-t^2/2}[/itex] and unless I'm completely crazy, [itex]v=-\xi^{-1}\cos(\xi t)[/tex], so that
[tex]\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[/tex]
HELP!