Integration by parts

1. Mar 12, 2009

St41n

I found this passage in a book:
http://img8.imageshack.us/img8/1452/75717730.jpg [Broken]
where Φ(x) is the c.d.f. of the normal distribution.
However, using integration by parts I also get this term which is not included in the passage:

$$[ e^{itx} \phi(x) ]^{+ \infty}_{- \infty}$$
where i is the imaginary unit and $$\phi(x)$$ is the normal p.d.f.
So,
$$e^{itx} \phi(x) = \frac{1}{\sqrt{2\pi}} \exp \left( -0.5x^2 + itx \right)$$
So, is this term:
$$\left[ \frac{1}{\sqrt{2\pi}} \exp \left( -0.5x^2 + itx \right) \right]^{+ \infty}_{- \infty}$$
equal to zero and why? It is not clearly evident to me

Last edited by a moderator: May 4, 2017
2. Mar 12, 2009

HallsofIvy

Yes, as x go to infinity,
$$e^{-x^2}$$
goes to 0 very quickly. Quickly enough so that even
$$e^{-x^2}e^{x}$$ goes to 0 as x goes to positive infinity.

3. Mar 12, 2009

St41n

I kept doing the same fault, but I figured it out now, thanks