St41n
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I found this passage in a book:
http://img8.imageshack.us/img8/1452/75717730.jpg
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:
[tex][ e^{itx} \phi(x) ]^{+ \infty}_{- \infty}[/tex]
where i is the imaginary unit and [tex]\phi(x)[/tex] is the normal p.d.f.
So,
[tex]e^{itx} \phi(x) = \frac{1}{\sqrt{2\pi}} \exp \left( -0.5x^2 + itx \right)[/tex]
So, is this term:
[tex]\left[ \frac{1}{\sqrt{2\pi}} \exp \left( -0.5x^2 + itx \right) \right]^{+ \infty}_{- \infty}[/tex]
equal to zero and why? It is not clearly evident to me
http://img8.imageshack.us/img8/1452/75717730.jpg
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:
[tex][ e^{itx} \phi(x) ]^{+ \infty}_{- \infty}[/tex]
where i is the imaginary unit and [tex]\phi(x)[/tex] is the normal p.d.f.
So,
[tex]e^{itx} \phi(x) = \frac{1}{\sqrt{2\pi}} \exp \left( -0.5x^2 + itx \right)[/tex]
So, is this term:
[tex]\left[ \frac{1}{\sqrt{2\pi}} \exp \left( -0.5x^2 + itx \right) \right]^{+ \infty}_{- \infty}[/tex]
equal to zero and why? It is not clearly evident to me
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