- #1

ghotra

- 53

- 0

What terrible mistake have I made.

[tex]

I = \int_{-\infty}^\infty \mathrm{e}^{-x^2} \mathrm{d}x

[/tex]

I consider the following integral:

[tex]

\int_C \mathrm{e}^{-z^2} \mathrm{d}z

[/tex]

where [itex]C[/itex] is the half-circle (of infinite radius) in the upper-half plane.

There are no singularities in the upper-half plane. So,

[tex]

\int_C \mathrm{e}^{-z^2} \mathrm{d}z = 0

[/tex]

Now, I can break this integral up into two parts.

[tex]

0 = \int_C \mathrm{e}^{-z^2} \mathrm{d}z = \int_{-\infty}^\infty \mathrm{e}^{-x^2} \mathrm{d}x + \int_{\theta=0}^{\theta=\pi} \mathrm{e}^{-z^2} \mathrm{d}z

[/tex]

Or...

[tex]

\int_{-\infty}^\infty \mathrm{e}^{-x^2} \mathrm{d}x &= - \int_{\theta=0}^{\theta=\pi} \mathrm{e}^{-z^2} \mathrm{d}z =

-\lim_{R\rightarrow \infty} \int_{0}^{\pi} \mathrm{e}^{-R^2 \exp(2\mathrm{i}\theta)}\cdot \mathrm{i} R \mathrm{e}^{\mathrm{i}\theta} \mathrm{d}\theta \stackrel{?}{=} \sqrt{\pi}

[/tex]

I know the answer

*should*be [itex]\sqrt{\pi}[/itex]...so the RHS must give this result! But it does not...at least, I have convinced myself that the RHS actually goes to zero. The exponential dominates the linear term, so in the limit as R goes to infinity, the RHS equals zero.

How can this be...it seems I have shown that the gaussian integral is zero.

I should cry.