- #1
mjordan2nd
- 177
- 1
I know that
[tex]
\int_{- \infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}.
[/tex]
What if I tried evaluating this over the complex plane? If I created a contour that ran along the real axis, then circled around from positive to negative infinity this should not change things as along the curve at infinity my complex function should be 0. So using that contour, I should have
[tex]
\oint_C e^{-z^2} dz = 0,
[/tex]
unless there are some poles in the upper half of the complex plane. But if there aren't, wouldn't this imply [itex]\int_{- \infty}^{\infty} e^{-x^2}dx = 0[/itex], since the integral over the other part of my contour is 0? I'm forced to conclude I am missing something here, but can't quite put my finger on what that might be.
[tex]
\int_{- \infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}.
[/tex]
What if I tried evaluating this over the complex plane? If I created a contour that ran along the real axis, then circled around from positive to negative infinity this should not change things as along the curve at infinity my complex function should be 0. So using that contour, I should have
[tex]
\oint_C e^{-z^2} dz = 0,
[/tex]
unless there are some poles in the upper half of the complex plane. But if there aren't, wouldn't this imply [itex]\int_{- \infty}^{\infty} e^{-x^2}dx = 0[/itex], since the integral over the other part of my contour is 0? I'm forced to conclude I am missing something here, but can't quite put my finger on what that might be.