80 reputation
7
bio website contraflo.ws
location Sydney, Australia
age 24
visits member for 1 year, 10 months
seen Mar 26 at 6:10

I am currently studying both Physics and Mathematics at the University of Sydney. My main interest areas are Differential Geometry and Particle Physics.


Oct
2
revised Proving $\frac{e^{z^2}}{\sqrt{\pi}}\int_{-\infty}^{z}e^{-t^2}dt$ is bounded for $\Re(z) \leq 0$
improving formatting
Oct
2
comment Proving $\frac{e^{z^2}}{\sqrt{\pi}}\int_{-\infty}^{z}e^{-t^2}dt$ is bounded for $\Re(z) \leq 0$
Integration over the straight line contour ${t+z:-\infty \le t \le 0}$
Oct
2
revised Proving $\frac{e^{z^2}}{\sqrt{\pi}}\int_{-\infty}^{z}e^{-t^2}dt$ is bounded for $\Re(z) \leq 0$
Clearer title
Oct
2
comment Proving $\frac{e^{z^2}}{\sqrt{\pi}}\int_{-\infty}^{z}e^{-t^2}dt$ is bounded for $\Re(z) \leq 0$
@GiovanniDeGaetano, My claim came about when entering the function into wolfram alpha and seeing the exact result. I've proved this to myself and am quite sure of it. In regards to your upper bound, I feel that as there is still the exponential term out the front which will explode at -$\infty$, that it isn't as simple as this.
Oct
2
revised Proving $\frac{e^{z^2}}{\sqrt{\pi}}\int_{-\infty}^{z}e^{-t^2}dt$ is bounded for $\Re(z) \leq 0$
Adding tag
Oct
2
awarded  Editor
Oct
2
revised Proving $\frac{e^{z^2}}{\sqrt{\pi}}\int_{-\infty}^{z}e^{-t^2}dt$ is bounded for $\Re(z) \leq 0$
Changed title
Oct
2
awarded  Student
Oct
2
asked Proving $\frac{e^{z^2}}{\sqrt{\pi}}\int_{-\infty}^{z}e^{-t^2}dt$ is bounded for $\Re(z) \leq 0$