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Oct
3
awarded  Autobiographer
Oct
3
comment Proving $\frac{e^{z^2}}{\sqrt{\pi}}\int_{-\infty}^{z}e^{-t^2}dt$ is bounded for $\Re(z) \leq 0$
I was just wanting clarification on the statement of the theorem. The limit calculation is fine.
Oct
3
comment Proving $\frac{e^{z^2}}{\sqrt{\pi}}\int_{-\infty}^{z}e^{-t^2}dt$ is bounded for $\Re(z) \leq 0$
Thank you Giovanni, are you able to clarify what L'Hôpital's rule implies?
Oct
3
awarded  Scholar
Oct
3
accepted Proving $\frac{e^{z^2}}{\sqrt{\pi}}\int_{-\infty}^{z}e^{-t^2}dt$ is bounded for $\Re(z) \leq 0$
Oct
2
revised Proving $\frac{e^{z^2}}{\sqrt{\pi}}\int_{-\infty}^{z}e^{-t^2}dt$ is bounded for $\Re(z) \leq 0$
improving formatting
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 $\Re(z) \leq 0$