- #1
bdforbes
- 152
- 0
Show that
[tex] \stackrel{lim}{\alpha \rightarrow \infty} \int^{\infty}_{-\infty}g(x)\sqrt{\frac{\alpha}{\pi}}e^{-\alpha x^2} dx = g(0) [/tex]
where g(x) is continuous.
To use the continuity of g(x) I started from
[tex] \left|g(x)-g(0)\right|<\epsilon [/tex]
and tried to put it in into the integral:
[tex]
\left| \int^{\delta}_{-\delta}g(x)\sqrt{\frac{\alpha}{\pi}}e^{-\alpha x^2} dx - \int^{\delta}_{-\delta}g(0)\sqrt{\frac{\alpha}{\pi}}e^{-\alpha x^2} dx \right|
\leq \int^{\delta}_{-\delta}\left|g(x)-g(0)\right|\sqrt{\frac{\alpha}{\pi}}e^{-\alpha x^2} dx
< \epsilon \int^{\delta}_{-\delta}\sqrt{\frac{\alpha}{\pi}}e^{-\alpha x^2} dx
[/tex]
But I'm not sure where this gets me.
[tex] \stackrel{lim}{\alpha \rightarrow \infty} \int^{\infty}_{-\infty}g(x)\sqrt{\frac{\alpha}{\pi}}e^{-\alpha x^2} dx = g(0) [/tex]
where g(x) is continuous.
To use the continuity of g(x) I started from
[tex] \left|g(x)-g(0)\right|<\epsilon [/tex]
and tried to put it in into the integral:
[tex]
\left| \int^{\delta}_{-\delta}g(x)\sqrt{\frac{\alpha}{\pi}}e^{-\alpha x^2} dx - \int^{\delta}_{-\delta}g(0)\sqrt{\frac{\alpha}{\pi}}e^{-\alpha x^2} dx \right|
\leq \int^{\delta}_{-\delta}\left|g(x)-g(0)\right|\sqrt{\frac{\alpha}{\pi}}e^{-\alpha x^2} dx
< \epsilon \int^{\delta}_{-\delta}\sqrt{\frac{\alpha}{\pi}}e^{-\alpha x^2} dx
[/tex]
But I'm not sure where this gets me.