Recent content by mfengwang

Thanks a lot! It's really a great help to me. :smile::smile::smile:

Hi, We know that the Gaussian integral is \int_{-\infty}^{+\infty}e^{-\frac{x^2}{a^2}}dx=a\sqrt{\pi} However, if the gaussian function is discrete in x, what is the result of \sum_{n=0}^{+\infty}e^{-\frac{n^2}{a}} = \\? where n is natural number, that is n=0,1,2,3....