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Theta Jacobi limit

By Tolaso in Uncategorized on September 2, 2019.

Prove that

$\lim_{x \rightarrow 1-} \sqrt{1-x} \sum_{n=0}^{\infty} x^{n^2} = \frac{\sqrt{\pi}}{2}$

Solution

We are working near $1$ . It follows from the Integral Comparison Test that

$\int_0^\infty x^{t^2}\,\mathrm{d}t \leq \sum_{n=0}^\infty x^{n^2} \leq 1 + \int_0^\infty x^{t^2}\,\mathrm{d}t$

However,

$\begin{align*} \int_0^\infty x^{t^2}\,\mathrm{d}t &= \int_0^\infty \exp\left[-\left(t\sqrt{-\log x}\right)^2\right]\,\mathrm{d}t \\ &= \frac{1}{\sqrt{-\log x}} \int_0^\infty e^{-u^2}\,\mathrm{d}u \\ &= \frac{1}{2} \sqrt{\frac{\pi}{-\log x}} \end{align*}$