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I know that for $D\geq 2$, there is no bound state for a Dirac potential $V=-\alpha \delta(\textbf{x})$ unless we use an ultraviolet cutoff $k_{max}=1/a$. I showed this by solving the Schrodinger's equation in $k$-space.

But I need to show that even with such a cutoff, for $D>3$ there cannot be a bound state as $\alpha \rightarrow 0$. I am stuck because the integral \begin{equation} \int_{0}^{1/a}\frac{x^{D-1}}{x^2+K^2}dx \end{equation} is a complicated hypergeometric function in $D$-dimensions and its not obvious how to show for general $D$ and not by taking specific values $D=2,3,4 \dots$.

There is a paper https://arxiv.org/abs/quant-ph/9801033 which uses regularization and renormalization to show this for $D=4$ and extrapolates to $D >4$ but I would like to know of a different argument.

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