One definition of the Yukawa potential on $R^n$ is the solution $G$ in the sense of distributions to $(-\Delta + \mu^2)G = \delta$. This 'green's function' is given by \begin{align*} G(x) = \int_0^\infty (4 \pi t)^{-n/2} \text{Exp}\left( -\frac{|x|^2}{4t} - \mu^2 t\right) dt \end{align*} Which is finite for all $x \in \mathbb{R}^n$ such that $x \ne 0$. If we also define the newtonian potential as \begin{align*} \Gamma(x) = k_n |x|^{2-n} \ , \ \ \ k_n = \left( (n-2)|\mathbb{S}^{n-1}|\right)^{-1} \end{align*} Where we have assumed that $n\ne2$ (else we use a logarithm). It would then seem that in the case of small $|x|$, these two would be about the same, up to a constant. I would like to evaluate \begin{align*} \lim_{x \to 0} \frac{G(x)}{\Gamma(x)} = \lim_{r \to 0} \int_{0}^\infty k_n (4 \pi t)^{-n/2} r^{2-n} \text{Exp}\left( -\frac{r^2}{4t} - \mu^2 t\right) dt \end{align*} In Lieb's "Analysis", it is claimed without proof that this limit exists (for $n>2$) and is equal to $1$. The problem is, I can't seem to verify this. The limit cannot be moved inside the integral (in fact if one does this, the limit of the integrand becomes zero, clearly contrary to the stated goal). I've also tried L'Hôpital's rule. I can't quite see how to proceed.
Appreciate any help!
You've made an algebraic mistake: since $\Gamma$ is in the denominator, it contributes $k_n^{-1}r^{n-2}$ rather than $k_n r^{2-n}$ to the integral.
As you observed, the integrand converges to zero pointwise. But it also acquires a sharp peak near $0$: something like this, only one-sided. To handle the limit, it helps to undo the horizontal shrinkage. A look at the function reveals the pattern $r^2/t$. So, change the variable of integration as $t=r^2s$. Then $$ \begin{split} &\int_{0}^\infty k_n^{-1} (4 \pi t)^{-n/2} r^{n-2} \exp\left( -\frac{r^2}{4t} - \mu^2 t\right) dt \\ &=\int_{0}^\infty k_n^{-1} (4 \pi r^2 s)^{-n/2} r^{n-2} \exp\left( -\frac{1}{4s} - \mu^2 r^2 s\right) r^2\,ds \\ &=\int_{0}^\infty k_n^{-1} (4 \pi s)^{-n/2} \exp\left( -\frac{1}{4s} - \mu^2 r^2 s\right)\,ds \end{split} $$ As $r\to 0$, the integrand increases to $k_n^{-1} (4 \pi s)^{-n/2} \exp\left( -\frac{1}{4s}\right)$. The monotone convergence theorem allows passing to the limit inside of the integral. Thus, the limit of the integral as $r\to 0$ is $$\int_{0}^\infty k_n^{-1} (4 \pi s)^{-n/2} \exp\left( -\frac{1}{4s} \right)\,ds \color{Green}{= 1} $$