How can the limit of this function be a Dirac delta function?

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I am currently reading a paper of Hugh Montgomery on Pair Correlation Conjecture. In his paper, he introduces a function $F(\alpha, T)$ which is defined for any real $\alpha$ and $T \geq 2$. He shows that for any fixed $0 \leq \alpha < 1$ $$ F(\alpha) = (1 + o(1))T^{-2\alpha} \log T + \alpha + o(1) $$ as $T \to \infty$.

He remarks that the first term on the right hand side behaves in the limit as a Dirac $\delta$-function. I do not quite understand what he means by that. Like Dirac $\delta$-function at what point?

I understand that as $\alpha \to \infty$, the first term behaves like $\infty \chi_{[0, 1]}$ but $T \geq 2$ and the aforementioned limiting function is not even a Dirac $\delta$-function. Any help will be highly appreciated.

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Since

$$ \int_0^1 T^{-2\alpha}\log T\mathrm d\alpha=\frac12-\frac1{2T^2} $$

and

$$ \lim_{T\to\infty}T^{-2\alpha}\log T=\begin{cases}0&\alpha\gt0\\\infty&\alpha=0\;,\end{cases} $$

in the limit $T\to\infty$ this function is increasingly concentrated at $\alpha=0$ with constant integral (except for the term $-\frac1{2T^2}$ that goes to zero). Such behaviour is often informally described as “behaving as a delta ‘function’”.

Note that this only holds for the integral up to $0$ from the right; obviously the function diverges with $T\to\infty$ for $\alpha\lt0$.