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I need this for a problem I am working on.

Define $h_n(y,t) = -\sqrt{n}y + (\sqrt{n} + t)^2\ln\left( 1 + \dfrac{y}{\sqrt{n}} \right).$

Then $\lim_{n \to \infty} h_n(y,t) = \dfrac{-y^2}{2} + 2yt.$

Is $\eqref{eq:1}$ true for all $t \in \mathbb{R}?$ $$\lim_{n\to\infty}\int_{0}^{\infty} \exp(h_n(y,t)) dy = \int_{0}^{\infty} \exp\bigg(\dfrac{-y^2}{2} + 2yt\bigg)dy \tag{a} \label{eq:1} .$$

The above does hold for $t=0$, in that case one can show $h_n(y,0) \leq \ln(1+y) - y$ and since $\int_0^\infty\exp(-y)(1+y)dy < \infty$ the interchange of limits and integration is justified by the dominated convergence theorem. However I cannot find a suitable dominating function for a general $t$.

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We can write

$$ h_n(y,t) = h_n(y,0) + (2t\sqrt{n} + t^2) \log\left(1 + \frac{y}{\sqrt{n}}\right). $$

Now by a direct computation, we find that

$$ \frac{\partial^2}{\partial n^2} h_n(y,0) = \frac{y^3}{4 n^{3/2} \left( \sqrt{n} + y \right)^2} > 0 \qquad \text{and} \qquad \lim_{n\to\infty} \frac{\partial}{\partial n} h_n(y,0) = 0. $$

This shows that $\frac{\partial}{\partial n} h_n(y,0) < 0$ and thus $h_n(y,0)$ is decreasing in $n$. Together with the inequality $\log(1+x) \leq x$, it follows that for $n \geq N$,

$$ h_n(y,t) \leq h_N(y,0) + 2|t|y + t^2 \log\left(1 + \frac{y}{\sqrt{N}}\right). $$

Now choose $N$ so that $\sqrt{N} > 2|t|$. Then

$$ \mathrm{e}^{h_n(y,t)} \leq \mathrm{e}^{-(\sqrt{N}-2|t|)y}\left(1 + \frac{y}{\sqrt{N}}\right)^{N+t^2}. $$

This bound serves as a dominating function, and hence we can apply the dominated convergence theorem to obtain the desired equality.

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