From Gardiner, it is known that if the correlation function of the operators $X(t)$ such that
$$ \left<X(t)X^{\dagger}(t^{\prime})\right>=\frac{\gamma}{2}e^{-\gamma|t-t^{\prime}|} $$
It is further mentioned that the limit of $\gamma\rightarrow\infty$ corresponds to a delta function $\delta(t-t^{\prime})$, since
$$ \int_{-\infty}^{\infty}\frac{\gamma}{2}e^{-\gamma|t-t^{\prime}|}dt^{\prime} =1 $$
and
$$ \lim_{\gamma\longrightarrow\infty}\frac{\gamma}{2}e^{-\gamma|t-t^{\prime}|}dt^{\prime}=0 $$
for $t\neq t^{\prime}$. I simply do not see how these two conditions lead to the delta function $\delta(t-t^{\prime})$. What is going on here?
Edit: I understand that the first condition is simply the normalization requirement for the delta function. However, I still do not see how the delta function arises. The manifestation of the delta function implies that $X(t)$ is a gaussian white noise
Let $\phi\in C^\infty_C$. Then clearly we have
$$\int_{-\infty}^\infty \phi(t')\frac\gamma2 e^{-\gamma|t-t'|}\,dt'\overbrace{=}^{\gamma(t-t')\mapsto t'}\frac12 \int_{-\infty}^\infty \phi(t-t'/\gamma)e^{-|t'|}\,dt'$$
Using the Dominated Convergence Theorem, we find that
$$\lim_{\gamma\to\infty}\int_{-\infty}^\infty \phi(t')\frac\gamma2 e^{-\gamma|t-t'|}\,dt'=\phi(0)$$
which implies that in distribution
$$\lim_{\gamma\to\infty}\frac\gamma2 e^{-\gamma|t-t'|}=\delta(t-t')$$
as was to be shown!
Interestingly there is not anything special about the function $\gamma2 e^{-\gamma|t-t'|}$. In fact, for any function $d( t)\in L^1$ such that $\int_{-\infty}^\infty d(t)\,dt=1$, then the Dominated Convergence Theorem guarantees that for any $\phi\in C_C^\infty$ and $\gamma>0$
$$\lim_{\gamma\to \infty}\int_{-\infty}^\infty \phi(t') \gamma d(\gamma(t-t'))\,dt'=\lim_{\gamma\to \infty}\int_{-\infty}^\infty \phi(t-t'/\gamma)d(t')\,dt'=\phi(t)$$