For $\int_0^\infty e^{-u-e^{-ku}}~du$ , where $k>0$ ,
$\int_0^\infty e^{-u-e^{-ku}}~du$
$=\int_0^\infty e^{-u}e^{-e^{-ku}}~du$
$=\int_1^0u^\frac{1}{k}e^{-u}~d\left(-\dfrac{\ln u}{k}\right)$
$=\dfrac{1}{k}\int_0^1u^{\frac{1}{k}-1}e^{-u}~du$
$=\dfrac{1}{k}\gamma\left(\dfrac{1}{k},1\right)$
For $\int_{-\infty}^\infty e^{-\theta^2-e^{-k\theta^2}}~d\theta$ , where $k>0$ ,
$\int_{-\infty}^\infty e^{-\theta^2-e^{-k\theta^2}}~d\theta$
$=\int_{-\infty}^\infty e^{-\theta^2}e^{-e^{-k\theta^2}}~d\theta$
$=\int_{-\infty}^0e^{-\theta^2}e^{-e^{-k\theta^2}}~d\theta+\int_0^\infty e^{-\theta^2}e^{-e^{-k\theta^2}}~d\theta$
$=\int_\infty^0e^{-(-\theta)^2}e^{-e^{-k(-\theta)^2}}~d(-\theta)+\int_0^\infty e^{-\theta^2}e^{-e^{-k\theta^2}}~d\theta$
$=\int_0^\infty e^{-\theta^2}e^{-e^{-k\theta^2}}~d\theta+\int_0^\infty e^{-\theta^2}e^{-e^{-k\theta^2}}~d\theta$
$=2\int_0^\infty e^{-\theta^2}e^{-e^{-k\theta^2}}~d\theta$ , which is similar to What is $\int_0^{\infty}\!e^{-x^2}e^{-ae^{bx^2}}\,dx$?
$=2\int_0^\infty e^{-\theta^2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(e^{-k\theta^2})^n}{n!}d\theta$
$=2\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^ne^{-(kn+1)\theta^2}}{n!}d\theta$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sqrt\pi}{n!\sqrt{kn+1}}$
Since $\theta{}>{}0$, use the delta method . That is, given that, for $X \sim \mbox{exp}\left(\theta\right)$ i.i.d samples, the sample mean $\bar{X}_n$ is asymptotically normally distributed, so that
$$
\sqrt{n}\left(\bar{X}_n{}-{}\theta^{-1}\right){}\to{}N\left(0,\theta^{-2}\right)\,\mbox{as }n{}\to{}\infty\,.
$$
Therefore, a first-order Taylor expansion of the function $\displaystyle \frac{1}{\bar{X}_n}$, in the "vicinity" of the asymptotic mean $\displaystyle \frac{1}{\theta}$, justifies
$$
\sqrt{n}(\hat{\theta}_{MLE} - \theta){}={}\sqrt{n}(\frac{1}{\bar{X}_n} - \theta)\approx\theta^2\sqrt{n}\left(\bar{X}_n{}-{}\theta^{-1}\right){}\sim{}N\left(0,\theta^{2}\right)\,\mbox{, as }n{}\to{}\infty\,.
$$
The "$\approx$" means that the random variables on either side have distributions that, with arbitrary precision, better approximate each other as $n{}\to{}\infty$. This approximation can be made rigorous.
Best Answer
The $t$ density is \begin{eqnarray*} f \left( x \right) & = & \frac{\Gamma \left( \frac{\nu + 1}{2} \right)}{\sqrt{\nu \pi} \Gamma \left( \frac{\nu}{2} \right)} \left( 1 + \frac{x^2}{\nu} \right)^{- \frac{\nu + 1}{2}} \end{eqnarray*} If you write your integrand (for $\nu = 1$) \begin{eqnarray*} \frac{1}{1 + x^2} & = & \left( 1 + \frac{x^2}{\nu} \right)^{- \frac{\nu + 1}{2}} \end{eqnarray*} then you notice that \begin{eqnarray*} \int_0^{\infty} \frac{1}{1 + x^2} \mathrm{d} x & = & \frac{\sqrt{\pi} \Gamma \left( \frac{1}{2} \right)}{\Gamma \left( 1 \right)}\\ & = & \frac{\pi}{2} \end{eqnarray*}