In addition to changing the "upper bound" $t$ of integral,
you can make the integrand depends on $t$. This allows you
to get rid of the tail of the integrand for finite $t$.
For any $n \in \mathbb{Z}_{+}$, let $f_n : [0,\infty) \to \mathbb{R}$ be the function
$$f_n(x) = \begin{cases} (1 - \frac{x^2}{n})^n, & x \le \sqrt{n}\\0, & x \ge \sqrt{n}\end{cases}$$
For any fixed $x \in [0,\infty)$, it is easy to see $f_n(x) \le f_{n+1}(x)$ whenever $x \ge \sqrt{n}$.
When $x < \sqrt{n}$, we can apply AM $\ge$ GM to $n$ copies of $1 - \frac{x^2}{n}$ and one copy of $1$ and get
$$1 - \frac{x^2}{n+1} = \frac{n}{n+1}\left(1 - \frac{x^2}{n}\right) + \frac{1}{n+1} \ge \left( 1 - \frac{x^2}{n}\right)^{n/n+1} \implies f_{n+1}(x) \ge f_n(x)$$
This means $f_1(x), f_2(x), \ldots$ is a sequence of pointwise non-decreasing,
non-negative functions. Since its pointwise limit equals to $e^{-x^2}$, we can use Monotone convergence theorm to convert the integral on $e^{-x^2}$ to a limit of single variable:
$$\begin{align}\frac{\sqrt{\pi}}{2}
= \int_0^\infty e^{-x^2} dx
&= \int_0^\infty \lim_{n\to\infty} f_n(x) dx
\stackrel{\rm MCT}{=}
\lim_{n\to\infty} \int_0^\infty f_n(x) dx\\
&= \lim_{n\to\infty} \int_0^{\sqrt{n}}\left(1 - \frac{x^2}{n}\right)^n dx
= \lim_{n\to\infty} \sqrt{n} \int_0^1 \left(1 - t^2\right)^n dt\\
&= \lim_{n\to\infty} \sqrt{n}\sum_{k=0}^n \frac{(-1)^k}{2k+1}\binom{n}{k}
\end{align}$$
I do not know in which context you faced this function (for mutual conveniency, I reuse the formula as @Mycroft wrote it).
$$A(a) = \frac{2 a(a-1) K\left(\frac{a+1}{2 a+1}\right)+4(2 a+1) E\left(\frac{a+1}{2 a+1}\right)}{3 \sqrt{a (2 a+1)}}$$
We faced a very similar one years ago in thermodynamics for $a>1$ and, for obvious computing reasons, we develop it as a series
$$A(a)=\sum_{n=0}^p \frac{\alpha_n}{\beta_n} a^{1-n}$$ The table below reproduces the values for the first $n$'s
$$\left(
\begin{array}{ccc}
n & \alpha_n & \beta_n \\
0 & \sqrt{2} \left(24 \pi ^2+\Gamma \left(\frac{1}{4}\right)^4\right) & 16
\sqrt{\pi } \Gamma \left(\frac{1}{4}\right)^2 \\
1 & \sqrt{2} \Gamma \left(\frac{1}{4}\right) \Gamma \left(\frac{5}{4}\right)^2+\pi
\Gamma \left(\frac{7}{4}\right) & 2 \sqrt{\pi } \Gamma \left(\frac{1}{4}\right)
\\
2 & -\sqrt{2} \left(3 \pi ^2+40 \Gamma \left(\frac{1}{4}\right) \Gamma
\left(\frac{5}{4}\right)^3\right) & 96 \sqrt{\pi } \Gamma
\left(\frac{1}{4}\right)^2 \\
3 & 5 \sqrt{2} \Gamma \left(\frac{1}{4}\right) \Gamma \left(\frac{5}{4}\right) &
512 \sqrt{\pi } \\
4 & 3 \sqrt{2} \left(14 \pi ^2-15 \Gamma \left(\frac{1}{4}\right)^3 \Gamma
\left(\frac{5}{4}\right)\right) & 10240 \sqrt{\pi } \Gamma
\left(\frac{1}{4}\right)^2 \\
5 & \sqrt{2} \left(-77 \pi ^2+45 \Gamma \left(\frac{1}{4}\right)^3 \Gamma
\left(\frac{5}{4}\right)\right) & 20480 \sqrt{\pi } \Gamma
\left(\frac{1}{4}\right)^2 \\
6 & 3 \sqrt{2} \left(154 \pi ^2-65 \Gamma \left(\frac{1}{4}\right)^3 \Gamma
\left(\frac{5}{4}\right)\right) & 163840 \sqrt{\pi } \Gamma
\left(\frac{1}{4}\right)^2
\end{array}
\right)$$
A few results for $a=10^k$
$$\left(
\begin{array}{ccc}
k & \text{approximation} & \text{exact} \\
0 & 2.91216151929910 & 2.91258419032827 \\
1 & 10.3487069573255 & 10.3487069573835 \\
2 & 88.9615534510584 & 88.9615534510584 \\
3 & 875.573856650175 & 875.573856650175 \\
4 & 8741.74602244500 & 8741.74602244500 \\
5 & 87403.4726014927 & 87403.4726014927 \\
6 & 874020.738884157 & 874020.738884157 \\
7 & 8740193.40176002 & 8740193.40176002
\end{array}
\right)$$
Best Answer
The integral converges to $\sqrt{\pi}/2$. Indeed, let
$$ F(x) = \int_{0}^{x} (1- e^{-1/t^2}) \, dt. $$
By the Abel's test, the series
$$ \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1} (1 - e^{-(2m+1)^2/x^2}) $$
converges uniformly on $[0,\infty)$ (with the convention $e^{-\infty} = 0$). So we can switch the integration and summation in the following computation:
\begin{align*} I_R &:= \int_{0}^{R} \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1} (1 - e^{-(2m+1)^2/x^2}) \, dx \\ &\hspace{3em}= \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1} \int_{0}^{R} (1 - e^{-(2m+1)^2/x^2}) \, dx\\ &\hspace{6em}= \sum_{m=0}^{\infty} (-1)^m F\left(\frac{R}{2m+1}\right). \end{align*}
Proceeding,
\begin{align*} I_R &= \sum_{m=0}^{\infty} \left\{ F\left(\frac{R}{4m+1}\right) - F\left(\frac{R}{4m+3}\right) \right\} \\ &\hspace{3em}= \sum_{m=0}^{\infty} \int_{\frac{R}{4m+3}}^{\frac{R}{4m+1}} (1 - e^{-1/t^2}) \, dt \\ &\hspace{6em}= \sum_{m=0}^{\infty} \int_{\frac{4m+1}{R}}^{\frac{4m+3}{R}} \frac{1 - e^{-x^2}}{x^2} \, dx, \end{align*}
where we applied the substitution $x = 1/t$ in the last line. (This substitution is not essential for our argument, but I adopted this step to make clear how monotonicity works.)
Now using the fact that the integrand is decreasing, we can bound $2I_R$ from below by
$$ 2I_R \geq \sum_{m=0}^{\infty} \left( \int_{\frac{4m+1}{R}}^{\frac{4m+3}{R}} \frac{1 - e^{-x^2}}{x^2} \, dx + \int_{\frac{4m+3}{R}}^{\frac{4m+5}{R}} \frac{1 - e^{-x^2}}{x^2} \, dx \right) = \int_{\frac{1}{R}}^{\infty} \frac{1 - e^{-x^2}}{x^2} \, dx. $$
Similar idea shows that
$$ 2I_R \leq \int_{\frac{1}{R}}^{\infty} \frac{1 - e^{-x^2}}{x^2} \, dx + \int_{\frac{1}{R}}^{\frac{3}{R}} \frac{1 - e^{-x^2}}{x^2} \, dx. $$
Finally, taking $R \to \infty$ proves
$$ \int_{0}^{\infty} \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1} (1 - e^{-(2m+1)^2/x^2}) \, dx = \lim_{R\to\infty} I_R = \frac{1}{2} \int_{0}^{\infty} \frac{1 - e^{-x^2}}{x^2} \, dx = \frac{\sqrt{\pi}}{2}. $$