What is the value of that integral? The Maple code 
int(exp(-z^2*sin(z)^2), z = 0 .. infinity, numeric, epsilon = 0.1e-1) 

outputs $2.835068335 $. However, I am not sure if the answer is correct.
$$ I = \int_0^{\infty}{e^{ - z^2  \sin^2(z)}} \text{d}z $$
PS. Does that integral converge at all?
 A: When integrating
$$
\int_0^\infty e^{-z^2\sin^2(z)}\,\mathrm{d}z
$$
the problem points are near $z=k\pi$ where $\sin^2(z)$ vanishes.
On $[(k-1/2)\pi,(k+1/2)\pi]$
$$
\begin{align}
\int_{(k-1/2)\pi}^{(k+1/2)\pi} e^{-z^2\sin^2(z)}\,\mathrm{d}z
&=\int_{-\pi/2}^{\pi/2}e^{-(z+k\pi)^2\sin^2(z)}\,\mathrm{d}z\tag{1}\\
&\ge\int_{-\pi/2}^{\pi/2}e^{-(k+1/2)^2\pi^2z^2}\,\mathrm{d}z\tag{2}\\
&=\frac1{(k+1/2)\pi}\int_{-(k+1/2)\pi^2/2}^{(k+1/2)\pi^2/2} e^{-z^2}\,\mathrm{d}z\tag{3}\\
&\sim\frac1{(k+1/2)\sqrt\pi}\tag{4}
\end{align}
$$
Explanation:
$(1)$: substitute $z\mapsto z+k\pi$
$(2)$: $(z+k\pi)^2\le(k+1/2)^2\pi^2$ and $\sin^2(z)\le z^2$
$(3)$: substitute $z\mapsto\frac{z}{(k+1/2)\pi}$
$(4)$: $\lim\limits_{k\to\infty}\int_{-(k+1/2)\pi^2/2}^{(k+1/2)\pi^2/2} e^{-z^2}\,\mathrm{d}z=\sqrt\pi$
Thus, adding these up in $k$ diverges like the harmonic series. Therefore, the integral diverges.
A: This will only be an illustration of Robjohn's excellent answer.
There is large numerical instability while evaluating this integral (the numerical results returned by CAS will depend of the precision required and the method used : I got values ranging from $1.5$ to $5$).
The problem is that the exponent will take the value $0$ for $\,z=\pi n\;$ as illustrated :

So let's rather evaluate :
$$f_n=\int_{\pi n}^{\pi (n+1)} e^{- z^2\sin^2(z)} \,dz$$
I got :
\begin{array} {c|c}
n&n\,f_n\\
\hline
10&0.538590916089835479\\
100&0.561397879233431915\\
1000&0.563907784742113993\\
10000&0.564161377032037476\\
\end{array}
confirming Robjohn's $\dfrac 1{\sqrt{\pi}}\,$ coefficient and the divergence of the integral.
