As you wrote
$$I=\int_{-\infty }^{\log( \pi)} \sqrt{\sin (e^t)} \, dt=\int_{0 }^{\pi}\frac{\sqrt{\sin (x)}}{x}\,dx$$
Using series expansion
$$\sqrt{\sin (x)}=\sum_{n=0}^\infty \frac {a_n}{b_n} x^{2 n+\frac{1}{2}} $$ The $a_n$ form the sequence $A008991$ and the $b_n$ form the sequence $A008992$ is $OEIS$.
So
$$\int\frac{\sqrt{\sin (x)}}{x}\,dx=2\sum_{n=0}^\infty \frac {a_n}{(4n+1)\,b_n} x^{2 n+\frac{1}{2}}$$
The convergence is very slow
Edit
None of the CAS I was able to use gave the antiderivative or the integral. However, using the $\large 1,400$ years old approximation
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician we can write
$$\frac{\sqrt{\sin (x)}}{x} \simeq 2 \sqrt{\frac{(\pi -x)}{x (x-a) (x-b)}}\quad \text{with} \quad a=\left(\frac{1}{2}-i\right) \pi\quad \text{and} \quad b=\left(\frac{1}{2}+i\right) \pi$$ which can be integrated using elliptic integrals of the first and third kinds.
Just a few numbers for
$$I_k=\int_0^{k\frac{\pi}{12}}\frac{\sqrt{\sin (x)}}{x}\,dx$$
$$\left(
\begin{array}{ccc}
k & \text{approximation}& \text{solution} \\
1 & 1.02920 & 1.02216 \\
2 & 1.44820 & 1.44060 \\
3 & 1.76172 & 1.75428 \\
4 & 2.01665 & 2.00943 \\
5 & 2.23049 & 2.22336 \\
6 & 2.41171 & 2.40459 \\
7 & 2.56500 & 2.55789 \\
8 & 2.69317 & 2.68612 \\
9 & 2.79790 & 2.79094 \\
10 & 2.87997 & 2.87305 \\
11 & 2.93881 & 2.93182 \\
12 & 2.96882 & 2.96169
\end{array}
\right)$$