= Solution
For this real integral choose $0<\arg z<2\pi$, so the upper bank of the positive real axis has $z^{1/4}=x^{1/4}>0$. Use a counterclockwise <keyhole contour> with inner radius $\varepsilon$ and outer radius $R$, enclosing $\pm i$. The upper-bank contribution tends to $J$. On the lower bank, $z^{-1/4}=e^{-i\pi/2}x^{-1/4}=-ix^{-1/4}$, and the reversed orientation makes its contribution $+iJ$.
The large arc is $O(R^{-5/4})$ because $g=O(R^{-9/4})$, while the small arc is $O(\varepsilon^{3/4})$. Both vanish, giving
$$
\oint g(z)\,dz\longrightarrow(1+i)J.
$$
The <residues> on this branch are
$$
\operatorname{Res}(g,i)=\frac{e^{-i\pi/8}}{2i},\qquad
\operatorname{Res}(g,-i)=-\frac{e^{-3i\pi/8}}{2i}.
$$
The <residue theorem> yields
$$
(1+i)J=\pi\left(e^{-i\pi/8}-e^{-3i\pi/8}\right)
=2\pi i e^{-i\pi/4}\sin\frac\pi8.
$$
Therefore
$$
\boxed{J=\pi\sqrt2\sin\frac\pi8=\frac{\pi}{2\sin(3\pi/8)}=\frac{\pi}{\sqrt{2+\sqrt2}}.}
$$
The branch cut in (a) was arbitrary; here placing it on the integration axis makes the bank contributions directly proportional to the desired positive-real integral.
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