 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 $\epsilon$ and outer radius $R$, enclosing $\pm i$. The upper-bank contribution tends to $J$. On the lower bank, $z^{-1/4}=e^{-i\pi /2}{x}^{-1/4}=-i{x}^{-1/4}$, 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 $O({\epsilon }^{3/4})$. Both vanish, givingThe residues on this branch areThe residue theorem yieldsThereforeThe 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.