= Solution
{c}
For $\operatorname{Im}k>0$, put $s=-ik$, so $\operatorname{Re}s>0$. The <Laplace transform> for the <Gamma function>, analytically continued from positive real $s$, gives
$$
\int_0^\infty x^{-1/4}e^{ikx}\,dx
=\Gamma(3/4)(-ik)^{-3/4}.
$$
Choose $\arg k\in(0,\pi)$ in the <upper half-plane>. Since $\arg(-ik)=\arg k-\pi/2$,
$$
\boxed{\int_0^\infty x^{-1/4}e^{ikx}\,dx
=\Gamma(3/4)e^{3i\pi/8}k^{-3/4}.}
$$
This integral is absolutely convergent throughout the <upper half-plane>. On the nonzero <real axis> it converges conditionally by <Dirichlet test for improper integrals>, and its value is the upper boundary limit of the same branch. In particular negative real $k$ has argument $\pi$, not $-\pi$ in this formula.
At $k=0$ the integral diverges at infinity; in the lower <half-plane> its <exponential function> factor grows and the ordinary improper integral diverges. The origin itself causes no local-integrability problem for $x^{-1/4}$. A useful continuation of the chosen power uses a <branch cut> descending from $k=0$ with argument in $(-\pi/2,3\pi/2)$.
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