= Solution
Choose $\gamma>1/2$. The inverse integrand is $\lambda^{-s}/[2(s-1/2)]$. Because $-\log\lambda>0$, the exponential factor decays towards the left. Closing the <Bromwich contour> to the left and using the <residue theorem> picks up just the <simple pole> at $s=1/2$, with <residue> $\lambda^{-1/2}/2$. Hence
$$
\boxed{h(\lambda)=\frac1{2\sqrt\lambda},\qquad0<\lambda<1.}
$$
Its defining <integral> checks both the answer and its convergence condition:
$$
\int_0^1\lambda^{s-1}\frac1{2\sqrt\lambda}\,d\lambda=\frac1{2(s-1/2)}=\frac1{2s-1},\qquad\operatorname{Re}s>\tfrac12.
$$
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