= Solution
Now the inverse integrand can be written
$$
\lambda^{-s}H(s)=\frac{e^{s\log(1/(2\lambda))}}{2(s+1)},\qquad\gamma>-1.
$$
For $0<\lambda<1/2$, the exponent has a positive real time parameter. Close the <Bromwich contour> to the left; the <residue> at $s=-1$ is $\tfrac12(2\lambda)=\lambda$. For $1/2<\lambda<1$, close to the right, where there are no <poles>. The reversed orientation there still gives zero. Consequently
$$
\boxed{h(\lambda)=\begin{cases}\lambda,&0<\lambda<\tfrac12,\\0,&\tfrac12<\lambda<1.\end{cases}}
$$
The same result follows from the <Laplace transform time-shift rule>: $H(s)=\tfrac12e^{-s\log2}/(s+1)$ corresponds to $f(t)=e^{-t}\Theta(t-\log2)$, where $\Theta$ is the <Heaviside step function>.
The original transform cannot fix a value at the single point $\lambda=1/2$. The <Bromwich inversion formula> chooses the average of the limiting values $1/2$ and zero, giving $\boxed{h(1/2)=1/4}$ under that convention. Indeed,
$$
\int_0^1\lambda^{s-1}h(\lambda)\,d\lambda=\int_0^{1/2}\lambda^s\,d\lambda=\frac{2^{-s-1}}{s+1},\qquad\operatorname{Re}s>-1,
$$
regardless of the assigned value at the jump.
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