 Solution

Now the inverse integrand can be writtenFor $0<\lambda <1/2$, the exponent has a positive real time parameter. Close the Bromwich contour to the left; the residue at $s=-1$ is $\frac{1}{2}(2\lambda )=\lambda$. For $1/2<\lambda <1$, close to the right, where there are no poles. The reversed orientation there still gives zero. ConsequentlyThe same result follows from the Laplace transform time-shift rule: $H(s)=\frac{1}{2}{e}^{-s\log 2}/(s+1)$ corresponds to $f(t)=e^{-t}\Theta (t-\log 2)$, 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,regardless of the assigned value at the jump.