Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2002 / Paper 2 / 5

For a piecewise continuous function of exponential order on $t\ge 0$, its unilateral Laplace transform isThe Bromwich inversion formula, at a point of continuity, isThe Bromwich contour is a straight vertical line, traversed upwards, with $\gamma$ to the right of the abscissa of convergence and hence of all singularities of the transform. Under the usual piecewise smoothness assumptions, at a jump the symmetric inversion limit gives the mean of the two one-sided values.

In the transform in the question, put $\lambda =e^{-t}$. Then $d\lambda =-e^{-t}dt$ andThus the restricted Mellin transform on the unit interval is the Laplace transform of $f(t)=h(e^{-t})$. Apply the Bromwich inversion formula with $t=-\log \lambda$, where $0<\lambda <1$, to obtainHere $\gamma$ lies in the convergence half-plane of $H$. At a discontinuity this formula has the same mean-value interpretation. The real natural logarithm defines ${\lambda }^{-s}$ unambiguously.

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 ${\lambda }^{-1/2}/2$. HenceIts defining integral checks both the answer and its convergence condition:

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.