Inverting Shifted Laplace function

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The discussion focuses on solving a beam deflection problem using the Laplace transform. The user has derived the Laplace transform of the beam displacement but struggles with inverting the first term due to its shifted nature. They reference the general shift property of the Laplace transform and the Mellin inversion integral but seek guidance on the initial steps for inversion. The main challenge lies in handling the term involving the shift, which complicates the inversion process. Assistance is requested to clarify the inversion steps for the specific term in question.
danmel413
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Homework Statement


A beam is supported at one end, as shown in the diagram (PROBLEM 11 page 281 of Lea, 159 of the course pack). A block of mass M and length l is placed on the beam, as shown. Write down the known conditions at x = 0. Use the Laplace transform to solve for the beam displacement.

Homework Equations


I've gotten almost the entire way through the problem using the differential equations for beam deflections (most relevant ones given here: http://www.me.berkeley.edu/~lwlin/me128/BeamDeflection.pdf (8.4 and 8.50), to get to a Laplace transform I now have to invert:

L{y(x)} = Y(x) = [(Mge-sx0)/(EILs5)] * (1-e-sL) - (Mg(x0 + L/2))/(EIs3) - (Mg)/EIs4). (This is correct, I have the solutions).

The general shift of a Laplace transform: L{S(t-t0)f(t-t0)}=e-st0)F(s) for t0 > 0

The Mellin inversion integral: f(t) = 1/(2πi) ∫F(s)estds from γ-i∞ to γ+i∞

The Attempt at a Solution


So up in my first equation, the only part that can't be inverted easily is the first term which has to be shifted. The solution doesn't walk through the inversion, just gives the answer, which is annoying. It's this part:[(Mge-sx0)/(EILs5)] * (1-e-sL). I'm working in x instead of t, of course, and my F(s) would be [(Mg)/(EILs5)] * (1-e-sL), and of course the other part of the integral would be e-sx0. Can someone put me on the first step to getting through this?
 
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Multiply it out and you get two terms, one with a shift X0 and one with a shift (X0+L).
 

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