Final answer:
The Laplace transform of sinh(bt) is determined by integrating e^-st multiplied by sinh(bt), which simplifies to b / (s^2 - b^2), assuming that the real part of s is greater than the absolute value of b for the integral to converge.
Step-by-step explanation:
To find the Laplace transform of sinh(bt), we use the definition of the Laplace transform:
L{f(t)} = ∫0∞ e-stf(t) dt.
Applying this to sinh(bt) gives:
L{sinh(bt)} = ∫0∞ e-st × \left(\frac{e^{bt} - e^{-bt}}{2}\right) dt.
This can be split into two integrals:
ßrac{1}{2} × (\int e^{(b-s)t} dt - \int e^{-(b+s)t} dt).
We then integrate each part:
For the first part:
\int e^{(b-s)t} dt = 1/(b-s) × e^{(b-s)t} + C,
and similarly for the second part:
\int e^{-(b+s)t} dt = -1/(b+s) × e^{-(b+s)t} + C.
When we evaluate these from 0 to infinity and subtract, assuming Re(s) > |b| for convergence, we get:
L{sinh(bt)} = ßrac{1}{2} × ( \frac{1}{b-s} - \frac{1}{b+s} ) = ßrac{1}{2} × \frac{2b}{(b-s)(b+s)}.
The final expression simplifies to:
L{sinh(bt)} = b / (s2 - b2)