I'm trying to find the Laplace transform of $\cosh(t)$ from first principles. My work is as follows:
$$\begin{align} \mathcal{L}\{\cosh(t)\} &= \int_0^\infty e^{-st} \cosh(t) \ dt \\ &= \dfrac{1}{2} \int_0^\infty e^{-st} \left( \dfrac{e^t + e^{-t}}{2} \right) \ dt \ \ \text{(Using the definition of $\cosh(t)$.)} \\ &= \dfrac{1}{2} \int_0^\infty e^{t – st} \ dt + \dfrac{1}{2} \int_0^\infty e^{-t – st} \ dt \\ &= \dfrac{1}{2(1 – s)} \lim_{t_1 \to \infty – s\infty} \int_0^{t_1} e^{u_1} \ du_1 + \dfrac{1}{2(-1 – s)} \lim_{t_2 \to -\infty – s\infty} \int_0^{t_2} e^{u_2} \ du_2 \\ &= \dfrac{1}{2(1 – s)} \lim_{t_1 \to \infty – s\infty} \left[ e^{u_1} \right]^{t_1}_0 + \dfrac{1}{2(-1 – s)} \lim_{t_2 \to -\infty – s\infty} \left[ e^{u_2} \right]^{t_2}_0 \\ &= \dfrac{1}{2(1 – s)} \left[ e^{\infty – s\infty} – 1 \right] + \dfrac{1}{2(-1 – s)} \left[ e^{-\infty – s\infty} – 1 \right] \\ &= \dfrac{1}{2(1 – s)} \left[ e^{\infty} e^{-s \infty} – 1 \right] + \dfrac{1}{2(-1 – s)} \left[ e^{-\infty} e^{- s\infty} – 1 \right] \\ &= \dfrac{1}{2(1 – s)} \left[ e^{\infty} e^{-\infty (x + iy)} – 1 \right] + \dfrac{1}{2(-1 – s)} \left[ e^{-\infty} e^{-\infty (x + iy)} – 1 \right] \\ &= \dfrac{1}{2(1 – s)} \left[ e^{\infty} e^{-\infty x} e^{-\infty iy} – 1 \right] + \dfrac{1}{2(-1 – s)} \left[ e^{-\infty} e^{-\infty x} e^{-\infty iy} – 1 \right] \\ &= \dfrac{1}{2(1 – s)} \left\{ e^{\infty} e^{-\infty x} \cos[(- \infty y) + i \sin(- \infty y)] – 1 \right\} + \dfrac{1}{2(-1 – s)} \left\{ e^{-\infty} e^{-\infty x} \cos[(- \infty y) + i \sin(- \infty y)] – 1 \right\} \\ &= \dfrac{1}{2(1 – s)} \left\{ e^{\infty} e^{-\infty x} \cos[(- \infty y) + i \sin(- \infty y)] – 1 \right\} + \dfrac{1}{2(-1 – s)} \left\{ \dfrac{1}{e^{\infty x} e^{\infty}} \cos[(- \infty y) + i \sin(- \infty y)] – 1 \right\} \\ &= \dfrac{1}{2(1 – s)} \left\{ e^{\infty} e^{-\infty x} \cos[(- \infty y) + i \sin(- \infty y)] – 1 \right\} – \dfrac{1}{2(-1 – s)} \\ &= \dfrac{1}{2(1 – s)} \left\{ \dfrac{e^{\infty}}{e^{\infty x}} \cos[(- \infty y) + i \sin(- \infty y)] \right\} – \dfrac{1}{2(-1 – s)} \end{align}$$
I know that, by the theory of Laplace transforms, we require that $\Re(s) > 0$. However, I'm still unsure of how to deal with the term with $\dfrac{e^{\infty}}{e^{\infty x}}$? How does this lead to a solution?
I would greatly appreciate it if people would please take the time to clarify this.
Best Answer
You are OK up to here: $$ \dfrac{1}{2} \int_0^\infty e^{t - st} \ dt + \dfrac{1}{2} \int_0^\infty e^{-t - st} \ dt $$ Now go like this: $$ \int_0^M e^{t-st}\;dt = \frac{1}{1-s}e^{t-st}\big|_{t=0}^M =\frac{e^{M(1-s)}}{1-s}-\frac{1}{1-s} \\ \int_0^\infty e^{t-st}\;dt = \lim_{M \to \infty} \left(\frac{e^{M(1-s)}}{1-s}-\frac{1}{1-s}\right) = \frac{-1}{1-s} \qquad\text{(assuming $\mathrm{Re}\;s>1$)} $$ And similarly for the other one $$ \int_0^\infty e^{-t-st}\;dt = \lim_{M \to \infty} \left(\frac{-e^{-M(s+1)}}{1+s}+\frac{1}{1+s}\right) = \frac{1}{1+s}\qquad\text{(assuming $\mathrm{Re}\;s>-1$)} $$