Laplace transform of $\cosh(t)$ from first principles: How to deal with infinities? 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.
 A: 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
$$
\int_0^\infty e^{-t-st}\;dt = \lim_{M \to \infty} \left(\frac{-e^{-M(s+1)}}{1+s}+\frac{1}{1+s}\right) = \frac{1}{1+s}\qquad\text{(assuming $\mathrm{Re}\;s>-1$)}
$$
A: If $f$ is bounded or integrable then $\int_0^\infty f(t)e^{-st}dt$ converges and is analytic for $\Re(s) > 0$. For other kind of functions everything can happen. In most cases you'll need to find the least $c$ such that $\int_0^\infty |f(t)e^{-(c+\epsilon)t}|dt<\infty$, then you'll consider $\int_0^\infty f(t)e^{-st}dt$ only for $\Re(s) > c$.

Here $c=1$

The abscissa of convergence theorem (which follows from an integration by parts) is that if $f$ is locally integrable and $\lim_{T\to \infty}\int_0^T f(t)e^{-\sigma t}dt$ converges then $\lim_{T\to \infty}\int_0^T f(t)e^{-s t}dt$ converges and is analytic for $\Re(s) > \sigma$.
