MHB Seth's question via email about a Laplace Transform

[Prove It]

Find the Laplace Transform of $\displaystyle 36\left[ \frac{\cosh{\left( 4\,t \right) } - 1}{t} \right] $.

Since this is of the form $\displaystyle \frac{f\left( t \right)}{t} $ we should use $\displaystyle \mathcal{L}\,\left\{ \frac{f\left( t \right) }{t} \right\} = \int_s^{\infty}{F\left( u \right) \,\mathrm{d}u } $.

Here $\displaystyle f\left( t \right) = \cosh{\left( 4\,t \right) } - 1 $ and so

$\displaystyle F\left( s \right) = \frac{s}{s^2 - 16} - \frac{1}{s} $

Therefore

$\displaystyle \begin{align*} \mathcal{L}\,\left\{ \frac{\cosh{\left( 4\,t \right) } - 1}{t} \right\} &= \int_s^{\infty}{ \left( \frac{u}{u^2 - 16} - \frac{1}{u} \right) \,\mathrm{d}u } \\
&= \lim_{b \to \infty}\int_s^b{ \left( \frac{u}{u^2 - 16} - \frac{1}{u} \right) \,\mathrm{d}u } \\
&= \lim_{b \to \infty} \left[ \frac{1}{2}\,\ln{\left| u^2 - 16 \right| } - \ln{\left| u \right| } \right] _s^b \\
&= \frac{1}{2} \lim_{b \to \infty} \left[ \ln{ \left| u^2 - 16 \right| } - 2\,\ln{ \left| u \right| } \right] _s^b \\
&= \frac{1}{2} \lim_{b \to \infty} \left[ \ln{ \left| u^2 - 16 \right| } - \ln{ \left| u^2 \right| } \right] _s^b \\
&= \frac{1}{2} \lim_{b \to \infty} \left[ \ln{ \left| \frac{u^2 - 16}{u^2} \right| } \right] _s^b \\
&= \frac{1}{2} \lim_{b \to \infty} \left[ \ln{ \left| 1 - \frac{16}{u^2} \right| } \right] _s^b \\
&= \frac{1}{2} \left\{ \lim_{b \to \infty} \left[ \ln{ \left| 1 - \frac{16}{b^2} \right| } \right] - \ln{ \left| 1 - \frac{16}{s^2} \right| }\right\} \\ &= \frac{1}{2} \left( \ln{ \left| 1 - 0 \right| } - \ln{ \left| 1 - \frac{16}{s^2} \right| } \right) \\
&= \frac{1}{2} \left( 0 - \ln{ \left| 1 - \frac{16}{s^2} \right| } \right) \\ &= -\frac{1}{2} \, \ln{ \left| 1 - \frac{16}{s^2} \right| } \end{align*} $

 

[benorin]

That seems correct to me, I check the tables, you forgot the factor of 36 out front tho.