# Seth's question via email about a Laplace Transform

• MHB
• Prove It
In summary, the Laplace Transform of $\displaystyle 36\left[ \frac{\cosh{\left( 4\,t \right) } - 1}{t} \right]$ is $\displaystyle -\frac{36}{2} \, \ln{ \left| 1 - \frac{16}{s^2} \right| }$. The process involves using the formula $\displaystyle \mathcal{L}\,\left\{ \frac{f\left( t \right) }{t} \right\} = \int_s^{\infty}{F\left( u \right) \,\mathrm{d}u }$, where $\displaystyle Prove It Gold Member MHB Find the Laplace Transform of$\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.

## 1. What is a Laplace Transform?

A Laplace Transform is a mathematical tool used to convert a function of time into a function of complex frequency. It is commonly used in engineering and physics to solve differential equations and analyze systems.

## 2. Why is a Laplace Transform useful?

A Laplace Transform allows us to analyze complex systems and differential equations in a simpler way. It also helps in solving initial value problems and finding solutions to differential equations that cannot be solved using traditional methods.

## 3. How is a Laplace Transform calculated?

The Laplace Transform is calculated by integrating the function of time multiplied by the exponential function of negative time. The result is a function of complex frequency.

## 4. What are the applications of a Laplace Transform?

A Laplace Transform has many applications in engineering, physics, and mathematics. It is used in control systems, signal processing, circuit analysis, and solving differential equations in various fields.

## 5. Are there any limitations to using a Laplace Transform?

While a Laplace Transform is a powerful tool, it has limitations in solving certain types of differential equations and functions with discontinuities. It also requires knowledge of complex analysis and can be challenging to interpret the results in some cases.

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