Finding the Laplace Transform of sin(t)/t

Question

I'm in a Differential Equations class, and I'm having trouble solving a Laplace Transformation problem.

This is the problem: Consider the function

$$f(t) = \{\begin{align}&\frac{\sin(t)}{t} \;\;\;\;\;\;\;\; t \neq 0\\& 1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; t = 0\end{align}$$

a) Using the power series (Maclaurin) for $\sin(t)$ - Find the power series representation for $f(t)$ for $t > 0.$

b) Because $f(t)$ is continuous on $[0, \infty)$ and clearly of exponential order, it has a Laplace transform. Using the result from part a) (assuming that linearity applies to an infinite sum) find $\mathfrak{L}\{f(t)\}$. (Note: It can be shown that the series is good for $s > 1$)

There's a few more sub-problems, but I'd really like to focus on b).

I've been able to find the answer to a):

$$ 1 - \frac{t^2}{3!} + \frac{t^4}{5!} - \frac{t^6}{7!} + O(t^8)$$

The problem is that I'm awful at anything involving power series. I have no idea how I'm supposed to continue here. I've tried using the definition of the Laplace Transform and solving the integral

$$\int_0^\infty e^{-st}*\frac{sin(t)}{t} dt$$

However, I just end up with an unsolvable integral.

Any ideas/advice?

Answer 1

The point of the question is to find the Laplace Transform of the Taylor series. Then try to use that to find the Laplace transform of the original function. As you rightly say:

$$\frac{\sin t}{t} \sim 1 - \frac{t^2}{3!} + \frac{t^4}{5!} - \frac{t^6}{7!} \pm \cdots $$

The claim then is that

$$\mathcal{L}\left(\frac{\sin t}{t}\right)(s) \sim \mathcal{L}\left(1 - \frac{t^2}{3!} + \frac{t^4}{5!} - \frac{t^6}{7!} \pm \cdots\right)(s)$$

The Laplace transform is linear, so we need to find:

$$\mathcal{L}(1)(s) - \frac{1}{3!}\mathcal{L}(t^2)(s) + \frac{1}{5!}\mathcal{L}(t^4)(s) - \frac{1}{7!}\mathcal{L}(t^6)(s) \pm \cdots $$

Hopefully, you remember that $\mathcal{L}(t^n)(s) = n!/s^{n+1}$. So we get:

\begin{array}{ccc} \mathcal{L}\left( \frac{\sin t}{t}\right)(s) &\sim& \frac{1}{s} - \frac{1}{3!}\frac{2!}{s^3} + \frac{1}{5!}\frac{4!}{s^5} - \frac{1}{7!}\frac{6!}{s^7} \pm \cdots \\ &\equiv& \frac{1}{s} - \frac{1}{3s^3} + \frac{1}{5s^5} - \frac{1}{7s^7} \pm \cdots \\ \\ &\equiv& \tan^{-1}\left(\frac{1}{s}\right) \end{array}

In the last step I just recognised that

$$\tan^{-1} x \sim x - \frac{1}{3}x^3 + \frac{1}{5}x^5 - \frac{1}{7}x^7 \pm \cdots $$

Answer 2

Just an small hint:

Theorem: If $\mathcal{L}\{f(t)\}=F(s)$ and $\frac{f(t)}{t}$ has a laplace transform, then $$\mathcal{L}\left(\frac{f(t)}{t}\right)=\int_s^{\infty}F(u)du$$