# Finding the Laplace Transform of sin(t)/t

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.

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

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

• Thank you very much. Using the linearity of the Laplace Transform completely slipped my mind. Of course, then I would've had to remember the Taylor Series of arctan. Commented Apr 23, 2013 at 17:38
• @Kupiakos You're very welcome. Make sure to always be on the look-out for little tricks like linearity. One of the key properties of integration and differentiation is its linearity. Without it, we'd be lost. Of course, one might ask: why is the world constructed in such a way that they are linear? I'll leave that to the philosophers. Commented Apr 23, 2013 at 17:44

Just an small hint:

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

• @FlybyNight: Yes. In fact, it is $\tan^{-1}{1/s}$ Commented Apr 23, 2013 at 17:15