Finding Laplace Transform of $f(t) = \frac{\sin(t-2)}{t-2}$

Question

Let $$f(t) = \frac{\sin(t-2)}{t-2}\tag{1}$$ Find the Laplace Transform of $f(t)$.

My try: If there was a unit step function in $f(t)$, it was easy to compute the transformation but I don't know how to tackle this problem. Also definition gives us $$L(f(t)) = \int_0^\infty e^{-st}\frac{\sin(t-2)}{t-2}dt\tag{2}$$ Is this integration solvable?

Answer 1

You can simply combine the power series expansion of $\dfrac{\sin(t)}{t}$: $$ \dfrac{\sin(t)}{t}=\sum_{k=0}^\infty \frac{(-1)^kt^{2k}}{(2k+1)!} $$ with the Laplace identity $$ \mathcal{L}\left\{t^{2k}\right\}(s)=\dfrac{(2k)!}{s^{2k+1}} $$ to show that $$\mathcal{L}\left\{\dfrac{\sin(t)}{t}\right\}(s)=\sum_{k=0}^\infty \dfrac{(-1)^k}{(2k+1)s^{2k+1}}.$$

Some close identities by means of special functions may be obtained if one notice that $\dfrac{\sin(t)}{t}$ (spherical Bessel function $j_0(t)$) admits the hypergeometric series representation $-i{~}_0F_1\left(\frac{3}{2};-\frac{t^2}{4}\right)$. For the Mittag-Leffler lovers, one can also use the identity $\dfrac{\sin(t)}{t}=E_{2,2}(-t^2)$ to show that the underlying result is a Fox-Wright function, in disguise -- see, for instance, https://en.wikipedia.org/wiki/Fox%E2%80%93Wright_function.

Answer 2

The answer is:

$$ \frac{1}{2} i \exp(-2 w) (Ei[2 (-i + w)] - Ei[2 (i + w)]) $$

Where: $Ei(z)$ - ExpIntegralEi is defined as in here. (and $i$ imaginary unit).

Answer 3

Hint.

As

$$ \int_0^{\infty}e^{-st}f(t-a)dt = \int_{-a}^{\infty}e^{-s(a+\xi)}f(\xi)d\xi=e^{-as}\int_{-a}^{\infty}e^{-s\xi}f(\xi)d\xi $$

if $f(x) = 0$ for $x < -a$ we have

$$ \int_0^{\infty}e^{-st}f(t-a)dt =e^{-a s}F(s) $$

Remember that the usual Laplace transform, with limits $0\le t\lt\infty$ assumes that the function to transform is $\theta (t)f(t)$ where $\theta(t)$ is the Heaviside step function.

Answer 4

I am assuming that the intended question was to compute the Laplace transform of the function $$ f(t)=\begin{cases}\frac{\sin(t-2)}{t-2},&t>2\\1,&t=2\\0,&t<2.\end{cases} $$ (At any rate, this question has a more interesting answer in my opinion...)

Then by change of variables from $t-2$ to $t$ we compute that $$ Lf(s)=e^{-2s}\int_0^{\infty}e^{-st}\frac{\sin t}{t}\ dt. $$ Now we observe that $$ \frac{d}{ds}\int_0^{\infty}e^{-st}\frac{\sin t}{t}\ dt=-\int_0^{\infty}e^{-st}\sin t\ dt. $$ Furthermore, $$ \int_0^{\infty}e^{-st}\sin t\ dt=\int_0^{\infty}\textrm{Im}\ e^{(i-s)t}\ dt=\textrm{Im} \frac{1}{s-i}=\textrm{Im} \frac{s+i}{s^2+1}=\frac{1}{s^2+1}. $$ Thus, using the famous integral $\int_0^{\infty}\frac{\sin t}{t}\ dt=\frac{\pi}{2}$ we obtain that $$ \int_0^{\infty}e^{-st}\frac{\sin t}{t}\ dt=\frac{\pi}{2}-\tan^{-1}s. $$ Finally, we obtain the answer $$ Lf(s)=e^{-2s}\Bigl(\frac{\pi}{2}-\tan^{-1}s\Bigr). $$