Laplace transform of delays Sinc function I am trying to compute the Laplace transform of the following Sinc function,
$$\operatorname{sinc}(a(t-t_0)) = \frac{\sin(a(t-t_0))}{a(t-t_0)} $$
Where, $t_0$ is the delay and $a$ is constant value.
I know the Laplace transform for $\frac{\sin(at)}{at}$ without delay is $\frac{1}{a}\left(\frac{\pi}{2} - \operatorname{atan}\left(\frac{s}{a}\right)\right)$. I am concerned whether we can use the phase shift property (i.e. multiply by $\exp(-st_0)$) or not, as the function has non-zero values on the negative axis as well?
 A: Well, we have the following Laplace transform:
$$\mathscr{L}_x\left[\text{sinc}\left(\alpha\left(x-x_0\right)\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty\text{sinc}\left(\alpha\left(x-x_0\right)\right)\exp\left(-\text{s}x\right)\space\text{d}x\tag1$$
It is not hard to show that (using Taylor series) $\forall\text{n}$:
$$\text{sinc}\left(\text{n}\right)=\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}\text{n}^{2\text{k}}}{\left(2\text{k}+1\right)!}\tag2$$
So, using that we can rewrite $(1)$ as follows:
$$\mathscr{L}_x\left[\text{sinc}\left(\alpha\left(x-x_0\right)\right)\right]_{\left(\text{s}\right)}=\int_0^\infty\left\{\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}\left(\alpha\left(x-x_0\right)\right)^{2\text{k}}}{\left(2\text{k}+1\right)!}\right\}\cdot\exp\left(-\text{s}x\right)\space\text{d}x=$$
$$\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}\alpha^{2\text{k}}}{\left(2\text{k}+1\right)!}\cdot\left\{\int_0^\infty\left(x-x_0\right)^{2\text{k}}\exp\left(-\text{s}x\right)\space\text{d}x\right\}\tag3$$
Now, let's take a look at a slightly different integral:
$$\mathcal{I}_\beta:=\int_0^\beta\left(x-x_0\right)^{2\text{k}}\exp\left(-\text{s}x\right)\space\text{d}x\tag4$$
First, let $\text{u}:=\left(x-x_0\right)^{2\text{k}+1}$, this leads to:
$$\mathcal{I}_\beta=\frac{\exp\left(-\text{s}x_0\right)}{2\text{k}+1}\int_{\left(-x_0\right)^{2\text{k}+1}}^{\left(\beta-x_0\right)^{2\text{k}+1}}\exp\left(-\text{s}\cdot\text{u}^\frac{1}{2\text{k}+1}\right)\space\text{du}\tag5$$
Now, let $\text{v}:=\text{us}^{2\text{k}+1}$, this leads to:
$$\mathcal{I}_\beta=\frac{\exp\left(-\text{s}x_0\right)}{\left(2\text{k}+1\right)\text{s}^{2\text{k}+1}}\int_{\left(-x_0\text{s}\right)^{2\text{k}+1}}^{\left(\left(\beta-x_0\right)\text{s}\right)^{2\text{k}+1}}\exp\left(-\text{v}^\frac{1}{2\text{k}+1}\right)\space\text{dv}\tag6$$
And this is a special function, called the Incomplete Gamma function:
$$\mathcal{I}_\beta=\frac{\exp\left(-\text{s}x_0\right)}{\left(2\text{k}+1\right)\text{s}^{2\text{k}+1}}\cdot\left[-\left(2\text{k}+1\right)\Gamma\left(2\text{k}+1,\text{v}^\frac{1}{2\text{k}+1}\right)\right]_{\left(-x_0\text{s}\right)^{2\text{k}+1}}^{\left(\left(\beta-x_0\right)\text{s}\right)^{2\text{k}+1}}=$$
$$\frac{-\left(2\text{k}+1\right)\exp\left(-\text{s}x_0\right)}{\left(2\text{k}+1\right)\text{s}^{2\text{k}+1}}\cdot\left[\Gamma\left(2\text{k}+1,\text{v}^\frac{1}{2\text{k}+1}\right)\right]_{\left(-x_0\text{s}\right)^{2\text{k}+1}}^{\left(\left(\beta-x_0\right)\text{s}\right)^{2\text{k}+1}}=$$
$$\frac{\exp\left(-\text{s}x_0\right)}{\text{s}^{2\text{k}+1}}\cdot\left[\Gamma\left(2\text{k}+1,\text{v}^\frac{1}{2\text{k}+1}\right)\right]_{\left(\left(\beta-x_0\right)\text{s}\right)^{2\text{k}+1}}^{\left(-x_0\text{s}\right)^{2\text{k}+1}}\tag7$$
