Laplace transform of $\displaystyle x^b\frac{\sin\pi(x+a)}{x+a}$ I would like to obtain the Laplace transform of 
$$f(x)= \frac{\sin\pi(x+a)}{x+a}\,x^b$$
that is, the following integral
$$\int_0^{\infty}e^{-s\,x} \frac{\sin\pi(x+a)}{x+a}\,x^b\,dx$$
where $b>-1$ and $a\in\mathbb{R}$.
I 've checked several books but I've not luck.
Any help will be welcomed.
 A: There exists an analytic formula for this integral involving the lower incomplete gamma function, at least for $s\in\mathbb{R}$.
The following forms of the integral are equivalent:
$$\begin{align}
\mathcal{Lf(s)}&=\int_{0}^{\infty}x^b e^{-sx}\frac{\sin\pi(x+a)}{x+a}dx\\&=\pi\int_{0}^{1}dy\int_{0}^{\infty}x^b e^{-sx}\cos\pi y(x+a)dx\\&=\pi\Re\int_{0}^1dy\int_{0}^{\infty}x^b e^{-sx}e^{i\pi y(x+a)}dx\\&=\pi\Re\int_{0}^1dy~\Gamma(b+1)(s-i\pi y)^{-b-1}e^{i\pi a y}\\&=\Re~i\Gamma(b+1)e^{sa}\int_{s-i\pi }^{s-i0}z^{-b-1}e^{-az}dz=\\&=\Re~i\Gamma(b+1)a^be^{sa}\int_{as-i\pi a}^{as-i0}z^{-b-1}e^{-z}dz\end{align}$$
In the last expression the contour is taken to be the straight line segment between the two endpoints. We change the contour trivially by deforming it to the rectangle connecting the endpoints and $M, M-i\pi a$ where $M$ is taken to infinity. It is then easy to show that 
$$\mathcal{L}f(s)=\Gamma(b+1)a^be^{sa}~\Re[i\Gamma(-b,as-i\pi a)-i\Gamma(-b,as)]$$
NOTE: If $s\in\mathbb{C}, \Re s>0$, then one should rewrite the cosine as a sum of imaginary exponentials and repeat the calculation for the result:
$$\mathcal{L}f(s)=i\frac{\Gamma(b+1)a^b e^{sa}}{2}\Big[\Gamma(-b,as-i\pi a)-\Gamma(-b,as+i\pi a)]$$
