Notice, using property of Laplace transform as follows
$$L\left(\frac{1}{t}f(t)\right)=\int_{s}^{\infty}L(f(t))dt$$
$$L(\sin bt)=\int_{0}^{\infty}e^{-st}\sin t dt=\frac{b}{b^2+s^2}$$
Now, we have
$$\int_{0}^{\infty} \frac{e^{-ax}\sin bx}{x}dx$$
$$=\int_{a}^{\infty} L(\sin bx)dx$$
$$=\int_{a}^{\infty}\frac{b}{b^2+x^2} dx$$
$$=b\int_{a}^{\infty}\frac{dx}{b^2+x^2} $$
$$=b\left[\frac{1}{b}\tan^{-1}\left(\frac{x}{b}\right)\right]_{a}^{\infty} $$
$$=\left[\tan^{-1}\left(\infty\right)-\tan^{-1}\left(\frac{a}{b}\right)\right] $$ $$=\frac{\pi}{2}-\tan^{-1}\left(\frac{a}{b}\right)$$ Hence, we have
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\int_{0}^{\infty} \frac{e^{-ax}\sin bx}{x}dx=\frac{\pi}{2}-\tan^{-1}\left(\frac{a}{b}\right)}}$$
I will not give step-by-step calculations of integrals here since they are rather long and tedious.
As @BenedictWilliamJohnIrwin suggested, the integral is of the form $$\int f(x)f^{-1}(y)\, dx \tag1$$ where $y=f(x)=(1-x^{2008})^\frac1{2007}$.
Now it is known that $$\int f^{-1}(y)\, dy=xf(x)-\int f(x) \, dx \tag2$$
(see http://www.med.umich.edu/schnell-lab/assets/sm6.pdf for details) so on integrating $(1)$ by parts and combining it with $(2)$, we get $$\left[xf(x)^2-f(x)\int f(x)\, dx\right]-\int \left(xf(x)f'(x)-\int f(x)\, dx\right) dx \tag3$$ We want to write it in this form since we do not need to deal with $(1-x^{2007})^\frac1{2008}$.
The remainder of the calculations relies heavily on the use of WolframAlpha. The final result is very messy since it contains many hypergeometric functions, so I will not display it.
By WolframAlpha, $$\int f(x) \, dx=\frac{x}{4015}\left(2008\, _2F_1\left(\frac1{2008}, \frac{2006}{2007}; \frac{2009}{2008}; x^{2008}\right)+2007(1-x^{2008})^\frac1{2007}\right)+C_1 \tag4$$ where $C_1$ is a constant.
$(3)$ can be written as $$g_1(x)-\int \left(g_2(x)-\int f(x) \, dx\right)dx$$ where $g_1(x)$ is $$\left[x(1-x^{2008})^\frac2{2007}-\frac{1}{4015}x(1-x^{2008})^\frac1{2007}\left(2008\, _2F_1\left(\frac1{2008}, \frac{2006}{2007}; \frac{2009}{2008}; x^{2008}\right)+2007(1-x^{2008})^\frac1{2007}\right)\right]=x(1-x^{2008})^\frac1{2007}\left[\frac{2008}{4015}\left((1-x^{2008})^\frac1{2007}-\, _2F_1\left(\frac1{2008}, \frac{2006}{2007}; \frac{2009}{2008}; x^{2008}\right)\right)\right]$$ and $g_2(x)$ is $$x(1-x^{2008})^\frac1{2007}\cdot\left(-\frac{1}{2007}\right)(1-x^{2008})^{-\frac{2006}{2007}}\cdot 2008x^{2007}=-\frac{2008}{2007}x^{2008}(1-x^{2008})^{-\frac{2005}{2007}}.$$
Again using WolframAlpha we find that $\int g_2(x)\, dx$ equates to $$\frac{2008}{6023}x\left((1-x^{2008})^\frac2{2007}-\, _2F_1\left(\frac1{2008}, \frac{2005}{2007}; \frac{2009}{2008}; x^{2008}\right)\right)+C_2$$ where $C_2$ is a constant.
Finally, we integrate $(4)$. To do this, we split the integrand into two functions; that is, $\int f(x)\, dx = g_3(x)+ g_4(x)+C_1$ where $g_4(x)=\frac{2007}{4015}xf(x)$ and $g_3(x)$ is the term with the hypergeometric function.
Using the second entry of http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/21/01/02/01/ with $\alpha = \frac{2009}{2008}$ (and WolframAlpha), the integral of $g_3(x)$ is $$-\frac{2008^2}{2009(4015)}x^{\frac{2009}{2008}}\, _3F_2\left(\frac1{2008}, \frac{2006}{2007}, \frac{2009}{2008}; \frac{2009}{2008}, \frac{4017}{2008}; x^{2008}\right)+C_3$$ and on integrating by parts (using $(4)$), the integral of $g_4(x)$ is $$-\frac{2007}{4015(6022)}x^2\left(2007(1-x^{2008})^\frac1{2007}+1004\, _2F_1\left(\frac1{1004}, \frac{2006}{2007}; \frac{1005}{1004}; x^{2008}\right)\right)+C_4$$ wherre $C_3$ and $C_4$ are constants.
Combining the relevant results gives the integral and some constant $C$. Don't forget to integrate $C_1$!
Best Answer
I record this here as it might be useful.
Let $-\infty<a<b\le\infty$ and $\lambda\in\mathbb C$ such that $b<\infty$ or $\Re\lambda>0$. Let $P$ be a polynomial. Then by integration by parts, \begin{align*} &&\int_a^bP(x)\,\mathrm e^{-\lambda x}\,\mathrm dx &=-\left[P(x)\,\frac{\mathrm e^{-\lambda x}}\lambda\right]_a^b+\frac1\lambda\int_a^bP'(x)\,\mathrm e^{-\lambda x}\,\mathrm dx, \end{align*} where $P'$ is again a polynomial. Applying the same with $P'(x)/\lambda$ in place of $P(x)$ yields $$\int_a^bP(x)\,\mathrm e^{-\lambda x}\,\mathrm dx=-\left[P(x)\,\frac{\mathrm e^{-\lambda x}}\lambda+P'(x)\,\frac{\mathrm e^{-\lambda x}}{\lambda^2}\right]_a^b+\frac1{\lambda^2}\int_a^bP''(x)\,\mathrm e^{-\lambda x}\,\mathrm dx.$$ Iterating, since eventually $P^{(k)}=0$ we can write $$\int_a^bP(x)\,\mathrm e^{-\lambda x}\,\mathrm dx=-\left[\sum_{k=0}^\infty P^{(k)}(x)\,\frac{\mathrm e^{-\lambda x}}{\lambda^{k+1}}\right]_a^b,$$ or in indefinite form, $$\int P(x)\,\mathrm e^{-\lambda x}\,\mathrm dx=-\sum_{k=0}^\infty\frac{P^{(k)}(x)\,\mathrm e^{-\lambda x}}{\lambda^{k+1}}+C.$$
Apply with $\lambda=-n$ and $P(x)=x^n$. The $P^{(k)}(x)$ are easy to compute.