Let $a,b >0$ and consider the following integral: $$ I(a,b) = \int_{0}^{\infty} e^{ - i a x } \sin( b x )\ dx $$
Is there a way to evaluate this integral? This looks very similar to the Fourier transform of the function $\sin(bx)$, but the limits of integration are different.
Looking at the wikipedia link, it seems I should be getting something involving dirac delta functions. The limits of integration are really throwing me off though, since I can't use the $\delta(a-b) = \int_{0}^{\infty} e^{i(a-b)x}dx$ in this form.
$$\int_{0}^{\infty} e^{-iax}\sin(bx)dx $$ $$ = \lim_{\sigma\to 0} \int_{0}^{\infty} e^{-(\sigma + ia)x}\sin(bx)dx$$ $$ = \lim_{\sigma\to 0} \int_{0}^{\infty} \dfrac{e^{ibx} - e^{-ibx} }{2i}e^{-(\sigma + ia)x}dx$$ $$ = \lim_{\sigma\to 0} \left[\int_{0}^{\infty} \dfrac{1}{2i}e^{ibx}e^{-(\sigma + ia)x}dx -\int_{0}^{\infty} \dfrac{1}{2i}e^{-ibx}e^{-(\sigma + ia)x}dx\right]$$ $$ = \lim_{\sigma\to 0} \left[\int_{0}^{\infty} \dfrac{1}{2i}e^{-(\sigma + ia -ib)x}dx -\int_{0}^{\infty} \dfrac{1}{2i}e^{-(\sigma + ia+ib)x}dx\right]$$ $$ = \lim_{\sigma\to 0} \dfrac{1}{2i} \left[-\dfrac{e^{-(\sigma + ia -ib)x}}{\sigma+ia-ib}\bigg\rvert_{0}^{\infty} +\dfrac{e^{-(\sigma + ia+ib)x}}{\sigma+ia+ib}\bigg\rvert_{0}^{\infty}\right]$$ $$ = \lim_{\sigma\to 0} \dfrac{1}{2i}\left[\dfrac{1}{\sigma+ia-ib} - \dfrac{1}{\sigma+ia+ib}\right]$$ $$ = \lim_{\sigma\to 0} \dfrac{1}{2i}\left[\dfrac{\sigma + ia +ib}{(\sigma+ia)^2+b^2} - \dfrac{\sigma +ia -ib}{(\sigma+ia)^2+b^2}\right]$$ $$ = \lim_{\sigma\to 0} \dfrac{1}{2i}\left[\dfrac{2ib}{(\sigma+ia)^2+b^2}\right]$$ $$ = \lim_{\sigma\to 0} \dfrac{b}{(\sigma+ia)^2+b^2}$$ $$ = \dfrac{b}{b^2-a^2}$$
Strictly speaking, the region of convergence for the integration is for a real number $\sigma > 0$. I don't have a rigorous justification for the limit as $\sigma \rightarrow 0$ to be valid; but the limit appears to be unambiguous and consistent with the result for $\sigma > 0$.
Update to evaluate the integral as a Fourier Transform (vs. Laplace Transform)
$$I(a, b) = \int_{0}^{\infty} e^{-iax}\sin(bx)dx$$ $$I(2\pi f, b) = \int_{0}^{\infty} \sin(bx) e^{-2\pi ifx}dx$$ $$ = \int_{-\infty}^{\infty} H(x) \sin(bx) e^{-2\pi ifx}dx$$ where $H(x)$ is the Heavyside unit step, which for this answer I'll choose as $H(x) = \dfrac{1}{2} + \dfrac{1}{2} \text{sgn}(x)$.
The integral is now in the form of a Fourier Transform of the product of $H(x)$ and $\sin(bx)$ $$ = \mathscr{F}\{H(x)\sin(bx)\} = \mathscr{F}\{H(x)\} * \mathscr{F}\{\sin(bx)\} = \mathscr{F}\left\{\dfrac{1}{2} + \dfrac{1}{2} \text{sgn}(x)\right\} * \mathscr{F}\{\sin(bx)\}$$
Without going through the derivation of those individual Fourier transforms, as presented in The Fourier Transform and Its Applications by Ronald N. Bracewell, I'll just substitute them in here without proof:
$$ = \left[\dfrac{1}{2}\delta(f)-\dfrac{1}{2}\dfrac{i}{\pi f}\right] *\dfrac{i}{2}\dfrac{\pi}{|b|}\left[\delta\left(\dfrac{\pi}{b}f + \dfrac{1}{2}\right)-\delta\left(\dfrac{\pi}{b}f - \dfrac{1}{2}\right)\right]$$ $$ = \left[\dfrac{1}{2}\delta(f)-\dfrac{1}{2}\dfrac{i}{\pi f}\right] *\dfrac{i}{2}\dfrac{\pi}{|b|}\left[\dfrac{\delta\left(f + \dfrac{b}{2\pi}\right)}{\left|\dfrac{\pi}{b}\right|}-\dfrac{\delta\left(f - \dfrac{b}{2\pi}\right)}{\left|\dfrac{\pi}{b}\right|}\right]$$ $$ = \left[\dfrac{1}{2}\delta(f)-\dfrac{1}{2}\dfrac{i}{\pi f}\right] *\dfrac{i}{2}\left[\delta\left(f + \dfrac{b}{2\pi}\right)-\delta\left(f - \dfrac{b}{2\pi}\right)\right]$$ $$ = \dfrac{i}{4}\left[\delta\left(f + \dfrac{b}{2\pi}\right)-\delta\left(f - \dfrac{b}{2\pi}\right)\right] + \dfrac{1}{4\pi}\left[\dfrac{1}{f+\dfrac{b}{2\pi}}-\dfrac{1}{f-\dfrac{b}{2\pi}}\right]$$ $$ = \dfrac{i}{4}\left[\delta\left(f + \dfrac{b}{2\pi}\right)-\delta\left(f - \dfrac{b}{2\pi}\right)\right] + \left[\dfrac{b}{b^2-(2\pi f)^2}\right]$$ $$ = \dfrac{i}{4}\left[\delta\left(\dfrac{a}{2\pi} + \dfrac{b}{2\pi}\right)-\delta\left(\dfrac{a}{2\pi} - \dfrac{b}{2\pi}\right)\right] + \left[\dfrac{b}{b^2-a^2}\right]$$
Which is the correct answer on the imaginary axis in the s-plane (my earlier Laplace transform answer is the analytic continuation to the right hand side of the s-plane).
So the integral blows up at $b = \pm a$ still, just differently than in my first answer.
I'll note that the Dirac delta functions in this answer arise from the shift of $1/2$ (to convert $\text{sgn}(x)$ to $H(x)$) being multiplied with the $\sin(bx)$ term. The other term is due to $\text{sgn}(x)$ being multiplied with the $\sin(bx)$ term.