I am interested in finding the solution to the following integral:
$$ \int \frac{1}{ca - b} e^{iax} da$$
where $c$ and $b$ are simply constants (or variables independent of $a$). I know in the case that $c$ = 1 and $b = 0$ that the above integral reduces to sgn$(x)$. But for general $b$ and $c$ independent of $a$, is there a general solution?
The improper integral of Riemann $$ \int_{-\infty}^\infty\frac{e^{itx}}{ct-b}\,\mathrm d t $$ doesn't converge for any values of $c$ and $b$. Assuming that $c\neq 0$, we have that the Cauchy principal value is $$ \begin{align*} \operatorname{PV}\int_{-\infty}^\infty\frac{e^{itx}}{ct-b}\,\mathrm d t&=\frac1c\operatorname{PV}\int_{-\infty}^\infty\frac{e^{itx}}{t-b/c}\,\mathrm d t \\&=\frac1c\operatorname{PV}\int_{-\infty}^\infty\frac{e^{i(s+b/c)x}}{s}\,\mathrm d s\\&=\frac{e^{ibx/c}}{c}\operatorname{PV}\int_{-\infty}^\infty\frac{e^{isx}}{s}\,\mathrm d s \end{align*}\tag1 $$ From here two cases: if $x=0$ then $\operatorname{PV}\int_{-\infty}^\infty\frac1{t}\,\mathrm d t=0$, and when $x\neq 0$ we have that $$ \begin{align*} \operatorname{PV}\int_{-\infty}^\infty\frac{e^{isx}}{s}\,\mathrm d s&=\operatorname{sign}(x)\operatorname{PV}\int_{-\infty}^\infty\frac{e^{ir}}{r}\,\mathrm d r\\&=i\operatorname{sign}(x)\int_{-\infty}^\infty\frac{\sin r}{r}\,\mathrm d r\\&=i\pi\operatorname{sign}(x) \end{align*}\tag2 $$ Therefore $$ \operatorname{PV}\int_{-\infty}^\infty\frac{e^{itx}}{ct-b}\,\mathrm d t=\frac{i\pi }{c}e^{ibx/c}\operatorname{sign}(x)\tag3 $$