I want to calculate the inverse Fourier transform of the following statement: $$(-ika)^{-B}(ik+A)^{-c}$$ I used convolution theorem.$F^{-1}((-ika)^{-B}*F^{-1}((ik+A)^{-c})$.
$$F^{-1}((ik+A)^{-c})=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{ikx}(ik+A)^{-c}dk=\sqrt{2\pi}\frac{(-x)^{c-1}e^{Ax}}{Γ(c)}$$ What are you suggesting for this phrase $F^{-1}(-ika)^{-B}$
The inverse Fourier Transform of $F(k)\equiv \text{sgn}(k)$, is given by
$$\begin{align} \mathscr{F}\{\text{sgn}\}(x)&=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty \text{sgn}(k)e^{-ikx}\,dk\\\\ &=\lim_{a\to 0^+}\left(\frac1{\sqrt{2\pi}}\int_{-\infty}^0 (-1)e^{(a-ix)k}\,dk+\frac1{\sqrt{2\pi}}\int_0^\infty e^{(-a-ix)k}\,dk \right)\\\\ &=\sqrt{\frac2\pi}\frac1{ix} \end{align}$$
The $n$'th ordered derivative of the inverse Fourier Transform of $F(k)$ is given by
$$\frac{d^n}{dx^n}\mathscr{F}\{F\}(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty (-ik)^nF(k)e^{-ikx}\,dk$$
whence we find that the inverse Fourier Transform of $(-ik)^n\text{sgn}(k)$ is
$$\begin{align} \frac1{2\pi}\int_{-\infty}^\infty (-ik)^n\text{sgn}(k)e^{ikx}\,dk&=\frac{d^n}{dx^n}\left(\sqrt{\frac2\pi}\frac1x \right)\\\\ &=(-1)^n\sqrt{\frac2\pi}\frac{n!}{ix^{n+1}} \end{align}$$
Finally, we have
$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \left(i\sqrt{\frac\pi2}\frac{(ik)^{n-1}}{(n-1)!}\text{sgn}(k)\right)e^{-ikx}\,dk=\frac1{x^n}$$