basic exponential algebra

Question

Sorry but i have stupid question. Denote \begin{equation} X(\omega) = \mathrm{exp}\bigg(j\lambda\pi\omega\int_{0}^{\infty} (d^{-1}(x))^2\mathrm{e}^{j\omega x}\mathrm{d}x\bigg) \end{equation} where $d(x) = x^{-\alpha}$ and the result is

\begin{equation} X(\omega) = \mathrm{exp}\bigg(j\lambda\pi\omega\int_{0}^{\infty} x^{-2/\alpha}\mathrm{e}^{j\omega x}\mathrm{d}x\bigg) \end{equation}

I've try to emulate but cannot come to the same result on $x^{-2/\alpha}$ part

What i did was \begin{align} d(x)&=x^{-\alpha}\\ d(x)&=\frac{1}{x^{\alpha}}\\ d^{-\frac{1}{\alpha}}(x)&=x\\ (d^{-1}(x))^{1/\alpha}&=x\\ \bigg((d^{-1}(x))^{1/\alpha}\bigg)^2&=x^2\\ (d^{-1}(x))^{2/\alpha}&=x^2\\ (d^{-1}(x))^2&=x^{2\alpha}\\ \end{align} Do i missed something or forget some properties to get $(d^{-1}(x))^2=x^{-2/\alpha}$ ?

edit: got the answer from the keyword given by Eric is inverse function instead of reciprocal

Using inverse function method $d(x) = x^{-\alpha}$ thus interchange $x=d(x)^{-\alpha}$ and solve for $d(x)$ we get $d(x)=x^{-1/\alpha}$ which is the inverse function \begin{equation} \bigg(d^{-1}(x)\bigg)^2=x^{(-1/\alpha)2}=x^{-2/\alpha} \end{equation}

Answer 1

Here $d^{-1}$ means the inverse function of $d$, not the reciprocal of $d$. So $d^{-1}(x)$ is the number $y$ such that $d(y)=x$, rather than $1/d(x)$. Since $$d(x^{-\frac{1}{\alpha}})=(x^{-\frac{1}{\alpha}})^{-\alpha}=x^{-\frac{1}{\alpha}\cdot(-\alpha)}=x^1=x,$$ we have $d^{-1}(x)=x^{-\frac{1}{\alpha}}$ so $d^{-1}(x)^2=(x^{-\frac{1}{\alpha}})^2=x^{-\frac{2}{\alpha}}.$

Answer 2

It is actually quite simple. We have $$d (x) = x^{-\alpha} $$ $$(d (x))^{-\frac {1}{\alpha}} = x^{-\alpha \times (-\frac {1}{\alpha})} $$ $$(d (x))^{-\frac {1}{\alpha}} = x $$ Thus, $d^{-1}(x) = x^{-\frac {1}{\alpha}} $. Hope it helps.