Given a real, continuous signal $x(t)$, which has a Fourier Transform $X(\omega)$ satisfying the property;
$$\ln\vert X(\omega)\vert = -\vert \omega \vert.$$ Find the signal $x(t)$, if we assume it to be an even signal.
My approach
We know that if $$x(t) \leftrightarrow X(\omega), \text{then}$$ $$ x^{*}(t) \leftrightarrow X^{*}(-\omega)$$
Since $x(t) $ is real, we have $X(\omega) = X^{*}(-\omega)$. Moreover since $x(t)$ is even, we obtain that $X(\omega) = X(-\omega)$. Using these two relations, we get $X(-\omega)$, hence $X(\omega) \in \mathbb{R}.$
Next,
\begin{align*} x(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}\vert X(\omega)\vert e^{j\omega t}d\omega \\ &= \frac{1}{2\pi}( \int_{-\infty}^{0}e^{\omega}e^{j\omega t}d\omega+\int_{0}^{\infty}e^{-\omega}e^{j\omega t}d\omega)\\ \end{align*}
My question is regarding the second integral as $\omega \rightarrow \infty$, it tends to $\infty$.
Since $\int_{0}^{\infty}e^{-\omega}e^{j\omega t}d\omega=e^{(jt-1)\omega}\vert_{0}^{\infty}.$
We know, $e^{(jt-1)\omega}=e^{j\omega t}.e^{-\omega}$ As $\omega \rightarrow \infty$, the term $e^{-\omega}$ tends to $0.$
On the other hand, $e^{j\omega t} = \cos\omega t+ j \sin\omega t.$ As $\omega \rightarrow \infty$ , $\cos\; \omega t$, and $\sin\; \omega t$ don't have a limit. They oscillate rapidly. Hence we have divergence.
Can I kindly get help here, if this is the right approach to follow? Thanks
