Evaluating $\int_0^{\infty} \exp(-e^{ia\theta}x^a) \,dx$ for $-\frac{\pi}{2a}<\theta<\frac{\pi}{2a}$ Let $1<a<\infty$ be a fixed number and define $$f(x,\theta)=\exp(-e^{ia\theta}x^a)\quad\text{for }0\leq x<\infty\text{ and }\theta\in\mathbb{R}.$$ Then define $$F(\theta)=\int_0^{\infty} f(x,\theta) \,dx\quad\text{for}-\frac{\pi}{2a}<\theta<\frac{\pi}{2a}.$$
One of my exercises asks to show that using integration by parts, $F(\theta)$ can be written as $$F(\theta)=\int_0^{\infty} \frac{1-a}{a x^a e^{ia\theta}}[f(x,\theta)-1] \,dx$$ but I don't see why that is the case.
Can someone explain to me how to derive such an expression? Moreover, is the integrand $\frac{1-a}{a x^a e^{ia\theta}}[f(x,\theta)-1]$ absolutely integrable? Thank you in advance.
 A: Let's start at the end:

Moreover, is the integrand $\frac{1-a}{a x^a e^{ia\theta}}[f(x,\theta)-1]$ absolutely integrable?

Yes, it is. Since $\lvert\theta\rvert < \frac{\pi}{2a}$, we have $\operatorname{Re} e^{ia\theta} > 0$, and so $\lvert f(x,\theta)\rvert \leqslant 1$ for $x \in [0,+\infty)$. Furthermore, the integrand is continuous on $(0,+\infty)$, so we need only look at the behaviour near $0$ and near $+\infty$ to determine integrability. The boundedness of $f(x,\theta)$ shows the integrand is bounded by $C\cdot x^{-a}$ for $C = \frac{2\lvert 1-a\rvert}{a}$, and since $a > 1$, integrability over $[1,+\infty)$ follows. Near $0$, the Taylor expansion of the exponential function gives $f(x,\theta) - 1 = -e^{ia\theta} x^a + O(x^{2a})$, and thus the integrand is bounded near $0$.
Now let's integrate by parts, starting from the second expression.
\begin{align}
\int_0^{\infty} \frac{(1-a)x^{-a}}{ae^{ia\theta}}[f(x,\theta) - 1]\,dx
&= \frac{x^{1-a}}{ae^{ia\theta}}[f(x,\theta)-1]\biggr\rvert_0^{\infty} - \int_0^{\infty} \frac{x^{1-a}}{a e^{ia\theta}}\frac{d}{dx}[f(x,\theta)-1]\,dx \\
&= -\frac{1}{a e^{ia\theta}} \int_0^{\infty} x^{1-a}\bigl(-e^{ia\theta} a x^{a-1} f(x,\theta)\bigr)\,dx \\
&= \int_0^{\infty} f(x,\theta)\,dx \\
&= F(\theta).
\end{align}
