Evaluating $\int_0^{\infty} \exp(-e^{ia\theta}x^a) \,dx$ for $-\frac{\pi}{2a}<\theta<\frac{\pi}{2a}$

Question

Let $1

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.

Answer 1

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}