prove $ F(x)=\int_0^\infty {\sin(tx)\over(t+1)\sqrt t} dt \in C^\infty(\mathbb R^*) $

Question

prove that :

$$ F(x)=\int_0^\infty {\sin(tx)\over(t+1)\sqrt t} \, dt \in C^\infty(\mathbb R^*) $$

i end up proving that $F(x)\in C^ \infty(\mathbb R^{*+})$ not $\mathbb R^*$ , and i studied the case with :

$F(x)= \sqrt x G(x)$ where $G(x)=\int_0^\infty {\sin(t)\over(t+x)\sqrt t} \, dt$ ($\sqrt x , G(x) \in C^\infty(R^{*+}))$

is my answer wrong ? , wich is the correct interval for $C^\infty$ ?

Answer 1

Assume $x > 0$ and write

$$ F(x) = \sqrt{x} G(x), \quad \text{where} \quad G(x) = \int_{0}^{\infty} \frac{\sin t}{(t + x)\sqrt{t}} \, dt. $$

The proof is divided into two steps, where we first improve the decaying behavior of the integral representation of $G(x)$ and then we prove that $G(x)$ is infinitely differentiable.

Step 1. We introduce the function

$$ A(x) = \int_{0}^{x} \frac{\sin t}{\sqrt{t}} \, dt = \frac{1-\cos x}{\sqrt{x}} + \frac{1}{2} \int_{0}^{x} \frac{1-\cos t}{t^{3/2}} \, dt. $$

Clearly $A(x)$ converges as $x \to \infty$. In particular, $A(x)$ is uniformly bounded on $[0, \infty)$. Then we have

\begin{align*} G(x) = \int_{0}^{\infty} \frac{\sin t}{(t + x)\sqrt{t}} \, dt &= \int_{0}^{\infty} \frac{A'(t)}{t + x} \, dt = \left[ \frac{A(t)}{t+x} \right]_{0}^{\infty} + \int_{0}^{\infty} \frac{A(t)}{(t + x)^{2}} \, dt \\ &= \int_{0}^{\infty} \frac{A(t)}{(t + x)^{2}} \, dt. \end{align*}

Step 2. Now we introduce the notation

$$ G_{k}(x) = (-1)^{k} (k+1)! \int_{0}^{\infty} \frac{A(t)}{(t + x)^{k+2}} \, dt, $$

so that $G(x) = G_{0}(x)$. We claim that $G_{k}'(x) = G_{k+1}(x)$. Once this is proved, it follows from the induction that $G(x)$ is infinitely differentiable. Indeed, by the Mean Value Theorem,

$$ \frac{1}{h} \left( \frac{1}{(t + x + h)^{k+2}} - \frac{1}{(t + x)^{k+2}} \right) = -\frac{k+2}{(t + x + \theta h)^{k+3}} $$

for some $0 \leq \theta \leq 1$, with the choice of $\theta$ depending on $t$, $x$ and $h$. Thus for $h$ satisfying $0 < |h| \ll x$, we have

\begin{align*} &\left| \frac{G_{k}(x+h) - G_{k}(x)}{h} - G_{k+1}(x) \right| \\ &= (k+1)! \left| \int_{0}^{\infty} A(t) \left\{ \frac{1}{h} \left( \frac{1}{(t + x + h)^{k+2}} - \frac{1}{(t + x)^{k+2}} \right) + \frac{k+2}{(t+x)^{k+3}} \right\} \, dt \right| \\ &= (k+2)! \left| \int_{0}^{\infty} A(t) \left( \frac{1}{(t+x)^{k+3}}-\frac{1}{(t + x + \theta h)^{k+3}} \right) \, dt \right| \end{align*}

Since the integrand of the last integral is dominated by the integrable function

$$ |A(t)| \left( \frac{1}{(t+x)^{k+3}} + \frac{1}{(t + x - |h|)^{k+3}} \right), $$

we can apply the dominated convergence theorem to conclude that

$$ \lim_{h\to 0} \frac{G_{k}(x+h) - G_{k}(x)}{h} = G_{k+1}(x) $$

as desired. Therefore the proof is complete.

Answer 2

One would think that proving $F(x)\in C^ \infty(\mathbb R^{*+})$ would be sufficient since $F(-x) = -F(x)$.