This question comes from here (Does this integral have a closed form or asymptotic expansion? $\int_0^\infty \frac{\sin(\beta u)}{1+u^\alpha} du$) and has haunted me a while.
Some observations
stationary phase does not work (at least I can't figure it out), because the phase function $x$ does not have a stationary point to expand around.
at $x = 0$, there is no oscillation, but integrand is $0$. So the integration is from regions away from $x=0$, but those regions are "packed" with oscillations as $\lambda \rightarrow \infty$.
- $\int_0^1 1/x dx$ itself does not converge, but with $\sin(\lambda x)$ multiplied, we have the celebrated $sinc$ function. What's more, the integration $I(\lambda) = \pi/2$: a small, finite number independent of $\lambda$.
- Numerical results suggest the $1/\lambda$ behavior applies to many cases - details below.
Numerical experiment was done on 6 functions (reminder, $I(\lambda) = \int_0^1 f(x) sin(\lambda x)$). The $\textsf{label as}$ corresponds to the legend in the figure.
$$ f(x) = \frac{1}{\sqrt{x}} \quad \textsf{label as "s"} $$
$$ f(x) = \frac{1}{1+x} \quad \textsf{label as "1"} $$
$$ f(x) = \frac{1}{1+x^2} \quad \textsf{label as "2"} $$ $$ f(x) = \frac{1}{1+x^3} \quad \textsf{label as "3"} $$ $$ f(x) = \frac{1}{1+x^4} \quad \textsf{label as "4"} $$
$$ f(x) = \frac{1}{e^x} \quad \textsf{label as "e", for this case we have} \quad I = \frac{\lambda}{1+\lambda^2} $$
The figure is below (the two solid lines are for references). It can be seen 5 function converges to the $1/\lambda$ line, the behavior for the $f(x) = 1/\sqrt{x}$ is different, but we can focus on it later. Any suggestion?
