From now on, $f$ denotes an infinitely differentiable function from $[ 0,1 ]$ to $\mathbb { R }$. We assume that there exists a unique point $x _ { 0 } \in \left[ 0,1 \left[ \right. \right.$ where $f ^ { \prime }$ vanishes. We also assume that $f ^ { \prime \prime } \left( x _ { 0 } \right) > 0$. We are also given an infinitely differentiable function $g : [ 0,1 ] \rightarrow \mathbb { R }$. Show that we have, as $t \rightarrow + \infty$, $$\int _ { x _ { 0 } } ^ { 1 } g ( x ) \sin ( t f ( x ) ) \mathrm { d } x = g \left( x _ { 0 } \right) \int _ { x _ { 0 } } ^ { 1 } \sin ( t f ( x ) ) \mathrm { d } x + O \left( \frac { 1 } { t } \right)$$
From now on, $f$ denotes an infinitely differentiable function from $[ 0,1 ]$ to $\mathbb { R }$. We assume that there exists a unique point $x _ { 0 } \in \left[ 0,1 \left[ \right. \right.$ where $f ^ { \prime }$ vanishes. We also assume that $f ^ { \prime \prime } \left( x _ { 0 } \right) > 0$. We are also given an infinitely differentiable function $g : [ 0,1 ] \rightarrow \mathbb { R }$.
Show that we have, as $t \rightarrow + \infty$,
$$\int _ { x _ { 0 } } ^ { 1 } g ( x ) \sin ( t f ( x ) ) \mathrm { d } x = g \left( x _ { 0 } \right) \int _ { x _ { 0 } } ^ { 1 } \sin ( t f ( x ) ) \mathrm { d } x + O \left( \frac { 1 } { t } \right)$$