Two special cases. Let $d > 0$. Let $g \in \mathcal { C } ^ { 0 } ( [ 0 , d ] )$ such that $g ( 0 ) \neq 0$. (a) Show that $$\int _ { 0 } ^ { d } e ^ { - t x } g ( x ) d x \underset { t \rightarrow + \infty } { \sim } \frac { g ( 0 ) } { t }$$ Hint. For $t > 0$, one can construct a function $g _ { t }$ piecewise continuous on $[ 0 , + \infty [$, bounded, such that $$\int _ { 0 } ^ { d } e ^ { - t x } g ( x ) d x = \frac { 1 } { t } \int _ { 0 } ^ { + \infty } e ^ { - x } g _ { t } ( x ) d x$$ (b) Show similarly that $$\int _ { 0 } ^ { d } e ^ { - t x ^ { 2 } } g ( x ) d x \underset { t \rightarrow + \infty } { \sim } \frac { \sqrt { \pi } } { 2 } \frac { g ( 0 ) } { \sqrt { t } }$$ Hint. We recall the equality $\int _ { 0 } ^ { + \infty } e ^ { - x ^ { 2 } } d x = \frac { \sqrt { \pi } } { 2 }$.
Two special cases. Let $d > 0$. Let $g \in \mathcal { C } ^ { 0 } ( [ 0 , d ] )$ such that $g ( 0 ) \neq 0$.\\
(a) Show that
$$\int _ { 0 } ^ { d } e ^ { - t x } g ( x ) d x \underset { t \rightarrow + \infty } { \sim } \frac { g ( 0 ) } { t }$$
Hint. For $t > 0$, one can construct a function $g _ { t }$ piecewise continuous on $[ 0 , + \infty [$, bounded, such that
$$\int _ { 0 } ^ { d } e ^ { - t x } g ( x ) d x = \frac { 1 } { t } \int _ { 0 } ^ { + \infty } e ^ { - x } g _ { t } ( x ) d x$$
(b) Show similarly that
$$\int _ { 0 } ^ { d } e ^ { - t x ^ { 2 } } g ( x ) d x \underset { t \rightarrow + \infty } { \sim } \frac { \sqrt { \pi } } { 2 } \frac { g ( 0 ) } { \sqrt { t } }$$
Hint. We recall the equality $\int _ { 0 } ^ { + \infty } e ^ { - x ^ { 2 } } d x = \frac { \sqrt { \pi } } { 2 }$.