To each function $f \in E$, we associate the function $U ( f )$ defined for all $x > 0$ by $$U ( f ) ( x ) = \left\langle k _ { x } \mid f \right\rangle = \int _ { 0 } ^ { + \infty } \left( \mathrm { e } ^ { \min ( x , t ) } - 1 \right) f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$$ Show that for all functions $f \in E$ and for all $x > 0$, $$U ( f ) ( x ) = \int _ { 0 } ^ { x } \left( 1 - \mathrm { e } ^ { - t } \right) \frac { f ( t ) } { t } \mathrm {~d} t + \left( \mathrm { e } ^ { x } - 1 \right) \int _ { x } ^ { + \infty } f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$$
To each function $f \in E$, we associate the function $U ( f )$ defined for all $x > 0$ by
$$U ( f ) ( x ) = \left\langle k _ { x } \mid f \right\rangle = \int _ { 0 } ^ { + \infty } \left( \mathrm { e } ^ { \min ( x , t ) } - 1 \right) f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$$
Show that for all functions $f \in E$ and for all $x > 0$,
$$U ( f ) ( x ) = \int _ { 0 } ^ { x } \left( 1 - \mathrm { e } ^ { - t } \right) \frac { f ( t ) } { t } \mathrm {~d} t + \left( \mathrm { e } ^ { x } - 1 \right) \int _ { x } ^ { + \infty } f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$$