Let $f : \mathbb { R } _ { + } \rightarrow \mathbb { R }$, be a piecewise continuous function, strictly positive and integrable. For all $x > 0$, we define $$g ( x ) = \frac { 1 } { x } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t \quad \text { and } \quad h ( x ) = \frac { 1 } { x } g ( x ) = \frac { 1 } { x ^ { 2 } } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t .$$ Deduce that the integral $\int _ { 0 } ^ { + \infty } h ( x ) \mathrm { d } x$ converges and that $$\int _ { 0 } ^ { + \infty } f ( x ) \mathrm { d } x = \int _ { 0 } ^ { + \infty } h ( x ) \mathrm { d } x .$$ You may use integration by parts.
Let $f : \mathbb { R } _ { + } \rightarrow \mathbb { R }$, be a piecewise continuous function, strictly positive and integrable. For all $x > 0$, we define
$$g ( x ) = \frac { 1 } { x } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t \quad \text { and } \quad h ( x ) = \frac { 1 } { x } g ( x ) = \frac { 1 } { x ^ { 2 } } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t .$$
Deduce that the integral $\int _ { 0 } ^ { + \infty } h ( x ) \mathrm { d } x$ converges and that
$$\int _ { 0 } ^ { + \infty } f ( x ) \mathrm { d } x = \int _ { 0 } ^ { + \infty } h ( x ) \mathrm { d } x .$$
You may use integration by parts.