The mean mortality is
x ¯ = ∫ 0 ∞ t p ( t ) d t = 1 π ∫ 0 ∞ 2 t 1 + t 2 d t . {\displaystyle {\bar {x}}=\int _{0}^{\infty }tp(t)dt={\frac {1}{\pi }}\int _{0}^{\infty }{\frac {2t}{1+t^{2}}}dt.}
The right-most integral equals
lim r → ∞ 1 π ∫ 0 r 2 t 1 + t 2 d t . {\displaystyle \lim _{r\to \infty }{\frac {1}{\pi }}\int _{0}^{r}{\frac {2t}{1+t^{2}}}dt.}
And this proper integral can be evaluated by using the substitution u = 1 + t 2 {\displaystyle u=1+t^{2}} ,
∫ 2 t 1 + t 2 d t = ∫ d u u = ln | u | = ln ( 1 + t 2 ) {\displaystyle \int {\frac {2t}{1+t^{2}}}dt=\int {\frac {du}{u}}=\ln |u|=\ln(1+t^{2})}
∫ 0 r 2 t 1 + t 2 d t = ln ( 1 + r 2 ) − ln ( 1 + 0 2 ) = ln ( 1 + r 2 ) {\displaystyle \int _{0}^{r}{\frac {2t}{1+t^{2}}}dt=\ln(1+r^{2})-\ln(1+0^{2})=\ln(1+r^{2})}
where the last equality follows from the fact that ln ( 1 ) = 0 {\displaystyle \ln(1)=0} . It follows that
x ¯ = 1 π lim r → ∞ ln ( 1 + r 2 ) = ∞ {\displaystyle {\bar {x}}={\frac {1}{\pi }}\lim _{r\to \infty }\ln(1+r^{2})=\infty }
Hence,
x ¯ = ∞ {\displaystyle \color {blue}{\bar {x}}=\infty }
,
. It follows that
