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I'm going to elaborate on Pietro Majer's answer a bit.

Suppose $S_1,S_2$ are independent random variables for which for all Borel sets $A\subseteq [0,\infty),$ \begin{align} \Pr(S_1\in A) & = \left. \int_A s^n e^{-s}\, \frac{ds} s \right/ \Gamma(n), \\[10pt] \Pr(S_2\in A) & = \left.\int_A s^m e^{-s}\, \frac{ds} s\right/ \Gamma(m), \end{align} where $n,m$ are positive real numbers. Then $$ \Pr(S_1+S_2\in A) = \left.\int_A s^{n+m} e^{-s} \, \frac{ds} s \right/ \Gamma(n+m). $$

A concrete instance: Suppose the waiting time $T$ until the next phone call arrives at a switchboard as a memoryless probability distribution: for $s,t\ge 0,$ one has $\Pr(T>s+t\mid T>s) = \Pr(T>t).$ That implies that for some $\mu>0$ and all $t>0,\,\,\,$ $\Pr(T>t) = e^{-t/\mu},$ and $\mu$ is the expected value of $T,$ i.e. $\mu$ is the average waiting time.

Then the distribution of the time $T_n$ until the arrival of the $n$th phone call after the present time is given by $$ \Pr(T_n \in A) = \frac 1 {\Gamma(n)} \int_A \left( \frac t \mu \right)^n e^{-t/\mu} \, \frac{(dt/\mu)} {(t/\mu)} = \frac 1 {\Gamma(n)} \int_{A/\mu} u^n e^{-u}\, \frac{du} u. $$

I'm going to elaborate on Pietro Majer's answer a bit.

Suppose $S_1,S_2$ are independent random variables for which for all Borel sets $A\subseteq [0,\infty),$ \begin{align} \Pr(S_1\in A) & = \left. \int_A s^n e^{-s}\, \frac{ds} s \right/ \Gamma(n), \\[10pt] \Pr(S_2\in A) & = \left.\int_A s^m e^{-s}\, \frac{ds} s\right/ \Gamma(m), \end{align} where $n,m$ are positive real numbers. Then $$ \Pr(S_1+S_2\in A) = \left.\int_A s^{n+m} e^{-s} \, \frac{ds} s \right/ \Gamma(n+m). $$

A concrete instance: Suppose the waiting time $T$ until the next phone call arrives at a switchboard as a memoryless probability distribution: for $s,t\ge 0,$ one has $\Pr(T>s+t\mid T>s) = \Pr(T>t).$ That implies that for some $\mu>0$ and all $t>0,\,\,\,$ $\Pr(T>t) = e^{-t/\mu},$ and $\mu$ is the expected value of $T,$ i.e. $\mu$ is the average waiting time.

Then the distribution of the time $T_n$ until the arrival of the $n$th phone call after the present time is given by \begin{align} \Pr(T_n \in A) & = \frac 1 {\Gamma(n)} \int_A \left( \frac t \mu \right)^n e^{-t/\mu} \, \frac{(dt/\mu)} {(t/\mu)} \\[10pt] & = \frac 1 {\Gamma(n)} \int_{A/\mu} u^n e^{-u}\, \frac{du} u. \end{align}

I'm going to elaborate on Pietro Majer's answer a bit.

Suppose $S_1,S_2$ are independent random variables for which for all Borel sets $A\subseteq [0,\infty),$ \begin{align} \Pr(S_1\in A) & = \left. \int_A s^n e^{-s}\, \frac{ds} s \right/ \Gamma(n), \\[10pt] \Pr(S_2\in A) & = \left.\int_A s^m e^{-s}\, \frac{ds} s\right/ \Gamma(m), \end{align} where $n,m$ are positive real numbers. Then $$ \Pr(S_1+S_2\in A) = \left.\int_A s^{n+m} e^{-s} \, \frac{ds} s \right/ \Gamma(n+m). $$

A concrete instance: Suppose the waiting time $T$ until the next phone call arrives at a switchboard as a memoryless probability distribution: for $s,t\ge 0,$ one has $\Pr(T>s+t\mid T>s) = \Pr(T>t).$ That implies that for some $\mu>0$ and all $t>0,\,\,\,$ $\Pr(T>t) = e^{-t/\mu},$ and $\mu$ is the expected value of $T,$ i.e. $\mu$ is the average waiting time.

Then the distribution of the time $T_n$ until the arrival of the $n$th phone call after the present time is given by $$ \Pr(T_n \in A) = \frac 1 {\Gamma(n)} \int_A \left( \frac t \mu \right)^n e^{-t/\mu} \, \frac{(dt/\mu)} {(t/\mu)} = \frac 1 {\Gamma(n)} \int_A u^n e^{-u}\, \frac{du} u. $$

I'm going to elaborate on Pietro Majer's answer a bit.

Suppose $S_1,S_2$ are independent random variables for which for all Borel sets $A\subseteq [0,\infty),$ \begin{align} \Pr(S_1\in A) & = \left. \int_A s^n e^{-s}\, \frac{ds} s \right/ \Gamma(n), \\[10pt] \Pr(S_2\in A) & = \left.\int_A s^m e^{-s}\, \frac{ds} s\right/ \Gamma(m), \end{align} where $n,m$ are positive real numbers. Then $$ \Pr(S_1+S_2\in A) = \left.\int_A s^{n+m} e^{-s} \, \frac{ds} s \right/ \Gamma(n+m). $$

A concrete instance: Suppose the waiting time $T$ until the next phone call arrives at a switchboard as a memoryless probability distribution: for $s,t\ge 0,$ one has $\Pr(T>s+t\mid T>s) = \Pr(T>t).$ That implies that for some $\mu>0$ and all $t>0,\,\,\,$ $\Pr(T>t) = e^{-t/\mu},$ and $\mu$ is the expected value of $T,$ i.e. $\mu$ is the average waiting time.

Then the distribution of the time $T_n$ until the arrival of the $n$th phone call after the present time is given by $$ \Pr(T_n \in A) = \frac 1 {\Gamma(n)} \int_A \left( \frac t \mu \right)^n e^{-t/\mu} \, \frac{(dt/\mu)} {(t/\mu)} = \frac 1 {\Gamma(n)} \int_{A/\mu} u^n e^{-u}\, \frac{du} u. $$

I'm going to elaborate on Pietro Majer's answer a bit.

Suppose the waiting time $T$ until the next phone call arrives at a switchboard as a memoryless probability distribution: for $s,t\ge 0,$ one has $\Pr(T>s+t\mid T>s) = \Pr(T>t).$ That implies that for some $\mu>0$ and all $t>0,\,\,\,$ $\Pr(T>t) = e^{-t/\mu},$ and $\mu$ is the expected value of $T,$ i.e. $\mu$ is the average waiting time.

Then the distribution of the time $T_n$ until the arrival of the $n$th phone call after the present time is given by $$ \Pr(T_n \in A) = \frac 1 {\Gamma(n)} \int_A \left( \frac t \mu \right)^n e^{-t/\mu} \, \frac{(dt/\mu)} {(t/\mu)} = \frac 1 {\Gamma(n)} \int_A u^n e^{-u}\, \frac{du} u. $$ More generally, suppose $S_1,S_2$ are independent random variables for which for all Borel sets $A\subseteq [0,\infty),$ \begin{align} \Pr(S_1\in A) & = \left. \int_A s^n e^{-s}\, \frac{ds} s \right/ \Gamma(n), \\[10pt] \Pr(S_2\in A) & = \left.\int_A s^m e^{-s}\, \frac{ds} s\right/ \Gamma(m), \end{align} where $n,m$ are positive real numbers (unlike in the switchboard scenario, these need not be integers). Then $$ \Pr(S_1+S_2\in A) = \left.\int_A s^{n+m} e^{-s} \, \frac{ds} s \right/ \Gamma(n+m). $$ Trivial though this may seem, I can't help wondering whether a person who finds it puzzling that $ds/s$ is used in the definition of $\Gamma$ rather than just using $ds$ may be unaware of it. The other possibility is that they know some reason unknown to me, to fail to be impressed by this.

I'm going to elaborate on Pietro Majer's answer a bit.

Suppose $S_1,S_2$ are independent random variables for which for all Borel sets $A\subseteq [0,\infty),$ \begin{align} \Pr(S_1\in A) & = \left. \int_A s^n e^{-s}\, \frac{ds} s \right/ \Gamma(n), \\[10pt] \Pr(S_2\in A) & = \left.\int_A s^m e^{-s}\, \frac{ds} s\right/ \Gamma(m), \end{align} where $n,m$ are positive real numbers. Then $$ \Pr(S_1+S_2\in A) = \left.\int_A s^{n+m} e^{-s} \, \frac{ds} s \right/ \Gamma(n+m). $$

A concrete instance: Suppose the waiting time $T$ until the next phone call arrives at a switchboard as a memoryless probability distribution: for $s,t\ge 0,$ one has $\Pr(T>s+t\mid T>s) = \Pr(T>t).$ That implies that for some $\mu>0$ and all $t>0,\,\,\,$ $\Pr(T>t) = e^{-t/\mu},$ and $\mu$ is the expected value of $T,$ i.e. $\mu$ is the average waiting time.

Then the distribution of the time $T_n$ until the arrival of the $n$th phone call after the present time is given by $$ \Pr(T_n \in A) = \frac 1 {\Gamma(n)} \int_A \left( \frac t \mu \right)^n e^{-t/\mu} \, \frac{(dt/\mu)} {(t/\mu)} = \frac 1 {\Gamma(n)} \int_A u^n e^{-u}\, \frac{du} u. $$