Gaussian, gamma and beta product integrals

For $s>0$, the gamma integrand on $(0,1)$ is bounded by $t^{s-1}$, whose integral is $1\mathrm{/}s$. For $t\ge 1$, $t^{s-1}{e}^{-t\mathrm{/}2}$ is bounded, so $t^{s-1}{e}^{-t}\le C_s{e}^{-t\mathrm{/}2}$ and the tail converges. On a compact positive $s$-interval, the lower endpoint bounds the near-zero power and the upper endpoint bounds the exponential tail uniformly. Hence $\mathrm{\Gamma }(s)$ is finite and has controlled endpoint tails.

Let $I={\int }_{\mathbb{R}}{e}^{-x^2}dx$. By [F2], $I^2$ equals the nonnegative plane Gaussian integral, and [F3] evaluates that integral as $\pi$. Hence $I^2=\pi$.

By [F1], $I$ is finite and positive. The positive square root in step 1.2 is therefore $I=\sqrt{\pi }$.

Integration by parts on $[\epsilon ,R]$ gives ${\int }_{\epsilon }^R{t}^s{e}^{-t}dt=[-t^s{e}^{-t}{]}_{\epsilon }^R+s{\int }_{\epsilon }^R{t}^{s-1}{e}^{-t}dt$. The boundary terms vanish by step 1.1, so $\mathrm{\Gamma }(s+1)=s\mathrm{\Gamma }(s)$; direct integration gives $\mathrm{\Gamma }(1)=1$.

For $s,t>0$, the beta integrand is integrable near $0$ by comparison with $u^{s-1}$ and near $1$ by comparison with $(1-u{)}^{t-1}$. On each compact rectangle $[1\mathrm{/}j,j{]}^2$ the continuous product integrand $x^{s-1}{y}^{t-1}{e}^{-(x+y)}$ has integral equal to the product of its one-dimensional integrals, as follows by factoring the finite rectangular Riemann sums. These nested rectangles exhaust the positive quadrant and their integrals tend to $\mathrm{\Gamma }(s)\mathrm{\Gamma }(t)$ by step 1.1.

Substitution $t=x^2$ on compact half-intervals and passage to the convergent improper limits gives $\mathrm{\Gamma }(\frac{1}{2})=2{\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx=I=\sqrt{\pi }$ by steps 1.2 and 2.1.

On $A_j=[1\mathrm{/}j,j]\times [1\mathrm{/}j,1-1\mathrm{/}j]$ in coordinates $(r,u)$, the map $\mathrm{\Phi }(r,u)=(ru,r(1-u))$ is injective, has determinant $-r\mathrm{\ne }0$, and is a $C^1$ diffeomorphism on an open neighborhood. Its image is a compact Jordan set. The images exhaust the positive quadrant cofinally with the rectangles in step 2.3: each compact rectangle stays a positive distance from both axes and has bounded $x+y$, while each image is compact and stays away from the axes. Applying [F4] on $A_j$ and taking the increasing limits gives $\mathrm{\Gamma }(s)\mathrm{\Gamma }(t)={\int }_0^{\mathrm{\infty }}{r}^{s+t-1}{e}^{-r}drB(s,t)=\mathrm{\Gamma }(s+t)B(s,t)$. The limits are finite by step 1.1 and the endpoint bounds in step 2.3.

For integer $r\ge 0$, $x^{2r}{e}^{-a{x}^2\mathrm{/}2}$ is bounded on $\mathbb{R}$, so $x^{2r}{e}^{-a{x}^2}\le C{e}^{-a{x}^2\mathrm{/}2}$ and [F1] gives absolute convergence; odd moments converge absolutely and vanish by reflection. On compact intervals the substitution $y=\sqrt{a}x$ is covered by [F4], and its improper limit gives $I_0(a)=a^{-1\mathrm{/}2}{I}_0(1)=\sqrt{\pi \mathrm{/}a}$ by steps 1.2 and 2.1. The substitution $t=a{x}^2$ on both half-lines gives $I_r(a)=a^{-r-1\mathrm{/}2}\mathrm{\Gamma }(r+\frac{1}{2})$; finiteness follows from step 1.1, the half-integer starting value from step 3.1, and recurrence from step 2.2.

Let $C=[c,d]\subset (0,\mathrm{\infty })$ and set $Q_j=[1\mathrm{/}j,j{]}^2$. The event $A_j=Q_j\cap \{(x,y):x+y\in C\}$ is Jordan measurable. On each grid rectangle $I\times J\subset Q_j$, independence gives probability $P(X\in I)P(Y\in J)=({\int }_I{f}_s)({\int }_J{f}_t)$; because the marginal densities are continuous on $Q_j$, the finite rectangular Riemann sums factor and converge to the integral of $f_s(x)f_t(y)$ on that rectangle. The inner and outer grid unions for $A_j$ differ only in cells meeting its finitely many line-segment boundary pieces; their total area tends to zero, and $f_s{f}_t$ is bounded on $Q_j$. Thus $P((X,Y)\in A_j)={\int }_{A_j}{f}_s{f}_{t}dxdy$. As $Q_j$ increases to the positive quadrant, countable additivity gives probabilities increasing to $P(X+Y\in C)$, and the nonnegative improper integrals increase to the integral over $x,y>0$. The images of $C\times [1\mathrm{/}j,1-1\mathrm{/}j]$ under $(r,u)\mapsto (ru,r(1-u))$ increase to $A_C=\{x,y>0:x+y\in C\}$. On each such compact set the map is injective with absolute Jacobian $r$, so [F4] transforms the integral to $r^{s+t-1}{e}^{-r}{u}^{s-1}(1-u{)}^{t-1}\mathrm{/}(\mathrm{\Gamma }(s)\mathrm{\Gamma }(t))$. Its rectangular Riemann sums factor; as $j\to \mathrm{\infty }$ the $u$-integral tends to $B(s,t)$. By step 3.2, $P(X+Y\in C)={\int }_C{r}^{s+t-1}{e}^{-r}\mathrm{/}\mathrm{\Gamma }(s+t)dr$. A singleton C has probability zero by the same zero-area grid-boundary argument. The density integrates to one by step 1.1.

Fix the integer $r\ge 0$. Step 4.1 evaluated $I_r(a)=\mathrm{\Gamma }(r+\frac{1}{2})a^{-r-1\mathrm{/}2}$ for every $a>0$ using only convergent improper substitutions, and step 2.2 proved the Gamma recurrence by integration by parts; neither argument used parameter differentiation. By [F5], differentiation of this established explicit function gives $I_r^{\mathrm{\prime }}(a)=-(r+\frac{1}{2})\mathrm{\Gamma }(r+\frac{1}{2})a^{-r-3\mathrm{/}2}=-\mathrm{\Gamma }(r+\frac{3}{2})a^{-r-3\mathrm{/}2}=-I_{r+1}(a)\mathrm{.}$ The same positive-base power formula makes this derivative continuous on $(0,\mathrm{\infty })$, hence on a neighborhood of every compact $K\subset (0,\mathrm{\infty })$. This proves the full parameter clause without interchanging a derivative and an improper integral.

Let $\mu$ be the law of $X+Y$ on $(0,\mathrm{\infty })$ and let $\nu$ be the probability measure with the density in step 4.2. The collection $\mathcal{P}=\{\mathrm{\varnothing }\}\cup \{[a,b]:0<a\le b<\mathrm{\infty }\}$ is closed under intersections, allowing a singleton or the empty set. It generates the Borel sigma-algebra of the positive half-line: its members are closed there, and every relatively open interval is a countable union of closed subintervals with positive endpoints; unbounded intervals are obtained by an additional increasing union. Step 4.2 gives equality of $\mu$ and $\nu$ on every member of $\mathcal{P}$, including the singleton case. The sets $P_j=[1\mathrm{/}j,j]$ for integers $j\ge 1$ increase to $(0,\mathrm{\infty })$, lie in $\mathcal{P}$, and have common measure at most $1$. Thus [F6] proves $\mu =\nu$ on all Borel sets, which establishes the claimed gamma convolution law.