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Put $h(t,x)=e^{-x^2-t^2/x^2}$. For every $t\ge 0$, $0<h(t,x)\le e^{-x^2}$, and the upper bound is a Lebesgue integrable function by part (b)(i). Since $h(t,x)\to e^{-x^2}$ for each $x>0$ as $t\downarrow 0$, the dominated convergence theorem gives right continuity at zero andTo apply dominated differentiation under the integral sign, fix $t_0>0$ and restrict $t$ to $[t_0/2,3t_0/2]$. ThenFor $x\ge 1$, $D(x)\le 3t_0{e}^{-x^2}$. For $0<x<1$, writing $a=t_0^2/4>0$, the substitution rule $u=1/x$ givesThus $D$ is integrable. The mean value theorem bounds nearby difference quotients by $D$, so the dominated convergence theorem justifies the derivative at $t_0$. As $t_0$ was arbitrary,Now make the change of variables $u=t/x$. The reversal of limits cancels the negative differential and givesConsequently $F^{\prime }(t)=-2F(t)$. This also follows in sign from the fact that the positive integrand decreases strictly as $t$ increases: the printed equation with a positive sign cannot hold. Solving the linear differential equation, or differentiating $e^{2t}F(t)$, and using right continuity at zero givesThis is a half-line instance of the Gaussian integral with an inverse-square term.

The integrand is even and decays at both infinity and zero, since $a,b>0$. On $x>0$ set $u=\sqrt{a}x-\sqrt{b}/x$, a strictly increasing bijection onto the whole real line. ThenUsing evenness for the negative half-axis givesThe second summand is an integrable odd function, so its integral vanishes. The remaining Gaussian integral givesFor the second integral use $x=1/t$ separately on each half-axis. The factor $x^{-2}dx$ becomes $-dt$, and reversing the transformed limits yields $J(a,b)=I(b,a)$. ThusThese identities are instances of the Gaussian integral with an inverse-square term. Splitting at zero before either change of variables avoids treating a discontinuous transformation as a single bijection.