How that $\int_0^x\int_0^{\infty}\exp\left(-\frac{t^b+s^a}{2}\right) dtds \leq \int_0^x\int_{0}^{+\infty}\exp\left(-\frac{t^a+s^b}{2}\right)dtds$ How to show the following inequality
\begin{align}
\int_{0}^{x}\int_{0}^{+\infty}\exp\left(-\frac{t^b+s^a}{2}\right)\,dt\,ds \leq \int_{0}^{x}\int_{0}^{+\infty}\exp\left(-\frac{t^a+s^b}{2}\right)\,dt\,ds
\end{align}
where $0<a<b$ and all $x>0$.
I learned This is the inequality from this question. However, I was not able to follow the proof.  
 A: As reformuated by Jack D'Aurizio, this boils down to showing that $$\int_0^x e^{-t^a/2}\,dt\space  \int_0^\infty e^{-t^b/2} \,dt\le \int_0^x e^{-t^b/2}\,dt\space  \int_0^\infty e^{-t^a/2}\,dt,$$ equivalently $A(x) B(\infty) \le B(x) A(\infty)$ or $$ R(x) = \frac{A(x)}{B(x)} \le \frac{A(\infty)}{B(\infty)} = R(\infty) $$ where $A(x)=\int_0^x e^{-t^a/2}\,dt$ and $B(x) = \int_0^x e^{-t^b/2}\,dt.$ Here $0<a<b$ and $0<x$.
Here is a sketch of a proof; details will be filled in below.  I have tried to number the steps for future reference.  I am sure a more direct or thematically satisfying argument exists.
Step 1: For $x\ge 1$ it suffices to check that $\alpha(x)/\beta(x)\ge R(\infty),$ where $\alpha(x) = \int_x^\infty e^{-t^a/2}\,dt$ and $\beta(x)=\int_x^\infty e^{-t^b/2}\,dt.$ Step 2: A simple argument (which is what D'Aurizio probably had in mind originally) shows that $\alpha(x)/\beta(x)$ increases on $[1,\infty)$, leaving the case $\alpha(1)/\beta(1)\ge R(\infty)$ (that is, that $R(1)\le R(\infty)$ to be checked by a side argument given in Step 3.
Step 4: For the range $[0,1]$ we check first that $R(0) = a/b\le R(\infty),$ and then in Step 5 analyse the critial points of $R$ in $[0,1].$ Let $r = (a/b)^{1/(b-a)},$ clearly $r \in [0,1].$ If $x\le r $ and $R'(x) = 0,$ then $R''(x)\le0$ but if $x\ge r$, and $R'(x)=0$ then $R''(x)\ge 0$.  Step 6: Side calculations show that $R$ is decreasing near $0$ and increasing at $1$; these imply that $R$ has exactly one local minimum and no interior local maxima in $[0,1],$ and hence $\max_{[0,1]} R(x) = \max(R(0), R(1))$.  Which is already covered in Steps 3 and 4.
Only step 3 gives me pleasure.
Here are the details.
STEP 1:
Since $A(x)+\alpha(x)=A(\infty),$ etc, so $R(x)\le R(\infty)$ translates into
$\alpha(x)/\beta(x)\ge R(\infty).$
STEP 2:
For $1\le x\le t$, the inequality $x^a-t^a \ge x^b - t^b$ holds.  It obviously holds when $t=x$ and since $(\partial/\partial t) (t^b-t^a)\ge 0,$ it holds for $t\ge x$ as well.  The sign of $(\alpha(x)/\beta(x))'$ is the same as the sign of $$\alpha'(x)\beta(x)-\alpha(x)\beta'(x) = \int_x^\infty ( -e^{-x^a/2}e^{-t^b/2} + e^{-x^b/2}e^{-t^a/2})\,dt.$$  (Since $x$ is the lower limit of integration, $\alpha'<0$, etc.) On multiplying by $\exp(x^a/2+x^b/2)$ the integrand is seen to have the same sign as $ e^{x^a/2}e^{-t^a/2} - e^{x^b/2}e^{-t^b/2},$ which is evidently non-negative.  Hence $\alpha/\beta$ is increaing on $[1,\infty). $
STEP 3:
Here we write $\alpha(1) = 2^{1/a}\Gamma(1/a,1/2),$ $\beta(1) = 2^{1/b}\Gamma(1/b,1/2),$ and $R(\infty) = 2^{1/a-1/b} \Gamma(1/a)/\Gamma(1/b),$ and the reqirement that $\alpha(1)/\beta(1)\ge R(\infty)$ follows from the observation that $\Gamma(s,1/2)/\Gamma(s)$ is increasing in $s$.  This follows from the convolution semigroup property of the family of Gamma distribution functions: if $G_s$ and $G_h$ are   independent, Gamma distributed random variables, with parameters $s>0$ and $h>0$ repectively, then  $G_s+G_h$ is Gamma distributed with parameter $t=s+h$.  The application of this to our case is that $P(G_s>1/2) = \Gamma(s,1/2)/\Gamma(s)$.  Since the event $[G_s > 1/2]$ is a subset of $[G_t>1/2] = [G_s+G_h>1/2]$, we have $ \Gamma(s,1/2)/\Gamma(s) =P(G_s>1/2)\le P(G_t>1/2) = \Gamma(t,1/2)/\Gamma(t).$ Take $s=1/b$ and $t=1/a$ to finish the argument that $\alpha(1)/\beta(1)\ge R(\infty).$
STEP 4:
Here we write $A(x) = 2^{1/a} \gamma(1/a,x^a/2),$ etc, and appeal to A\&S 6.5.29 to write $A(x) = a x (1 - x^a/(2(a+1)) + o(x^a))$ as $x\to 0$, etc, so $R(0) = a/b$ and (since $x^b = o(x^a)$, $R$ is decreasing near $0$.  We now show $a/b \le R(\infty) = 2^{1/a - 1/b}\Gamma(1/a) /\Gamma(1/b) $ by observing that the function $s\mapsto s 2^s \Gamma(s) = 2^s \Gamma(s+1)$ is increasing.  Its logarithmic derivative is $ \log 2 + \psi(1+s)$.  Since $\psi$ is increasing, it suffices to check this at $s=0$, viz., that $\log 2 + \psi(1) \ge 0$.  But $\psi(1) = -\gamma,$ so $\log 2 + \psi(1)\approx 0.6931 - 0.5772 > 0.$ 
STEP 5:
At a critical point of $R$, we have $A'B=AB'$ and $R'' = (A'' B - A B'')/B^4.$ Hence, at a critical point of $R$, the sign of $R''$ is the same as the sign of $A'' /A' - B''/B'.$  This last expression is $bx^{b-1}/2 - ax^{a-1}/2$, which is negative if $x<r$ and positive if $r>r$, where $r=(a/b)^{1/(b-a)}.$
STEP 6: We saw at the beginning of Step 4 that $R$ decreases near $0$.  The sign of $R'(1)$ is that of $A'(1)B(1)-A(1)B'(1)$.  But $A'(1) = B'(1) = \exp(-1/2)$, so the sign of $R'(1)$ is that of $B(1)-A(1) = \int_0^1 \exp(-t^b/2)-\exp(-t^a/2)\,dt.$  Since $t^a> t^b$ on $(0,1)$, the integrand is positive a.e. and hence $R'(1)>0$.
