Is this function concave? Let
$$h(u):=u^3 \left|\int_u^\infty \frac{e^{-i t}}{t^3} \, dt\right|$$
for $u>0$. Is the function $h$ concave on $(0,\infty)$?
(For context, see Proposition 4.4.4 and formula (4.4.21) in this paper or, equivalently, Proposition 4.4 and formula (4.21) in the corresponding arXiv preprint. In particular, according to that proposition, $h(v^{1/6})^2$ is concave in $v>0$.)
Here is the graph $\{(u,h(u)):0<u<16\}$:

 A: Strange that I missed this one. On the other hand I haven't visited MO often lately. Anyway, here is the solution (It is always tempting to renew our old discussion on what problems a minimally intelligent AI should be able to solve, but I'll abstain from it for now).
Yes, it is concave.
Step 1: Convenient representation
$$
u^3\left|\int_u^\infty \frac{e^{is}}{s^3}\,ds\right|
=u^3\left|\int_1^\infty \frac{e^{ius}}{(us)^3}\,d(us)\right|
\\
=u\left|\int_1^\infty \frac{e^{ius}}{s^3}\,ds\right|
=u\left|\int_0^\infty \frac{e^{ius}}{(s+1)^3}\,ds\right|\,.
$$
Now
$$
\frac1{(s+1)^3}=c\int_0^\infty t^2e^{-(s+1)t}\,dt\,,
$$
so we need to consider
$$
u\left|\int_0^\infty t^2e^{-t}\,dt\int_0^\infty{e^{-s(t-iu)}}\,ds\right|
=u\left|\int_0^\infty \frac1{t-iu}t^2e^{-t}\,dt\right|=u|I(u)|\,.
$$
We have
$$
|I(u)|^2=I(u)\overline{I(u)}=\iint_{(0,\infty)^2}\Re\left[\frac{1}{t-iu}\frac{1}{\tau+iu}\right]t^2\tau^2e^{-(t+\tau)}\,dt d\tau
\\
=\iint_{(0,\infty)^2}\frac{u^2+t\tau}{(u^2+t^2)(u^2+\tau^2)}t^2\tau^2e^{-(t+\tau)}\,dt d\tau
\\
=\iint_{(0,\infty)^2}\frac 1{t+\tau}\left[\frac{t}{u^2+t^2}+\frac{\tau}{u^2+\tau^2}\right]t^2\tau^2e^{-(t+\tau)}\,dt d\tau
\\
=2\int_0^\infty\frac{t}{u^2+t^2} t^2\Phi(t)\,dt=2J(u)\,,
$$
where
$$
\Phi(t)=e^{-t}\int_0^\infty\frac{\tau^2}{t+\tau}e^{-\tau}\,d\tau\,.
$$
Step 2: Differentiation
$$
\frac{d}{du}[u\sqrt{J(u)}]=\frac 1{\sqrt{J(u)}}[J(u)+\tfrac 12uJ'(u)]\,.
$$
Also
$$
\frac 12uJ'(u)=-\int_0^\infty\frac{tu^2}{(u^2+t^2)^2} t^2\Phi(t)\,dt\,,
$$
so
$$
J(u)+\frac 12uJ'(u)=\int_0^\infty\frac{t^3}{(u^2+t^2)^2} t^2\Phi(t)\,dt\,.
$$
Step 3: Reduction to the ratio of integrals
We need to show that the last integral, when divided by $\sqrt{J(u)}$, becomes a decreasing function of $u$. Since $J(u)$ is clearly a decreasing function of $u$, it will suffice to prove the decreasing property for the integral divided by $J(u)$ instead, i.e., to show that
$$
\frac{\int_0^\infty\frac{t^5}{(u^2+t^2)^2}\Phi(t)\,dt}{\int_0^\infty\frac{t^3}{u^2+t^2}\Phi(t)\,dt}
$$
is a decreasing function of $u$.
Denoting $T=t^2$, $U=u^2$, $\Psi(T)=\Phi(\sqrt T)$, we can rewrite this ratio as
$$
\frac{\int_0^\infty\frac{T^2}{(U+T)^2}\Psi(T)\,dT}{\int_0^\infty\frac{T}{U+T}\Psi(T)\,dT}\,.
$$
Integrating by parts, we get
$$
\int_0^\infty\frac{T^2\Psi(T)}{(U+T)^2}\,dT=2\int_0^\infty\frac{T}{U+T}\Psi(T)\,dT+\int_0^\infty\frac{T}{U+T}T\Psi'(T)\,dT\,.
$$
Thus we just need to prove that
$$
\frac{\int_0^\infty\frac{T}{U+T}T\Psi'(T)\,dT}{\int_0^\infty\frac{T}{U+T}\Psi(T)\,dT}
$$
is decreasing in $U$.
Step 4: A standard Lemma
Let $\mu$ be any positive measure and let $f:(0,+\infty)\to\mathbb R$ be any decreasing function. Then
$$
\frac{\int_0^\infty\frac{T}{U+T}f(T)\,d\mu(T)}{\int_0^\infty\frac{T}{U+T}\,d\mu(T)}
$$
is decreasing in $U$.
Indeed, it is enough to consider $f=\chi_{(0,a)}-C$ (every decreasing function is an integral linear combination of functions of this kind with non-negative coefficients). $-C$, clearly, does not matter, and
$$
\frac{\int_0^a\frac{T}{U+T}\,d\mu(T)}{\int_0^\infty\frac{T}{U+T}\,d\mu(T)}=\frac{1}{1+\frac{\int_a^\infty\frac{T}{U+T}\,d\mu(T)}{\int_0^a\frac{T}{U+T}\,d\mu(T)}}\,
$$
so we just need to check that
$$
\frac{\int_0^a\frac{T}{U+T}\,d\mu(T)}{\int_a^\infty\frac{T}{U+T}\,d\mu(T)}
$$
is decreasing in $U$, which is quite obvious because when we replace $U$ by $U_1>U$, the integrand in the numerator is multiplied at every point by at most $\frac{U+a}{U_1+a}$ and the integrand in the denominator by at least that amount ($T\mapsto \frac{U+T}{U_1+T}$ is increasing in $T>0$).
We now have the
Moral
If $\frac{T\Psi'(T)}{\Psi(T)}$ is a decreasing function of $T$, then
$$
\frac{\int_0^\infty\frac{T^2}{(U+T)^2}\Psi(T)\,dT}{\int_0^\infty\frac{T}{U+T}\Psi(T)\,dT}
$$
is a decreasing function of $U$.
Step 5: The decreasing property of $\frac{T\Psi'(T)}{\Psi(T)}$
We have
$$
\frac{T\Psi'(T)}{\Psi(T)}=\frac12\frac{\sqrt T\Phi'(\sqrt T)}{\Phi(\sqrt T)}
$$
so it suffices to check that $\frac{t\Phi'(t)}{\Phi(t)}$ decreases in $t>0$.
Now,
$$
\frac{t\Phi'(t)}{\Phi(t)}=-t-\frac{\int_0^\infty\frac{\tau^2t}{(t+\tau)^2}e^{-\tau}\,d\tau}{\int_0^\infty\frac{\tau^2}{t+\tau}e^{-\tau}\,d\tau}=-t-1+\frac{\int_0^\infty\frac{\tau^2}{(t+\tau)^2}\tau e^{-\tau}\,d\tau}{\int_0^\infty\frac{\tau}{t+\tau}\tau e^{-\tau}\,d\tau}
$$
and it remains to show that the last term is a decreasing function of $t$. By the Moral (with $t$ instead of $U$ and $\tau$ instead of $T$), we just need to check that
$$
\frac{\tau\frac{d}{d\tau}(\tau e^{-\tau})}{\tau e^{-\tau}}=1-\tau
$$
is a decreasing function of $\tau$, which is sort of obvious.
The End
