Consider the following double integral depending on a parameter $x \in \mathbb{R}$: $$ I(x) := \int_{x}^{\infty} f ( z) e^{-(z-x)} \int_{0}^{\infty} \frac{e^{-(z-x)t}}{10+\ln^2{t}} \ dt \ dz, $$ where $f(z)$ is continuous and vanishes at infinity, meaning that $$ \lim_{ | z | \rightarrow \infty} f ( z ) = 0 $$ It is known that under this condition $I(x)$ is also continuous and vanishes at infinity.
I am trying to find out whether $I(x)$ is necessarily differentiable in $x$ without any additional assumptions on $f$.
If we denote by $$ J(x,z) := \int_{0}^{\infty} \frac{e^{-(z-x)t}}{10+\ln^2{t}} \ dt $$ then a naive "differentiation" of $J(x,z)$ w.r.t. $x$ may lead to $$ \frac{d}{dx} J(x,z) \overset{?}{=} \int_{ 0 }^{ \infty} \frac{te^{-(z-x)t}}{10 + \ln^2 t} \ dt, $$ which does not converge if we set $z=x$. Can this imply that $I(x)$ is not necessarily differentiable?
One could also try a change of variables through $y :=z-x$ and get $$ I( x ) := \int_{0}^{\infty} f( y + x ) e^{-y} \int_{0}^{\infty} \frac{e^{-yt}}{10+\ln^2{t}} \ dt \ dy, $$ and it seems that this would hardly lead anywhere, since we know nothing about the differentiability of $f$ w.r.t. $x$.
Another thing, which may or may not be helpful, is that $$ \int_{0}^{\infty} \int_{0}^{\infty} \frac{e^{-yt}}{10+\ln^2{t}} \ \color\red{dy} \ dt = \frac{ \pi }{ \sqrt{10}}, $$ as computed here by wolframalpha.
In summary, knowing that $I(x)$ is continuous and vanishing at infinity, are there any criteria which can be used to show that $I(x)$ is necessarily differentiable without additional assumptions on $f$, and, if so, how does the derivative look like? Alternatively, is there some $f$ which is continuous and vanishing at infinity, but such that $I(x)$ is not differentiable?
NOT AN ANSWER: Some (badly done) numerics and a conjecture
For the integral $I(x) := \int_{x}^{\infty} f ( z) e^{-(z-x)} \int_{0}^{\infty} \frac{e^{-(z-x)t}}{10+\ln^2{t}} \ dt \ dz$, the complete naive differentiation is
$$I'(x) =\\ -f ( x) \int_{0}^{\infty} \frac{1}{10+\ln^2{t}} dt + \int_{x}^{\infty} f ( z) e^{-(z-x)} \left(\int_{0}^{\infty} \frac{e^{-(z-x)t}}{10+\ln^2{t}} dt + \int_0^\infty\frac{te^{-(z-x)t}}{10+\ln^2{t}} dt\right)dz$$ which looks bad. So lets look for a counterexample.
The integrand is complicated in $t$, simpler in $z$. If $f$ is non-negative, Tonelli's theorem guarantees that $$I(x) = \int_0^\infty \frac{1}{10+\ln^2 t} \int_x^\infty f(z) e^{-(z-x)(t+1)} dz dt $$
Discovery phase 1
We could try the following continuous function on $[0,\infty)$ vanishing at infinity: $$ f(z) = (1-z)_+ = \max(1-z,0) = \mathbb1_{[0,1]}(1-z).$$ In fact, this $f$ is supported on $[0,1]$, so $I(x)=0$ for all $x\ge 1$. For $x<1$, we have $$ \int_{x}^1 (1-z) e^{-(z-x)(t+1)} dz = -\frac{x}{t+1} + \frac{t}{(t+1)^2} + \frac{e^{- (t + 1)(1-x)}}{(t+1)^2}$$ so $$I(x) = c_0-c_1 x+ \int_0^\infty \frac{e^{- (t + 1)(1-x)}}{(10+\ln^2 t)(t+1)^2}dt,\\ c_0 = \int_0^\infty \frac{t}{(10+\ln^2 t)(t+1)^2}dt<\infty,\\ c_1 =\int_0^\infty \frac{1}{(10+\ln^2 t)(t+1)} dt<\infty. $$ ($c_0$ isn't the $y-$intercept.) Each of these terms is differentiable away from $1$. At $x=1$ Numerically, it seems that the derivative is $C^\alpha$, maybe $\alpha\approx 0.2$...
Discovery phase 2
That worked better than expected; what if $f(z) = \mathbb1_{z\in[0,1]}$ vanishes at infinity but is not continuous? Then the inner integral is again $0$ for $x\ge 1$ and for $x<1$: $$ \int_x^1 e^{(z-x)(t+1)}dz= - \frac{e^{-\left(1 - x\right) \left( t + 1\right)}}{t + 1} + \frac{1}{t + 1}$$ Hm, following similar calculations, $I(x)$ now looks $C^\alpha$, and again $\alpha \approx 0.2$.
Conjecture.
Since the regularity only improves by $\approx 0.2$, it seems sensible that $(1-\sqrt z)_+$ would give rise to a $C^{\approx 0.7}$ function, which would be a counterexample. The important integral to be calculated is now (for $x<1$) $$I_{1/2}(x)= \int_x^1 \sqrt z e^{-(t+1)(z-x)}dz$$ I tried a CAS, which spits this out after some simplification (clearly using a substitution $w = z^2$): $$ I_{1/2}(x,t)= \frac{\left(\sqrt{x} \sqrt{t + 1} e^{t + 1} - \sqrt{t + 1} e^{t x + x} - \frac{\sqrt{\pi} e^{t x + t + x + 1} \operatorname{erf}{\left(\sqrt{x} \sqrt{t + 1} \right)}}{2} + \frac{\sqrt{\pi} e^{t x + t + x + 1} \operatorname{erf}{\left(\sqrt{t + 1} \right)}}{2}\right) e^{- t - 1}}{\left(t + 1\right)^{\frac{3}{2}}}$$ But I haven't attempted to compute estimates or numerics for the iterated integral $$ I(x) = \int_0^\infty \frac{I_{1/2}(x,t)}{10+\ln^2 t} dt=\dots ?$$