A simple example is given by $$ f(t,x)=\cases{\exp(-(x-t^{-2})^2),&$t\neq 0$,\\0,&$t=0.$} $$ For each fixed $t\neq 0$, $\int f(t,x)\,dx$ is a Gaussian integral equal to $\sqrt{\pi}$, while for $t=0$, the integral equals to zero. Therefore, $F(t)$ is not even continuous at $0$, let alone differentiable.

On the other hand, at $t\neq 0$, any partial derivative of $f$ has a form $P(t^{-1},x)\exp(-(x-t^{-2})^2)$ for some polynomial $P$. This tends to zero as $t\to 0$, uniformly in $x$ in compacts. Therefore, $f$ is smooth with all partial derivatives vanishing on at $\mathbb{R}$. Also, any expression of this form, and in particular $\partial_t f(t,x)$, is clearly integrable over $x$ for all $t$.

Update: The above example may feel like cheating, since in essence the discontinuity of $f$ has just been moved to infinity. Here's a better one in that respect:

$$ f(t,x)=\cases{\mathrm{sign}(t)|t|^\frac{1}{2}\exp\left(-\frac{(x-t)^2}{|t|}\right),&$t\neq 0$,\\0,&$t=0.$} $$ This has $F(t)=\sqrt{\pi}t$ while still $\partial_t f(t,x)|_{t=0}\equiv 0$. It has an additional feature that $$ \int_I\partial_t f(t,x)\,dx\neq\partial_t \int_I f(t,x)\,dx $$ for any non-empty interval $I$ containing the origin. It has smooth restrictions onto vertical and horisontal lines, but it is not overall smooth or even continuously differentiable: indeed, if it were, it would have its derivatives bounded on compacts, thus $\int_I$ and $\partial_t$ would commute for any compact interval $I$.

A simple example is given by $$ f(t,x)=\cases{\exp(-(x-t^{-2})^2),&$t\neq 0$,\\0,&$t=0.$} $$ For each fixed $t\neq 0$, $\int f(t,x)\,dx$ is a Gaussian integral equal to $\sqrt{\pi}$, while for $t=0$, the integral equals to zero. Therefore, $F(t)$ is not even continuous at $0$, let alone differentiable.

On the other hand, at $t\neq 0$, any partial derivative of $f$ has a form $P(t^{-1},x)\exp(-(x-t^{-2})^2)$ for some polynomial $P$. This tends to zero as $t\to 0$, uniformly in $x$ in compacts. Therefore, $f$ is smooth with all partial derivatives vanishing on at $\mathbb{R}$. Also, any expression of this form, and in particular $\partial_t f(t,x)$, is clearly integrable over $x$ for all $t$.

Update: The above example may feel like cheating, since in essence the discontinuity of $f$ has just been moved to infinity. Here's a better one in that respect:

$$ f(t,x)=\cases{\mathrm{sign}(t)|t|^\frac{1}{2}\exp\left(-\frac{(x-\sqrt[5]{t})^2}{|t|}\right),&$t\neq 0$,\\0,&$t=0.$} $$ This has $F(t)=\sqrt{\pi}t$ while still $\partial_t f(t,x)|_{t=0}\equiv 0$. It has an additional feature that $$ \int_I\partial_t f(t,x)\,dx\neq\partial_t \int_I f(t,x)\,dx $$ for any non-empty interval $I$ containing the origin. It has smooth restrictions onto vertical and horisontal lines, but it is not overall smooth or even continuously differentiable: indeed, if it were, it would have its derivatives bounded on compacts, thus $\int_I$ and $\partial_t$ would commute for any compact interval $I$.

A simple example is given by $$ f(t,x)=\cases{\exp(-(x-t^{-2})^2),&$t\neq 0$,\\0,&$t=0.$} $$ For each fixed $t\neq 0$, $\int f(t,x)\,dx$ is a Gaussian integral equal to $\sqrt{\pi}$, while for $t=0$, the integral equals to zero. Therefore, $F(t)$ is not even continuous at $0$, let alone differentiable.

On the other hand, at $t\neq 0$, any partial derivative of $f$ has a form $P(t^{-1},x)\exp(-(x-t^{-2})^2)$ for some polynomial $P$. This tends to zero as $t\to 0$, uniformly in $x$ in compacts. Therefore, $f$ is smooth with all partial derivatives vanishing on at $\mathbb{R}$. Also, any expression of this form, and in particular $\partial_t f(t,x)$, is clearly integrable over $x$ for all $t$.

A simple example is given by $$ f(t,x)=\cases{\exp(-(x-t^{-2})^2),&$t\neq 0$,\\0,&$t=0.$} $$ For each fixed $t\neq 0$, $\int f(t,x)\,dx$ is a Gaussian integral equal to $\sqrt{\pi}$, while for $t=0$, the integral equals to zero. Therefore, $F(t)$ is not even continuous at $0$, let alone differentiable.

On the other hand, at $t\neq 0$, any partial derivative of $f$ has a form $P(t^{-1},x)\exp(-(x-t^{-2})^2)$ for some polynomial $P$. This tends to zero as $t\to 0$, uniformly in $x$ in compacts. Therefore, $f$ is smooth with all partial derivatives vanishing on at $\mathbb{R}$. Also, any expression of this form, and in particular $\partial_t f(t,x)$, is clearly integrable over $x$ for all $t$.

Update: The above example may feel like cheating, since in essence the discontinuity of $f$ has just been moved to infinity. Here's a better one in that respect:

$$ f(t,x)=\cases{\mathrm{sign}(t)|t|^\frac{1}{2}\exp\left(-\frac{(x-t)^2}{|t|}\right),&$t\neq 0$,\\0,&$t=0.$} $$ This has $F(t)=\sqrt{\pi}t$ while still $\partial_t f(t,x)|_{t=0}\equiv 0$. It has an additional feature that $$ \int_I\partial_t f(t,x)\,dx\neq\partial_t \int_I f(t,x)\,dx $$ for any non-empty interval $I$ containing the origin. It has smooth restrictions onto vertical and horisontal lines, but it is not overall smooth or even continuously differentiable: indeed, if it were, it would have its derivatives bounded on compacts, thus $\int_I$ and $\partial_t$ would commute for any compact interval $I$.