Can integration spoil real-analyticity? Is there an example of a function $f:(a,b)\times(c,d)\to\mathbb{R}$, which is real analytic in its domain, integrable in the second variable, and such that the function 
$$ g:(a,b)\to\mathbb{R},\qquad g(x) = \int_c^d f(x,y) dy$$
is not real-analytic on $(a,b)$?
Edit: What about an example of bounded $f$ satisfying the above?
 A: Maybe this is too obvious to be interesting, but if we have bounds $\frac{1}{n!} \left| \frac{\partial^n}{(\partial x)^n} f(x,y) \right| < C_n$ on all of $(a,b) \times (c,d)$ such that $\sum C_n r^n$ converges for some $r>0$, then we get the conclusion. 
This is because the Taylor series with respect to $x$ then converges on circles of radius $r$ around any point of $(a,b)$. Write $U$ for the open subset of points in $\mathbb{C}$ which are within distance $r$ of some point of $(a,b)$. These Taylor series give an extension of $f$ to $U \times (c,d)$, holomorphic in the first variable. (They agree on overlaps, since they agree on the real interval where they overlap.) Moreover, $|f| \leq \sum C_n r^n$ everywhere on $U \times (c,d)$.
We want to claim that $\int_c^d f(x,y) dy$ is holmorphic on $U$. By Morera's theorem, it is enough to check that $\int_{x \in \gamma} \int_{y=c}^d f(x,y) dy dx=0$ for any closed curve $\gamma \subset U$. By Fubini, we may switch the integral to $\int_{y=c}^d \int_{x \in \gamma} f(x,y) dx dy$, and the inner integral is zero since $f$ is holomorphic. (We may apply Fubini because we have a uniform bound for $f$ throughout $U \times (c,d)$.)
A: $$\int_0^1 \sqrt{x^2+y}\; dy = \dfrac{2}{3} \left((x^2+1)^{3/2} - |x|^3\right)$$
for $x \in (-1,1)$.
A: Yes, if the integral is improper as in your other question. E.g.
$$
\int_{-1}^1\frac{\sin x\,dy}{(y-\cos x)^2 + \sin^2x}
=\begin{cases}
\frac\pi2&x\in(0,\pi)\\
0& x=0,\pi,2\pi\\
-\frac\pi2&x\in(\pi,2\pi)\\
\end{cases}
$$
— an example due to Hermite (1870).
