Find an integrable, positive, unbounded, analytic function Is there a standard example of a function $f \in L^1( \mathbb R)$ which is analytic, positive, integrable but not bounded?
An example which comes immediately to mind is to take the series of narrower but steeper gaussians centered at every integer. But do we obtain an analytic function this way?
 A: What do you mean by analytic? Analytic on the real line, or entire. The method you suggest can be used to obtain an entire example. Let $F$ be a Gaussian.
Consider a series 
$$\sum a_nF(b_n(z-c_n))$$
with positive $a_n, b_n$ and $c_n$. If $a_n\to+\infty$ does not grow fast, but $c_n$ does grow fast enough, the
series converges on every compact in the complex plane, and gives you an entire function with the properties you stated.
EDIT. I corrected according to R. Israel suggestion. With $b_n=1$ and $a_n\to\infty$
it will not be in $L^1$. 
In fact, Carleman's theorem says that for every continuous function $\phi$ on the real line, and every positive continuous function $\epsilon$ on the real line, one can find an
entire function $f$ such that $|f(x)-\phi(x)|<\epsilon(x)$.
A: Consider the function 
$$ f(t) = \frac{2e^{-2t^2}}{e^{-3t^2}(1+\cos(2t))+(1-\cos(2t))}. $$
I think that this does what you want.  It is certainly real analytic and strictly positive.  It satisfies $f(k\pi)=e^{k^2\pi^2}$, and so is unbounded.  Near $k\pi$ it has a hump of width approximately proportional to $e^{-3k^2\pi^2/2}$, so the area scales like $e^{-k^2\pi^2/2}$.  Away from these humps, the function is extremely small.  This suggests that it is integrable, but I have not proved that rigorously.
A: The function you suggest is perfectly fine. It's just a question of filling in the details.
For all positive $t$, define
$$
g_t(z) := \frac{t}{\sqrt{\pi}} e^{-t^2z^2}.  
$$
For all $t$, $g_t$ is a holomorphic function defined over the whole of $\Bbb{C}$ (in other words, it is entire), its restriction to the real line is positive, and its integral over the real line is equal to $1$. Furthermore, these functions all decay exponentially over the quadrants $\left\{-\pi/4<\text{Arg}(z)<\pi/4\right\}$ and $\left\{3\pi/4<\text{Arg}(z)<5\pi/4\right\}$ of the complex plane. In particular, for all $r>0$, if $\left|z\right|>r$ and if $\left|\text{Arg}(z)\right|<\frac{\pi}{6}$ (or if $\left|\text{Arg}(z)-\pi\right|<\frac{\pi}{6}$), then,
$$
\left|g_t(z)\right| \leq \frac{t}{\sqrt{\pi}}e^{-t^2 r^2/2}.
$$
We now define, for example
$$
f(z) = \sum_{n\in\Bbb{Z}\setminus\left\{0\right\}} \frac{1}{n^2}g_{n^3}(z-n).
$$
Observe that $f$ is unbounded. However, it follows from the above inequality that $f$ converges uniformly over every compact subset of $\Bbb{C}$. It is therefore also an entire function, and in particular is analytic over $\Bbb{R}$. Furthermore, since $\sum_{n\in\Bbb{Z}\setminus\left\{0\right\}}\frac{1}{n^2}$ is finite, $f$ also has finite integral, as desired.
