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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?

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3 Answers 3

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$$\int_0^1 \sqrt{x^2+y}\; dy = \dfrac{2}{3} \left((x^2+1)^{3/2} - |x|^3\right)$$ for $x \in (-1,1)$.

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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).

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  • $\begingroup$ This is a very nice example. Could there exist a bounded $f$ with given property? $\endgroup$ Aug 28, 2015 at 19:34
  • $\begingroup$ I am asking, because in my other question (link here) I am able to express the integral as a sum of two improper ones and one with bounded integrand. $\endgroup$ Aug 28, 2015 at 19:50
  • $\begingroup$ @H.Berbeleque No, I don't think this can be done with a bounded integrand. See e.g. Theorem 12.12 in Bartle. $\endgroup$ Aug 28, 2015 at 20:54
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    $\begingroup$ (Just to be clear, the question I first answered had no boundedness hypothesis. You may want to do your edits in such a way that existing answers still make sense.) $\endgroup$ Aug 28, 2015 at 21:05
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    $\begingroup$ Yes, indeed, I feel guilty for first accepting your answer, and consequently changing my mind. My sincere apologies for that. As for the edit: thank you very much for pointing this out. Indeed, I should have added a remark on boundedness instead of incorporating it into assumptions, so I do edit it the better way now. $\endgroup$ Aug 29, 2015 at 6:37
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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 $U \times (c,d)$.)

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