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May integration spoil real-analyticity?

up vote 14 down vote favorite

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?

2 Answers

up vote 4 down vote accept

Yes, if the integral is improper as in your other question. E.g. $$ \int_{-1}^1\frac{\sin x\,dy}{(y-\cos x)^2 + \sin^2x} =\begin{cases} \frac\pi2&x\in(0,\pi)\\ 0& x=0,\pi,2\pi\\ -\frac\pi2&x\in(\pi,2\pi).\\ \end{cases} $$

up vote 3 down vote

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


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May integration spoil real-analyticity?

up vote 14 down vote

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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up vote 9 down vote

Yes, if the integral is improper as in your other question. E.g. $$ \int_{-1}^1\frac{\sin x\,dy}{(y-\cos x)^2 + \sin^2x} =\begin{cases} \frac\pi2&x\in(0,\pi)\\ 0& x=0,\pi,2\pi\\ -\frac\pi2&x\in(\pi,2\pi).\\ \end{cases} $$

edit

(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.) - Francois Ziegler Aug 28 at 21:05

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