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I have the following equation

$$g(y)=\int_{0}^{\infty} f(x,y) dx$$

I know that $f$ is continuous in $x$ and $y$. But I would like to infer that $g$ is continuous in $y$. Can I do this?


I wont write down the function here, since it is huge, but I can guarantee that:

$f$ is differentiable in both $x$ and $y$

$\lim_{y\rightarrow \infty} f(x,y) = \infty$

I could try to solve the integral, but I do not know if I'm able too. Applying some theorem that guarantees continuity would be great, but now I see I would need further hypothesis.

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Yes, but it requires the notion of uniform continuity. Are you familiar with that? Edit: I didn't notice the infinite integration limit. In this case, you need a further hypothesis. – Harald Hanche-Olsen Jan 29 '13 at 16:20
Regarding extra hypotheses: See for example the Lebesgue's dominated convergence theorem. It is true for Riemann integrals, because the Riemann integral is a special case of the Lebesgue integral. But the proof is easier for the latter. – Harald Hanche-Olsen Jan 29 '13 at 16:37

Counterexample without further hypotheses: $$ f(x,y)=\begin{cases}ye^{-xy}&y>0,\\0&y\le0.\end{cases} $$ Then $$ g(y)=\begin{cases}1&y>0,\\0&y\le0.\end{cases} $$

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+1 Nice and simple – DonAntonio Jan 29 '13 at 16:39
I think Lebesgue's dominated convergence theorem does not help me, neither does Monotone convergence theorem. It is because, in my application I have $\lim_{y\rightarrow \infty} f(x,y) = \infty$, that is, my function does not converge pointwise to anything. Sorry for not mentioning that before. – user60111 Jan 29 '13 at 17:20
@user60111 Bad behaviour as $y\to\infty$ shouldn't be a problem for continuity, if you can use Lebesgue for any bounded set of $y$ values. – Harald Hanche-Olsen Jan 29 '13 at 17:39

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