Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

EDIT:

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.

share|improve this question
    
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

1 Answer 1

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

share|improve this answer
    
+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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.