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I'm struggling to analyze an improper integral of two variables.

$$ \lim_{y\rightarrow \infty}\int_1^y f(x)\sqrt{y-x}dx $$ where $f$ is a real valued function defined on $[0,\infty)$.

One simple example $f(x)= 1$ does not make the integral converge.

I want to find some sufficient condition for the improper integral to be well defined.

My expected answer is like

$$ \int_1^{\infty} \frac{1}{x^p} dx $$ is convergent $p>1$ and divergent $p<=1$.

Thank you for advance.

My question may be somewhat vague. Ask me freely for some points making you confused. Thank you!

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1 Answer 1

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If $f$ is any positive measurable function then the limit is $\infty$ by Monotone Convergence Theorem.

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  • $\begingroup$ Thank you for your reply. But, $f$ is not restricted to a positive function. :< $\endgroup$
    – user155214
    Jun 5, 2018 at 6:47
  • $\begingroup$ I doubt if any reasonable sufficient condition can be found. $\endgroup$
    – Kavi Rama Murthy
    Jun 5, 2018 at 7:30

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