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We're trying to prove the integral $$\int_0^1\frac{\cos x}{x^\frac12}\,dx$$ exists as an improper integral. My teacher says that in order to prove there exists the limit of $\int_a^1\frac{\cos x}{x^\frac12}\,dx$ as a goes to $0$, we must prove that the integral $\int_a^1\frac{\cos x}{x^\frac12}\,dx$ is continuous at $a = 0$.

Why is the existence of the limit the same thing as continuity of the integral? I don't understand at all.

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This is true because if one defines $$g(a) = \int_a^1 \frac{\cos x}{\sqrt{x}}dx$$, then the value of $g(0)$ is the improper integral, and only exists if the limit converges. When this occurs, $g$ is automatically continuous at $0$ because its value at $0$ is defined to be the limit. So the choices are:

  • The limit doesn't exist, and $0$ is not in the domain of $g$, or
  • The limit exists and $g$ is continuous there.
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  • $\begingroup$ Oh I see! Thanks! $\endgroup$ – Rainroad Apr 9 '16 at 16:02
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Now that I've finally read the question carefully, I would say - if we want to be precise - that it doesn't make sense to talk about continuity of $\int_a^1(\cos x)/\sqrt x\, dx$ at $0$ until a value of $\int_0^1(\cos x)/\sqrt x\, dx$ has been assigned. And assigning an appropriate value at $0$ is the same as showing $\lim_{a\to 0^+}\int_a^1(\cos x)/\sqrt x\, dx$ exists, so things are getting circular here.

It would be better to say $\int_0^1(\cos x)/\sqrt x\, dx$ exists iff the function $a\to \int_a^1(\cos x)/\sqrt x\, dx,$ initially only defined on $(0,1],$ has a continuous extension to $[0,1].$ I'm not sure how helpful that is in showing the integral converges.


Previous (unhelpful) answer: Hint: $\int_0^1|f|<\infty$ implies $\int_0^1f$ converges.

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    $\begingroup$ That doesn't address the question asked. $\endgroup$ – Paul Sinclair Apr 9 '16 at 5:14
  • $\begingroup$ I guess you're right. I'll delete it. $\endgroup$ – zhw. Apr 9 '16 at 17:59
  • $\begingroup$ I edited my answer after reading the question more carefully. $\endgroup$ – zhw. Apr 9 '16 at 18:45

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