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This question has been asked before here, but no one really dealt with the problem, and just critiqued the user's method, which seems to have been what the poster wanted. I'm looking for the full solution. In other words, is my full answer correct? If not, what should the changes be? I can find small details all over the place on Math.SE but for my own confidence (and reference), I would be very happy to have the full solution here. Perhaps others will benefit, too.

My answer seems too easy to be correct, and I am suspicious that the Monotone and/or Dominated convergence theorems don't make an appearance in my solution. We are in Royden's "Real Analysis", page 84, question 19. The question states:

"For a number $\alpha$, define $f(x)=x^\alpha$ for $0<x\leq 1$ and $f(0)=0$. Compute $\int\limits_{0}^{1}f.$"

Now, for $\alpha\geq0$, the problem is easy, as the function is Riemann integrable, and we obtain $\frac{1}{\alpha+1}$. But what about $\alpha<0$? It seems that the three cases are then $-1<\alpha<0$, $\alpha=-1$, and $\alpha<-1$. In the second, we obtain:

$$-\lim\limits_{\varepsilon\to0^+}\ln\varepsilon=\infty.$$

In the last, we obtain:

$$\frac{1}{\alpha+1}+\lim\limits_{\varepsilon\to0^+}\frac{\varepsilon^{\alpha+1}}{\alpha+1}=\infty,$$ since $\alpha+1<0$.

In the first, we obtain: $$\frac{1}{\alpha+1}+\lim\limits_{\varepsilon\to0^+}\frac{\varepsilon^{\alpha+1}}{\alpha+1}=\frac{1}{\alpha+1},$$ since $\alpha+1>0$.

So, is my answer correct when I say $\int\limits_{0}^{1}f=\frac{1}{1+\alpha}$ for $\alpha>-1$ and $\int\limits_{0}^{1}f=\infty$ otherwise?

Is the whole point of the problem that $x^\alpha$ is not Lebesgue integrable for $\alpha<-1$?

Community
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The Count
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    Doesn't the monotonicity of integration imply that for $\alpha < -1$, $\int_0^1 f \geq \int_0^1 x^{-1} = \infty$? – jodag Oct 15 '18 at 04:16

1 Answers1

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Your result looks correct. I would add that the monotone convergence theorem is in fact used to enable you to use your limit.

So you implicitly write $f(x)$ as the limit of the increasing sequence $f_n(x)=f(x)I_{x\ge1/n}$ and use the monotone convergence theorem to say that $\int f(x)=\lim_{n\rightarrow\infty}\int f_n(x)$, which you later calculate as being either $\infty$ for $\alpha\le-1$ or $\frac{1}{\alpha+1}$ for $\alpha>-1$

Momo
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  • Thanks very much! I just took the integrals and then took the limit, actually, but you make a good point. I could do it the other way. Excellent. Thank you! – The Count Dec 27 '16 at 02:02