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Let $f$ be a continuous function in [0,1] and $\alpha >0$, I'd love your help with finding the following limit: $\lim_{x \to 0}, x^{\alpha}\int_{x}^{1}\frac{f(t)}{t^{\alpha+1}}$.

First I tried to bound the function since it is continues in a closed interval

$\lim_{x \to 0}, |x^{\alpha}\int_{x}^{1}\frac{f(t)}{t^{\alpha+1}} |\leq \lim_{x \to 0},| x^{\alpha}\int_{x}^{1}\frac{M}{t^{\alpha+1}}|$ but I get a number depends on $\alpha$ and $M$, and it won't do. So, I assume that the integral is always diverges, since $f$ is continues and it is divided by $t$ with exponent bigger than one, so it is 0 multiplied by $\infty$, maybe we can use L'Hopital somehow?

Thanks!

2 Answers 2

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You said it yourself in the question: use L'Hospital's Rule applied to $\lim_{x\to0^+}\frac{\int_x^1\frac{f(t)}{t^{\alpha+1}}\,dt}{1/x^{\alpha}}$ The numerator and denominator each approach $\infty$, given your conditions on $f$.


EDIT: The denominator approaches $\infty$ simply because $\alpha$ is positive.

If $f(0)$ is nonzero, then $f$ is bounded below by some positive $\epsilon$ in a neighborhood of $0$. So the integral is bounded below by $\epsilon\int_x^\delta\frac{1}{t^{\alpha+1}}\,dt+\int_{\delta}^1\frac{f(t)}{t^{\alpha+1}}\,dt$ which diverges to infinity as $x$ approaches $0^+$, since the power of $t$ is greater than $1$.

(And if $f(0)=0$ the numerator might not approach $\infty$. But it still approaches something since $f$ is integrable on $[0,1]$ and $t^{\alpha+1}$ is monotonic. Let's call it $L$. Then L'Hospital's Rule is not needed - the OP's limit is $0\cdot L$, or just $0$. This is consistent with the formula given below when $f(0)\neq0$.)


The result is $\begin{align}\lim_{x\to0^+}\frac{\int_x^1\frac{f(t)}{t^{\alpha+1}}\,dt}{1/x^{\alpha}}&=\lim_{x\to0^+}\frac{\frac{d}{dx}\int_x^1\frac{f(t)}{t^{\alpha+1}}\,dt}{\frac{d}{dx}1/x^{\alpha}}\\ &=\lim_{x\to0^+}\frac{-\frac{d}{dx}\int_1^x\frac{f(t)}{t^{\alpha+1}}\,dt}{\frac{d}{dx}x^{-\alpha}}\\ &=\lim_{x\to0^+}\frac{-\frac{f(x)}{x^{\alpha+1}}}{-\alpha x^{-\alpha-1}}\\ &= \lim_{x\to0^+}\frac{f(x)}{\alpha}\\ &=\frac{f(0)}{\alpha} \end{align}$

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    @Peter If the numerator does _not_ approach infinity, the OP's limit is just $0\cdot L$ for some real $L$.2012-02-09
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Let $\varepsilon>0$ and $\delta$ such that if $|x|\leq\delta$ then $|f(x)-f(0)|\leq\varepsilon$. We have for $0: \begin{align*}\left|x^{\alpha}\int_x^1\frac{|f(t)-f(0)|}{t^{\alpha+1}}dt\right| &\leq x^{\alpha}\int_x^{\delta}\frac{\varepsilon}{t^{\alpha+1}}dt+x^{\alpha}\int_{\delta}^1\frac{|f(0)|}{t^{\alpha+1}}dt\\ &=x^{\alpha}\varepsilon \frac{-1}{\alpha}(\delta^{-\alpha}-x^{-\alpha})+x^{\alpha}\int_{\delta}^1\frac{|f(0)|}{t^{\alpha+1}}dt\\ &=\frac{(1-x^{\alpha}\delta^{-\alpha})}{\alpha}\varepsilon+x^{\alpha}\int_{\delta}^1\frac{|f(0)|}{t^{\alpha+1}}dt\\ &\leq \frac{(1+|x|^{\alpha}\delta^{-\alpha})}{\alpha}\varepsilon+|x|^{\alpha}\int_{\delta}^1\frac{|f(0)|}{t^{\alpha+1}}dt, \end{align*} so $\limsup_{x\to 0}\left|x^{\alpha}\int_x^1\frac{|f(t)-f(0)|}{t^{\alpha+1}}dt\right|\leq \varepsilon$ and $\lim_{x\to 0}x^{\alpha}\int_x^1\frac{f(t)-f(0)}{t^{\alpha+1}}dt=0$. Since $x^{\alpha}\int_x^1\frac{dt}{t^{\alpha+1}}=-\frac 1{\alpha}(1-x^{-\alpha})x^{\alpha}=\frac 1{\alpha}(1-x^\alpha),$ so finally $\lim_{x\to 0}x^{\alpha}\int_x^1\frac{f(t)}{t^{\alpha+1}}dt=\frac{f(0)}{\alpha}.$

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    How does that cha$n$ge in notation affect the proof?2012-02-08