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Let $ f$ be a continious function and $a>0$

calculate the limit $$ \lim \limits_{x \to 0} x^a \int_x^1 \frac{f(t)}{t^{a+1}} \, dt$$

i tried to divied to 2 cases, if $ \lim \limits_{x \to 0} \int_x^1 \frac{f(t)}{t^{a+1}} \, dt = L < \infty$

then the answer is clearly $0$, else we can use l'Hopital rule, but I'm not sure how to apply it on the integral,

I know that if $F(x) = \int_0^x \frac{f(t)}{t^{a+1}} \, dt$ then the derivative will be $\frac{f(x)}{x^{a+1}} $ but what do i do with $\int_x^1$

thx

Leucippus
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Daniel Katzan
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1 Answers1

0
  1. This integral limit already has had attention here
  2. A series expansion based upon integration by parts is provided in this solution.

Let \begin{align} I &= \int_{x}^{1} \frac{f(t)}{t^{a+1}} \, dt \\ &= \left[ - \frac{1}{a} \, t^{-a} \, f(t) \right]_{x}^{1} + \frac{1}{a} \, \int_{x}^{1} \frac{f'(t)}{t^{a}} \, dt \\ &= \left[ - \frac{1}{a} \, t^{-a} \, f(t) \right]_{x}^{1} + \left[ - \frac{1}{a \, (a-1)} \, t^{-a+1} \, f'(t) \right]_{x}^{1} + \frac{1}{a \, (a-1)} \, \int_{x}^{1} \frac{f''(t)}{t^{a-1}} \, dt \\ &= \left[ \frac{f(x)}{a \, x^{a}} + \frac{f'(x)}{a(a-1) \, x^{a-1}} + \cdots \right] - \left[\frac{f(1)}{a} + \frac{f'(1)}{a(a-1)} + \cdots \right] + \cdots + \frac{1}{a!} \, \int_{x}^{1} \frac{f^{(a)}(t)}{t} \, dt \end{align}

Now, taking the desired limit yields \begin{align} \lim_{x \to 0} \, x^{a} \, \int_{x}^{1} \frac{f(t)}{t^{a+1}} \, dt = \frac{f(0)}{a} \end{align} It is required that$a \neq 0$.

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