How can one show that $ f(0)\ln(\frac{b}{a})=\lim_{\epsilon\rightarrow 0}\int_{\epsilon a}^{\epsilon b} \frac{f(x)}{x}dx$?

Solution 1:

$$\int_{\epsilon a}^{ \epsilon b} \frac {f(x)}{x} dx = \int_{\epsilon a}^ {\epsilon b}\left[\frac {f(0)}{x} +\frac {f(x) - f(0)}{x}\right] dx = f(0)[\ln (\epsilon b) - \ln (\epsilon a)] + \int_{\epsilon a}^ {\epsilon b} \frac {f(x) - f(0)}{x}dx$$

The last integral goes to 0 as $\epsilon \to 0$ because $a > 0$ (and I am also assuming $b > 0$), so $x$ is not near $0$.

So the final answer is $f(0)[\ln (\epsilon) +\ln (b) -(\ln (\epsilon) + \ln(a))]$

and this goes to $f(0)\ln \dfrac{b}{a}$ as $\epsilon \to 0$.

Solution 2:

Hint: Consider first the case where $f(0)=0$. Prove that the integral in question is zero. Then apply this result to $f(x)-f(0)$.

Solution 3:

One way to look at this is with the mean value theorem for integration

since $a$ is positive, and I assume $b$ is too, then $\frac 1 x$ is positive in the interval $(\epsilon a, \ \epsilon b )$ ( or negative if $\epsilon $ aproches $0$ from the left), and it doesn't change sign, thus: $$ \lim_{\epsilon\rightarrow 0}\int_{\epsilon a}^{\epsilon b} \frac{f(x)}{x}dx = \lim_{\epsilon\rightarrow 0} f(c)\int_{\epsilon a}^{\epsilon b} \frac{1}{x}dx $$

for some $ c \in (\epsilon a, \ \epsilon b ) $, therfore $$ \lim_{\epsilon\rightarrow 0}\int_{\epsilon a}^{\epsilon b} \frac{f(x)}{x}dx = \lim_{\epsilon\rightarrow 0} f(c) \Big [\ln(\epsilon b)-\ln(\epsilon a) \Big ]\\=\lim_{\epsilon\rightarrow 0} f(c) \ln(\frac b a) $$ and because $\epsilon a<c<\epsilon b$ and both $\epsilon a $ and $\epsilon b$ aprach $0$ as $\epsilon $ aproches $0$, then, by the squeeze theorem, $c$ aproaches $0$, and because $f$ is continous, it doesn't matter how $c$ aproches $0$, thus:

$$ \lim_{\epsilon\rightarrow 0}\int_{\epsilon a}^{\epsilon b} \frac{f(x)}{x}dx = f(0) \ln(\frac b a) $$