$\lim_{h \to 0} \int_{x}^{x+h} \ln(t) dt$
Unless I'm missing something, isn't this just $0$ due to how the integral is just $\int_{x}^{x}=0$
I'm sure I could integrate the inside and then evaluate that as $h \to 0$, but this answer seems to deceptively easy if correct
$\endgroup$ 22 Answers
$\begingroup$Yes, you're correct. You can argue it that way, or even if you go as far as integrating first you'll find the same result:
\begin{eqnarray*} \lim_{h\to 0} \int_x^{x+h}\ln(t)dt & = & \lim_{h\to 0} \left . t\ln(t) - t \right |_x^{x+h} \\ & = & \lim_{h\to 0}(x+h)\ln(x+h) - (x+h) - x\ln(x) + x \\ & = & \lim_{h\to 0} x \ln\left (\frac{x+h}{x}\right ) + h\ln(x+h) - h \\ & = & x \ln(1) + 0 - 0 \;\; =\;\; 0. \end{eqnarray*}
$\endgroup$ $\begingroup$It is not that trivial for $x=0$, it is an improper integral:\begin{align*} \left|\int_{0}^{h}\log tdt\right|&=-\int_{0}^{h}\log tdt\\ &=\int_{0}^{h}-\log tdt\\ &=\int_{0}^{h}\log(1/t)dt\\ &\leq C\int_{0}^{h}\dfrac{1}{t^{1/2}}dt\\ &=2C\sqrt{h}\\ &\rightarrow 0, \end{align*}here $h>0$ is small.
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