含参积分$f(a) = \int_{0}^{1} \frac{\sin(ax)}{\ln(x)} \,dx$在$a=0$时取值的疑惑求解

Hey there! Great question—indeterminate forms at single points can feel confusing when working with integrals, but let's unpack this clearly:

First, let's hit the core point: the value of a function at a single point has no impact on its Riemann integral over an interval.

When $a=0$, your integrand becomes $\frac{\sin(0)}{\ln(x)} = \frac{0}{\ln(x)}$. For every $x \in [0,1)$ (all points except $x=1$), this is exactly 0. At $x=1$, we get the indeterminate form $\frac{0}{0}$, but here's the key: a single point contributes nothing to the area under the curve (which is what the integral measures). Even if we left the function undefined at $x=1$, or assigned it any arbitrary value, the integral over $[0,1]$ would still equal the integral of 0 over $[0,1)$, which is 0.

To make this more rigorous, we can also define the integrand at $x=1$ to match its limit as $x \to 1^-$, which removes the indeterminacy. When $a=0$, $\lim_{x \to 1^-} \frac{0}{\ln(x)} = 0$ (since the numerator is fixed at 0, and the denominator approaches 0 from the negative side—0 divided by any non-zero number is 0, and the limit stays 0). With this definition, the integrand is now continuous on all of $[0,1]$, and the integral is clearly $\int_{0}^{1} 0 ,dx = 0$.

Another way to confirm this is thinking about the continuity of $f(a)$ as a function of $a$. For small $|a|$, the integrand $\frac{\sin(ax)}{\ln(x)}$ is bounded on $[0,1]$ (we can safely handle the $x \to 0^+$ and $x \to 1^-$ limits), so $f(a) \to f(0)$ as $a \to 0$. But even without that, your initial thought was totally correct—the indeterminate form at $x=1$ is a red herring for the integral's value; single points don't change the result of a Riemann integral.

备注：内容来源于stack exchange，提问作者Water