이상적분

(a) We write the improper integral as a limit and use the substitution $u = \ln(x)$ and $du = \frac{1}{x} \ dx$ to evaluate the definite integral, then take the limit: $$\begin{aligned}\int_e^{\infty} \frac{\ln(x)}{x} \, dx &= \lim_{b \to \infty} \int_e^{b} \frac{\ln(x)}{x} \, dx \\ &= \lim_{b \to \infty} \int_1^{\ln(b)} u \, du \\ &= \lim_{b \to \infty} \left( \frac{u^2}{2} \right)\biggm|_1^{\ln(b)} \\ &= \lim_{b \to \infty} \left( \frac{(\ln(b))^2}{2} - \frac{1}{2} \right) \\ &= \infty\end{aligned}\,.$$ So the improper integral $\int_e^{\infty} \frac{\ln(x)}{x} \, dx$ diverges. (b) We write the improper integral as a limit and use the substitution $u = \ln(x)$ and $du = \frac{1}{x} \ dx$ to revaluate the definite integral, then take the limit:. $$\begin{aligned}\int_e^{\infty} \frac{1}{x\ln(x)} \, dx &= \lim_{b \to \infty} \int_e^{b} \frac{1}{x\ln(x)} \, dx \\ &= \lim_{b \to \infty} \int_1^{\ln(b)} \frac{1}{u} \, du \\ &= \lim_{b \to \infty} \left( \ln(|u|) \right)\biggm|_1^{\ln(b)} \\ &= \lim_{b \to \infty} \left( \ln(|\ln(b)|) - \ln(1) \right) \\ &= \infty\end{aligned}\,.$$ So the improper integral $\int_e^{\infty} \frac{1}{x\ln(x)} \, dx$ diverges. (c) We again write the improper integral as a limit and use the substitution $u = \ln(x)$ and $du = \frac{1}{x} \ dx$ to evaluate the definite integral, then take the limit: $$\begin{aligned}\int_e^{\infty} \frac{1}{x(\ln(x))^2} \, dx &= \lim_{b \to \infty} \int_e^{b} \frac{1}{x(\ln(x))^2} \, dx \\ &= \lim_{b \to \infty} \int_1^{\ln(b)} \frac{1}{u^2} \, du \\ &= \lim_{b \to \infty} \left( -u^{-1} \right)\biggm|_1^{\ln(b)} \\ &= \lim_{b \to \infty} \left( -\frac{1}{\ln(b)} - (-1) \right) \\ &= 1\end{aligned}\,.$$ So the improper integral $\int_e^{\infty} \frac{1}{x(\ln(x))^2} \, dx$ converges to $1$. (d) We again write the improper integral as a limit and use the substitution $u = \ln(x)$ and $du = \frac{1}{x} \ dx$ to rewrite the definite integral. If $p=1$, then the integral diverges by part (b). If $p \neq 1$, we have $$\begin{aligned}\int_e^{\infty} \frac{1}{x(\ln(x))^p} \, dx &= \lim_{b \to \infty} \int_e^{b} \frac{1}{x(\ln(x))^p} \, dx \\ &= \lim_{b \to \infty} \int_1^{\ln(b)} \frac{1}{u^p} \, du \\ &= \lim_{b \to \infty} \int_1^{\ln(b)} u^{-p} \, du \\ &= \lim_{b \to \infty} \left( \frac{u^{-p+1}}{-p+1} \right)\biggm|_1^{\ln(b)} \\ &= \frac{1}{1-p} \lim_{b \to \infty} \left( (\ln(b))^{1-p} - 1 \right) \\ &= 1\end{aligned}\,.$$ If $p < 1$, then $1-p > 0$ and $$\frac{1}{1-p} \lim_{b \to \infty} \left( (\ln(b))^{1-p} - 1 \right) = \left(\frac{1}{1-p}\right)\infty = \infty\,.$$ In this case the improper integral $\int_e^{\infty} \frac{1}{x(\ln(x))^p} \, dx$ diverges. If $p > 1$, then $1-p < 0$ and $$\frac{1}{1-p}\lim_{b \to \infty} \left( (\ln(b))^{1-p} - 1 \right) = \frac{1}{1-p}\lim_{b \to \infty} \left( \frac{1}{(\ln(b))^{p-1}} - 1 \right) = \left(\frac{1}{1-p}\right)(0-1) = \frac{1}{p-1}$$ In this case the improper integral $\int_e^{\infty} \frac{1}{x(\ln(x))^p} \, dx$ converges to $\frac{1}{p-1}$. Thus, the improper integral $\int_e^{\infty} \frac{1}{x(\ln(x))^p} \, dx$ diverges if $p \leq 1$ and converges to $\frac{1}{p-1}$ if $p > 1$. (e) This integral is improper since the integrand is undefined at $x = 0$. So we write the improper integral as a limit, use the substitution $u = \ln(x)$ and $du = \frac{1}{x} \ dx$ to evaluate the definite integral, then take the limit: $$\begin{aligned}\int_0^1 \frac{\ln(x)}{x} \, dx &= \lim_{a \to 0^+} \int_a^1 \frac{\ln(x)}{x} \, dx \\ &= \lim_{a \to 0^+} \int_{\ln(a)}^{0} u \, du \\ &= \lim_{a \to 0^+} \left( \frac{u^2}{2} \right)\biggm|_{\ln(a)}^{0} \\ &= \lim_{a \to 0^+} \left( -\frac{(\ln(a))^2}{2} \right) \\ &= -\infty\end{aligned}\,.$$ So the improper integral $\int_0^1 \frac{\ln(x)}{x} \, dx$ diverges. (f) This integral is improper since the integrand is undefined at $x=0$. So we write the improper integral as a limit, then use parts with $u = \ln(x)$, $du = \frac{1}{x} \, dx$, $dv = dx$, and $v = x$ to evaluate the definite integral, then take the limit: $$\begin{aligned}\int_0^1 \ln(x) \, dx &= \lim_{a \to 0^+} \int_a^1 \ln(x) \, dx \\ &= \lim_{a \to 0^+} \left( x\ln(x)\biggm|_a^1 - \int_{a}^{1} \ dx \right) \\ &= \lim_{a \to 0^+} \left( -a\ln(a) - x\biggm|_{\ln(a)}^{0} \right) \\ &= \lim_{a \to 0^+} \left( -a\ln(a) - (1-a) \right) \\ &= \lim_{a \to 0^+} \left( -a\ln(a) \right) - 1\end{aligned}\,.$$ The limit $\lim_{a \to 0^+} \left( -a\ln(a) \right)$ has the form $0 \times \infty$, which is indeterminate, so we need to rewrite it in a form in which we can apply L'Hopital's Rule. Now, $$\lim_{a \to 0^+} \left( -a\ln(a) \right) = \lim_{a \to 0^+} \left( \frac{-\ln(a)}{\frac{1}{a}} \right)\,,$$ and this latter limit is of the form $\frac{\infty}{\infty}$. Applying L'Hopital's Rule gives us $$\begin{aligned}\lim_{a \to 0^+} \left( \frac{-\ln(a)}{\frac{1}{a}} \right) &= \lim_{a \to 0^+} \left( \frac{-\frac{1}{a}}{-\frac{1}{a^2}} \right) \\ &= \lim_{a \to 0^+} a \\ &= 0\end{aligned}$$ Returning to our earlier established equality that $$\int_0^1 \ln(x) \, dx = \lim_{a \to 0^+} \left( -a\ln(a) \right) - 1\,,$$ this most recent limit shows that the improper integral $\int_0^1 \ln(x) \, dx$ converges to $-1$.