이상적분

$\pi/8$

If $b= \pm \infty$, then our integral is improper because one limit is not a real number.

Furthermore, our integral will be improper if its domain of integration contains either of its infinite discontinuities, $x=1$ and $x=-1$. Since one limit of integration is 0, the integral is improper if $b \geq 1$ or if $b \leq -1$.

Below, we've graphed $\frac{1}{x^2-1}$ to make it clearer why values of $b$ in $(-1,1)$ are the only values that don't result in an improper integral when the other limit of integration is $a=0$.

Since the integrand is continuous for all real $x$, the only kind of impropriety available to us is to set $b = \pm\infty$.

The denominator is zero when $x=1$, but the numerator is not, so the integrand has a singularity (infinite discontinuity) at $x=1$. Let's replace the limit $x=1$ with a variable that creeps toward 1.

$$\int_0^1\frac{x^4}{x^5-1}\,\mathrm{d}x =\lim_{t\rightarrow 1^-}\int_0^t\frac{x^4}{x^5-1}\,\mathrm{d}x$$

To evaluate this integral we use the substitution $u=x^5$, $\mathrm{d}u=5x^4\mathrm{d}x$. When $x=0$ we have $u=0$, and when $x=t$ we have $u=t^5$, so

$$\begin{aligned} \int_0^1\frac{x^4}{x^5-1}\,\mathrm{d}x&=\lim_{t \to 1^-}\int_{0}^{t}\frac{x^4}{x^5-1}\,\mathrm{d}x =\lim_{t \to 1^-}\int_{u=0}^{u=t^5}\frac{1}{5(u-1)}\,\mathrm{d}u\\ &=\lim_{t \to 1^-}\left( \left[\frac{1}{5}\log|u-1|\right]_0^{t^5}\right) =\lim_{t \to 1^-}\frac{1}{5}\log|t^5-1|=-\infty \end{aligned}$$

The limit does not exist, so the integral diverges.

The denominator of the integrand is zero when $x=-1$, but the numerator is not. So, the integrand has a singularity (infinite discontinuity) at $x=-1$. This is the only “source of impropriety" in this integral, so we only need to make one break in the domain of integration.

$$\int_{-2}^2\frac{1}{(x+1)^{4/3}}\,\mathrm{d}x =\lim_{t\rightarrow -1^-}\int_{-2}^t \frac{1}{(x+1)^{4/3}}\,\mathrm{d}x +\lim_{t\rightarrow -1^+}\int_t^2 \frac{1}{(x+1)^{4/3}}\,\mathrm{d}x$$

Let's start by considering the left limit.

$$\begin{aligned} \lim_{t \to -1^-}\int_{-2}^t \frac{1}{(x+1)^{4/3}}\,\mathrm{d}x &=\lim_{t \to -1^-}\left(\left[-\frac{3}{(x+1)^{1/3}}\right|_{-2}^t\right)\\ &=\lim_{t \to -1^-}\left(-\frac{3}{(t+1)^{1/3}}+\frac{3}{(-1)^{1/3}}\right)=\infty \end{aligned}$$

Since this limit diverges, the integral diverges. (A similar argument shows that the second integral diverges. Either one of them diverging is enough to conclude that the original integral diverges.)

There are two “sources of impropriety": the two (infinite) limits of integration. So, we break our integral into two pieces.

$$\begin{aligned} \int_{-\infty}^\infty \cos x~\mathrm{d}x&=\int_{-\infty}^0 \cos x~\mathrm{d}x+\int_{0}^\infty \cos x~\mathrm{d}x\\ &=\lim_{a \to \infty }\left[\int_{-a}^0 \cos x~\mathrm{d}x\right]+ \lim_{b \to \infty }\left[\int_{0}^b \cos x~\mathrm{d}x\right] \\ \text{These are easy enough to antdifferentiate.} \\ &=\lim_{a \to \infty }\left[\sin 0 - \sin(- a) \right]+ \lim_{b \to \infty }\left[\sin b - \sin 0 \right]\\ &=\text{DNE} \end{aligned}$$

Since the limits don't exist, the integral diverges. (It happens that both limits don't exist; even if only one failed to exist, the integral would still diverge.)

(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 $-1$.