이상적분

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.)