§6.5 Improper integrals (continued)

Improper Integrals Involving Unbounded Integrands

Summary

$\int_0^1 \frac{1}{x^{1/3}} \, dx$

$\int_0^2 e^{-x} \, dx$

$\int_1^4 \frac{1}{\sqrt{4-x}} \, dx$

$\int_{-2}^2 \frac{1}{x^2} \, dx$

$\int_0^{\pi/2} \tan(x) \, dx$

$\int_0^1 \frac{1}{\sqrt{1-x^2}} \, dx$

Determine, with justification, whether each of the following improper integrals converges or diverges. -$\int_e^{\infty} \frac{\ln(x)}{x} \, dx$-$\int_e^{\infty} \frac{1}{x\ln(x)} \, dx$-$\int_e^{\infty} \frac{1}{x(\ln(x))^2} \, dx$-$\int_e^{\infty} \frac{1}{x(\ln(x))^p} \, dx$, where $p$ is a positive real number -$\int_0^1 \frac{\ln(x)}{x} \, dx$-$\int_0^1 \ln(x) \, dx$

Sometimes we may encounter an improper integral for which we cannot easily evaluate the limit of the corresponding proper integrals. For instance, consider $\int_1^{\infty} \frac{1}{1+x^3} \, dx$. While it is hard (or perhaps impossible) to find an antiderivative for $\frac{1}{1+x^3}$, we can still determine whether or not the improper integral converges or diverges by comparison to a simpler one. Observe that for all $x \gt 0$, $1 + x^3 \gt x^3$, and therefore

.

It therefore follows that

for every $b \gt 1$. If we let $b \to \infty$ so as to consider the two improper integrals $\int_1^\infty \frac{1}{1+x^3} \, dx$ and $\int_1^\infty \frac{1}{x^3} \, dx$, we know that the larger of the two improper integrals converges. And thus, since the smaller one lies below a convergent integral, it follows that the smaller one must converge, too. In particular, $\int_1^\infty \frac{1}{1+x^3} \, dx$ must converge, even though we never explicitly evaluated the corresponding limit of proper integrals. We use this idea and similar ones in the exercises that follow. - Explain why $x^2 + x + 1 \gt x^2$ for all $x \ge 1$, and hence show that $\int_1^{\infty} \frac{1}{x^2 + x + 1} \, dx$ converges by comparison to $\int_1^{\infty} \frac{1}{x^2} \, dx$. - Observe that for each $x \gt 1$, $\ln(x) \lt x$. Explain why

for each $b \gt 2$. Why must it be true that $\int_2^\infty \frac{1}{\ln(x)} \, dx$ diverges? - Explain why $\sqrt{\frac{x^4+1}{x^4}} \gt 1$ for all $x \gt 1$. Then, determine whether or not the improper integral

converges or diverges.