Solutions - 0130-02

(a) Improper: integrand $\to \infty$ as $x\to 0^{+}$.

Note: this converges too, since it’s a $p$-integral to zero with $p<1$.

(b) Proper: no source of infinity.

Note: automatically converges.

(c) Improper: $\sec x\to \infty$ as $x\to \pi /2$.

Note: this diverges. Antiderivative is $\ln |\tan x+\sec x|\to \infty$ as $x\to \pi /2$.

(d) Improper: infinite upper bound.

Note: this diverges. Antiderivative is $-\cos x$ which has no limit as $x\to \infty$.

(e) Improper: infinite upper bound.

Note: this diverges. Antiderivative is $x\ln x-x\to -\infty$ as $x\to \infty$. (L’Hopital’s rule, indeterminate form.)

$$x\ln x-x=\frac{\ln x-1}{1/x}\stackrel{x\to \infty }{⟶}\frac{1/x}{-1/x^2}=-x\to -\infty$$

(f) Improper: infinite integrand at $x=0$.

Note: this converges. Antiderivative is $x\ln x-x\to 0$ as $x\to 0^{+}$. (Same indeterminate form as (e).)