Exercises — Improper integrals — Type I (infinite limits), Type II (discontinuous integrand)

${\int }_1^{\infty }\frac{dx}{x^p}\text{converges}⟺p>1,{\int }_0^1\frac{dx}{x^p}\text{converges}⟺p<1.$ "Big at infinity, small at zero." $p=1$ diverges in both cases.

The test: does the integrand blow up anywhere on the closed interval (Type II), and/or is a limit infinite (Type I)?

• (a) $\frac{1}{x-1}$ blows up at $x=1$, but $1\notin [2,5]$. Interval is finite. → Proper (ordinary integral, nothing improper).
• (b) Upper limit is $\infty$; integrand fine on $[1,\infty )$. → Type I, bad point at $+\infty$.
• (c) $\frac{1}{x-2}\to \pm \infty$ as $x\to 2$, and $2$ is the right endpoint of $[0,2]$. → Type II, bad point $x=2^{-}$.
• (d) Upper limit $\infty$ and $\frac{1}{\sqrt{x}}\to \infty$ as $x\to 0^{+}$ (left endpoint). → Both (Type I at $\infty$, Type II at $0$). Must split, e.g. ${\int }_0^1+{\int }_1^{\infty }$.

$x^{2/3}={\left(\sqrt[3]{x}\right)}^2\ge 0$ and equals $0$ only at $x=0$. So $\frac{1}{x^{2/3}}\to +\infty$ as $x\to 0$. Since $0$ is interior to $[-1,1]$, this is a Type II integral with an interior singularity at $x=0$. You must split at $0$ and check both halves. (Skipping the split is the classic error — see below.)

Type I. Replace $\infty$ by $t$: ${\int }_0^t{e}^{-2x}dx={\left[-\frac{1}{2}{e}^{-2x}\right]}_0^t=-\frac{1}{2}{e}^{-2t}+\frac{1}{2}.$ As $t\to \infty$, $e^{-2t}\to 0$, so the value is $\boxed{\frac{1}{2}}$. Converges.

Type I, $p=\frac{3}{2}>1$ → expect convergence. Directly: ${\int }_1^t{x}^{-3/2}dx={\left[\frac{x^{-1/2}}{-1/2}\right]}_1^t={\left[-\frac{2}{\sqrt{x}}\right]}_1^t=-\frac{2}{\sqrt{t}}+2.$ As $t\to \infty$, $\frac{2}{\sqrt{t}}\to 0$, giving $\boxed{2}$. (Cross-check with the formula $\frac{1}{p-1}=\frac{1}{3/2-1}=2$ ✓.)

The integrand $\frac{1}{\sqrt{4-x}}\to \infty$ as $x\to 4^{-}$ (right endpoint). Type II. Creep up with $t\to 4^{-}$: ${\int }_0^t(4-x{)}^{-1/2}dx.$ Let $u=4-x$, $du=-dx$. Antiderivative: $-(-2)(4-x{)}^{1/2}=...$ carefully — $\int (4-x{)}^{-1/2}dx=-2\sqrt{4-x}$. ${\left[-2\sqrt{4-x}\right]}_0^t=-2\sqrt{4-t}+2\sqrt{4}=-2\sqrt{4-t}+4.$ As $t\to 4^{-}$, $\sqrt{4-t}\to 0$, so value $=\boxed{4}$. Converges.

$\ln x\to -\infty$ as $x\to 0^{+}$: Type II at the left end. Creep with $t\to 0^{+}$. Antiderivative (by parts): $\int \ln xdx=x\ln x-x$. ${\int }_t^1\ln xdx={\left[x\ln x-x\right]}_t^1=(0-1)-(t\ln t-t)=-1-t\ln t+t.$ As $t\to 0^{+}$: $t\to 0$, and $t\ln t\to 0$ (the $t$ crushes the slow $\ln t\to -\infty$; see Limits at Infinity). So value $=\boxed{-1}$. Converges. (A negative answer is fine here — $\ln x<0$ on $(0,1)$, so the "area" is genuinely below the axis.)

Two routes. Route A (comparison, quick verdict): for $x\ge 1$, $x^2+x\ge x^2$, so $\frac{1}{x^2+x}\le \frac{1}{x^2}$, and ${\int }_1^{\infty }{x}^{-2}dx$ converges ($p=2>1$). By the Comparison Test for Integrals, the smaller non-negative integral converges. Route B (exact value): partial fractions $\frac{1}{x^2+x}=\frac{1}{x}-\frac{1}{x+1}$. ${\int }_1^t\left(\frac{1}{x}-\frac{1}{x+1}\right)dx={\left[\ln \frac{x}{x+1}\right]}_1^t=\ln \frac{t}{t+1}-\ln \frac{1}{2}.$ As $t\to \infty$, $\frac{t}{t+1}\to 1$ so $\ln \to 0$. Value $=0-\ln \frac{1}{2}=\ln 2\approx 0.6931$. So $\boxed{\ln 2}$.

Since $-1\le \sin x\le 1$, the numerator satisfies $2+\sin x\ge 1>0$. Thus $\frac{2+\sin x}{x}\ge \frac{1}{x}(x\ge 1).$ But ${\int }_1^{\infty }\frac{dx}{x}$ diverges ($p=1$). A non-negative function larger than a divergent one must also diverge (its area is even bigger). → Diverges. Direction matters: to prove divergence you bound below by something divergent; to prove convergence you bound above by something convergent.

Bad point at $x=0$ (denominator $\to 0$): Type II. Near $0$, $x^2$ is negligible next to $\sqrt{x}$, so $\sqrt{x}+x^2\ge \sqrt{x}$, giving $\frac{1}{\sqrt{x}+x^2}\le \frac{1}{\sqrt{x}}=x^{-1/2}.$ ${\int }_0^1{x}^{-1/2}dx$ converges ($p=\frac{1}{2}<1$, value $2$). By comparison, the smaller non-negative integral converges.

Both limits infinite → split at a convenient $c=0$ and check each half. Antiderivative: $\arctan x$. ${\int }_0^{\infty }\frac{dx}{1+x^2}={\lim }_{t\to \infty }[\arctan x{]}_0^t={\lim }_{t\to \infty }\arctan t-0=\frac{\pi }{2}.$ ${\int }_{-\infty }^0\frac{dx}{1+x^2}={\lim }_{s\to -\infty }[\arctan x{]}_s^0=0-\left(-\frac{\pi }{2}\right)=\frac{\pi }{2}.$ Both halves converge, so the whole integral converges to $\frac{\pi }{2}+\frac{\pi }{2}=\boxed{\pi }$.

Type II at $0$ ($\frac{1}{\sqrt{x}}\to \infty$) and Type I at $\infty$. Split at $1$: ${\int }_0^1+{\int }_1^{\infty }$. Substitution unifies both. Let $x=u^2$, $dx=2udu$, $\sqrt{x}=u$. Then $\int \frac{2udu}{u(1+u^2)}=\int \frac{2du}{1+u^2}=2\arctan u=2\arctan \sqrt{x}.$ As $x$ runs $0\to \infty$, $\sqrt{x}$ runs $0\to \infty$, and $\arctan$ runs $0\to \frac{\pi }{2}$: ${\int }_0^{\infty }\frac{dx}{\sqrt{x}(1+x)}={\left[2\arctan \sqrt{x}\right]}_0^{\infty }=2\cdot \frac{\pi }{2}-0=\boxed{\pi }.$ Both endpoint behaviours are tame under the substitution → converges cleanly.

This is a trap in disguise. Split at $1$: ${\int }_0^1{x}^{-p}dx+{\int }_1^{\infty }{x}^{-p}dx$.

• ${\int }_0^1{x}^{-p}$ converges $⟺p<1$ (Type II rule).
• ${\int }_1^{\infty }{x}^{-p}$ converges $⟺p>1$ (Type I rule). These two conditions are mutually exclusive — no single $p$ satisfies both. Therefore ${\int }_0^{\infty }{x}^{-p}dx$ diverges for every real $p$. The picture below shows why: whichever end you tame, the other end blows up.

Split at $0$. The right half ${\int }_0^{\infty }xdx={\lim }_{t\to \infty }\frac{t^2}{2}=\infty$ diverges. Since one piece diverges, the whole integral diverges — full stop. The tempting "symmetric limit" ${\lim }_{t\to \infty }{\int }_{-t}^{t}xdx={\lim }_{t\to \infty }{\left[\frac{x^2}{2}\right]}_{-t}^t={\lim }_{t\to \infty }\left(\frac{t^2}{2}-\frac{t^2}{2}\right)=0$. This clean $0$ is the Cauchy principal value, not convergence: it only works because the growing negative and positive areas are forced to cancel at the same rate. Genuine convergence demands each half settle on its own. Answer: diverges (PV $=0$ is a different, weaker notion).

Type I. Integrate by parts, $u=x,dv=e^{-x}dx$: $\int x{e}^{-x}dx=-x{e}^{-x}-e^{-x}=-(x+1)e^{-x}$. ${\int }_0^{t}x{e}^{-x}dx={\left[-(x+1)e^{-x}\right]}_0^t=-(t+1)e^{-t}+1.$ As $t\to \infty$, $(t+1)e^{-t}\to 0$ (exponential beats the linear factor). Value $=\boxed{1}$. Connection: the Gamma Function is $\Gamma (n)={\int }_0^{\infty }{x}^{n-1}{e}^{-x}dx$, and $\Gamma (n)=(n-1)!$. Here $n=2$, so this is $\Gamma (2)=1!=1$ ✓. This same integral also gives the mean of the exponential distribution with rate $1$.

${\int }_0^t{e}^{-sx}dx={\left[-\frac{1}{s}{e}^{-sx}\right]}_0^t=-\frac{1}{s}{e}^{-st}+\frac{1}{s}.$

• If $s>0$: $e^{-st}\to 0$ as $t\to \infty$, value $=\boxed{\frac{1}{s}}$. Converges.
• If $s=0$: integrand is $1$, ${\int }_0^{t}1dx=t\to \infty$. Diverges.
• If $s<0$: exponent $-sx>0$ grows, $e^{-st}\to \infty$. Diverges. So ${\int }_0^{\infty }{e}^{-sx}dx=\frac{1}{s}$ only for $s>0$ — this is exactly $\mathcal{L}\{1\}(s)=\frac{1}{s}$, the Laplace Transform of the constant function $1$, valid precisely on its region of convergence.

${\int }_1^{\infty }{x}^{-3/2}dx$ ::: converges to $2$ ($p=3/2>1$) ${\int }_0^1\ln xdx$ ::: converges to $-1$ ${\int }_{-\infty }^{\infty }\frac{dx}{1+x^2}$ ::: converges to $\pi$ ${\int }_0^{\infty }{x}^{-p}dx$ ::: diverges for all real $p$ ${\int }_0^{\infty }{e}^{-sx}dx$ ::: equals $1/s$ for $s>0$, diverges for $s\le 0$ ${\int }_{-\infty }^{\infty }xdx$ ::: diverges (PV $=0$ is not convergence)