10. 이상적분Improper Integrals

Solution.

(1) $\displaystyle\int_0^\infty e^{-x/2}\,dx = \lim_{b\to\infty}\int_0^b e^{-x/2}\,dx = \lim_{b\to\infty}\left[-2e^{-x/2}\right]_0^b = \lim_{b\to\infty}\left(2-2e^{-b/2}\right) = 2$.

(2) Use integration by parts with $u=\ln x$, $dv=x^{-2}\,dx$, so $du=dx/x$, $v=-1/x$: $$\int \frac{\ln x}{x^2}\,dx = -\frac{\ln x}{x} + \int \frac{dx}{x^2} = -\frac{\ln x}{x} - \frac1x + C.$$ Hence $$\int_1^\infty \frac{\ln x}{x^2}\,dx = \lim_{b\to\infty}\left[-\frac{\ln x}{x}-\frac1x\right]_1^b = \lim_{b\to\infty}\left(1-\frac{\ln b}{b}-\frac1b\right) = 1,$$ where $\lim_{b\to\infty}\dfrac{\ln b}{b}=0$ follows from L'Hôpital's Rule.

Solution.

(1) Split $\displaystyle\int_{-\infty}^\infty \frac{dx}{1+x^2}$ at $x=0$ and check that each half converges. $\displaystyle\int_0^\infty\frac{dx}{1+x^2}=\lim_{b\to\infty}[\arctan x]_0^b=\frac\pi2$ and $\displaystyle\int_{-\infty}^0\frac{dx}{1+x^2}=\lim_{a\to-\infty}[\arctan x]_a^0=\frac\pi2$, so both converge and the original integral converges to $\dfrac\pi2+\dfrac\pi2=\pi$.

(2) For $p\neq1$, $\displaystyle\int_1^b x^{-p}\,dx=\left[\frac{x^{1-p}}{1-p}\right]_1^b=\frac{b^{1-p}-1}{1-p}$. If $p\gt1$, then $1-p\lt0$, so $b^{1-p}\to0$ as $b\to\infty$, and the integral converges to $\dfrac{1}{p-1}$. If $p\lt1$, then $1-p\gt0$, so $b^{1-p}\to\infty$ and it diverges. If $p=1$, $\displaystyle\int_1^b\frac{dx}{x}=\ln b\to\infty$, so it diverges. Hence the integral converges only for $p\gt1$, with value $\dfrac1{p-1}$.

(3) $\displaystyle\int_{-\infty}^2\frac{2\,dx}{x^2+4}=\lim_{a\to-\infty}\left[\arctan\frac x2\right]_a^2=\arctan1-\lim_{a\to-\infty}\arctan\frac a2=\frac\pi4-\left(-\frac\pi2\right)=\frac{3\pi}4$.

(4) Split at $x=0$. Substituting $u=x^2+4$, $du=2x\,dx$, $$\int\frac{x\,dx}{(x^2+4)^{3/2}}=\frac12\int u^{-3/2}\,du=-u^{-1/2}=-\frac1{\sqrt{x^2+4}}.$$ Then $\displaystyle\int_0^\infty\frac{x\,dx}{(x^2+4)^{3/2}}=\lim_{b\to\infty}\left[-\frac1{\sqrt{x^2+4}}\right]_0^b=0-\left(-\frac12\right)=\frac12$ and $\displaystyle\int_{-\infty}^0\frac{x\,dx}{(x^2+4)^{3/2}}=\lim_{a\to-\infty}\left[-\frac1{\sqrt{x^2+4}}\right]_a^0=-\frac12-0=-\frac12$. Both halves converge, so the original integral converges, with value $\dfrac12+\left(-\dfrac12\right)=0$.

Proof.

For $b\gt0$, $$\int_{-b}^b \frac{2x\,dx}{x^2+1} = \left[\ln(x^2+1)\right]_{-b}^b = \ln(b^2+1)-\ln(b^2+1) = 0,$$ so $\displaystyle\lim_{b\to\infty}\int_{-b}^b\frac{2x\,dx}{x^2+1}=0$. This holds automatically on any symmetric interval because the integrand is an odd function; such a symmetric limit is called the Cauchy principal value.

On the other hand, $\displaystyle\int_0^\infty\frac{2x\,dx}{x^2+1}=\lim_{b\to\infty}\left[\ln(x^2+1)\right]_0^b=\lim_{b\to\infty}\ln(b^2+1)=\infty$, so it diverges.

The improper integral $\displaystyle\int_{-\infty}^\infty f(x)\,dx$ is defined to converge only when the two halves obtained by splitting at some point (here $x=0$), $\displaystyle\int_{-\infty}^0 f(x)\,dx$ and $\displaystyle\int_0^\infty f(x)\,dx$, each converge independently. Since $\displaystyle\int_0^\infty\frac{2x\,dx}{x^2+1}$ diverges (and by the same reasoning so does $\displaystyle\int_{-\infty}^0\frac{2x\,dx}{x^2+1}$), the integral $\displaystyle\int_{-\infty}^\infty\frac{2x\,dx}{x^2+1}$ diverges by definition. Thus the existence of a finite symmetric limit (the Cauchy principal value) does not guarantee convergence of the improper integral itself.

Solution.

(1) $\displaystyle\int_3^\infty e^{-3x}\,dx=\lim_{b\to\infty}\left[-\frac13e^{-3x}\right]_3^b=\lim_{b\to\infty}\left(-\frac13e^{-3b}+\frac13e^{-9}\right)=\frac{e^{-9}}3$.

(2) The error is $\displaystyle\int_0^\infty e^{-x^2}\,dx-\int_0^3e^{-x^2}\,dx=\int_3^\infty e^{-x^2}\,dx$. Since $x^2\geq3x$ for $x\geq3$, we have $e^{-x^2}\leq e^{-3x}$, so by the Comparison Test and part (1), $$\int_3^\infty e^{-x^2}\,dx\leq\int_3^\infty e^{-3x}\,dx=\frac{e^{-9}}3.$$ Hence the error bound is $\dfrac{e^{-9}}3$.

Solution. For large $x$, $\dfrac{ax}{x^2+1}\sim\dfrac ax$, so the integrand can be written as $$\frac{ax}{x^2+1}-\frac1{2x}=\frac{2ax^2-(x^2+1)}{2x(x^2+1)}=\frac{(2a-1)x^2-1}{2x(x^2+1)},$$ which is asymptotically comparable to $\dfrac{2a-1}{2x}$ as $x\to\infty$. If $a\neq\dfrac12$, then $2a-1\neq0$, and since $\displaystyle\int_1^\infty\frac{dx}{2x}$ diverges, the Limit Comparison Test shows the given integral also diverges. Hence convergence requires $a=\dfrac12$.

When $a=\dfrac12$, $$\int_1^\infty\left(\frac x{2(x^2+1)}-\frac1{2x}\right)dx=\lim_{b\to\infty}\left[\frac14\ln(x^2+1)-\frac12\ln x\right]_1^b=\lim_{b\to\infty}\left(\frac14\ln\frac{b^2+1}{b^2}-\frac14\ln2\right).$$ As $b\to\infty$, $\dfrac{b^2+1}{b^2}\to1$, so $\ln\dfrac{b^2+1}{b^2}\to0$, and the integral converges to $-\dfrac{\ln2}4$.

Solution. $\displaystyle\int_0^1\frac{dx}{\sqrt x}=\lim_{a\to0^+}\int_a^1 x^{-1/2}\,dx=\lim_{a\to0^+}\left[2\sqrt x\right]_a^1=\lim_{a\to0^+}\left(2-2\sqrt a\right)=2$.

Solution.

(1) $\displaystyle\int_0^1\frac{dx}{1-x}=\lim_{b\to1^-}\left[-\ln(1-x)\right]_0^b=\lim_{b\to1^-}\left(-\ln(1-b)\right)=\infty$, so it diverges.

(2) The integrand has a singularity at $x=1$, so split into $\displaystyle\int_0^1$ and $\displaystyle\int_1^3$. Since $\displaystyle\int(x-1)^{-2/3}\,dx=3(x-1)^{1/3}$, $$\int_0^1(x-1)^{-2/3}\,dx=\lim_{a\to1^-}\left[3(x-1)^{1/3}\right]_0^a=0-3(-1)=3,$$ $$\int_1^3(x-1)^{-2/3}\,dx=\lim_{a\to1^+}\left[3(x-1)^{1/3}\right]_a^3=3\sqrt[3]2-0=3\sqrt[3]2.$$ Both halves converge, so the original integral converges to $3+3\sqrt[3]2=3(1+\sqrt[3]2)$.

(3) Substituting $u=\sqrt x$, so $x=u^2$, $dx=2u\,du$, $$\frac{dx}{(1+x)\sqrt x}=\frac{2u\,du}{(1+u^2)u}=\frac{2\,du}{1+u^2}.$$ As $x$ ranges over $0\to\infty$, $u$ ranges over $0\to\infty$, so $$\int_0^\infty\frac{dx}{(1+x)\sqrt x}=\int_0^\infty\frac{2\,du}{1+u^2}=2\left[\arctan u\right]_0^\infty=2\cdot\frac\pi2=\pi.$$

(4) Substituting $u=r^2$, $du=2r\,dr$, $$\frac{4r\,dr}{\sqrt{1-r^4}}=\frac{2\,du}{\sqrt{1-u^2}}.$$ As $r$ ranges over $0\to1$, $u$ ranges over $0\to1$, so $$\int_0^1\frac{4r\,dr}{\sqrt{1-r^4}}=\int_0^1\frac{2\,du}{\sqrt{1-u^2}}=2\left[\arcsin u\right]_0^1=2\cdot\frac\pi2=\pi.$$

Solution.

(1) Substituting $u=\ln x$, $du=dx/x$; as $x$ ranges over $1\to2$, $u$ ranges over $0\to\ln2$, so $$\int_1^2\frac{dx}{x(\ln x)^p}=\int_0^{\ln2}u^{-p}\,du.$$ This is a $p$-integral with a singularity at $u=0$, and $\displaystyle\int_0^a u^{-p}\,du$ ($a\gt0$) converges only when $p\lt1$. Hence the original integral converges for $p\lt1$.

(2) Similarly, substituting $u=\ln x$, as $x$ ranges over $2\to\infty$, $u$ ranges over $\ln2\to\infty$, so $$\int_2^\infty\frac{dx}{x(\ln x)^p}=\int_{\ln2}^\infty u^{-p}\,du.$$ Since $\ln2\gt0$, the right-hand side is a $p$-integral over an infinite interval, which converges only when $p\gt1$. Hence the original integral converges for $p\gt1$.

Solution.

(1) For all $x$, $0\leq\sin^2x\leq1$, so $\displaystyle0\leq\frac{\sin^2x}{x^2}\leq\frac1{x^2}$ for $x\geq1$. Since $\displaystyle\int_1^\infty\frac{dx}{x^2}$ is a convergent $p$-integral ($p=2\gt1$), the Comparison Test shows $\displaystyle\int_1^\infty\frac{\sin^2x}{x^2}\,dx$ converges.

(2) As $x\to\infty$, $\dfrac1{\sqrt{x^2-0.1}}\sim\dfrac1x$. Since $\displaystyle\lim_{x\to\infty}\frac{1/\sqrt{x^2-0.1}}{1/x}=\lim_{x\to\infty}\frac x{\sqrt{x^2-0.1}}=1$ and $\displaystyle\int_1^\infty\frac{dx}x$ diverges, the Limit Comparison Test shows $\displaystyle\int_1^\infty\frac{dx}{\sqrt{x^2-0.1}}$ diverges.

(3) The integrand has a singularity at $x=0$. On $[0,\pi/2]$, $\cos x\geq0$ and $0\leq\cos x\leq1$, so $0\leq\dfrac{\cos x}{\sqrt x}\leq\dfrac1{\sqrt x}$ for $0\lt x\leq\pi/2$. Since $\displaystyle\int_0^{\pi/2}\frac{dx}{\sqrt x}$ is a convergent $p$-integral ($p=1/2\lt1$), the Comparison Test shows $\displaystyle\int_0^{\pi/2}\frac{\cos x}{\sqrt x}\,dx$ converges.

(4) Directly, $\displaystyle\int_1^\infty\frac{dx}{1+x^2}=\lim_{b\to\infty}[\arctan x]_1^b=\frac\pi2-\frac\pi4=\frac\pi4$, so it converges.

(5) As $x\to\infty$, $1-e^{-x}\to1$, so $\dfrac{1-e^{-x}}x\sim\dfrac1x$. Since $\displaystyle\lim_{x\to\infty}\frac{(1-e^{-x})/x}{1/x}=\lim_{x\to\infty}(1-e^{-x})=1$ and $\displaystyle\int_1^\infty\frac{dx}x$ diverges, the Limit Comparison Test shows $\displaystyle\int_1^\infty\frac{1-e^{-x}}x\,dx$ diverges.

Solution.

(1) The integrand has a singularity at $x=1$ (since $\ln1=0$). Since $\displaystyle\lim_{x\to1}\frac{1/(x\ln x)}{1/(x-1)}=\lim_{x\to1}\frac{x-1}{x\ln x}$, and by L'Hôpital's Rule this equals $\displaystyle\lim_{x\to1}\frac1{\ln x+1}=1$. Both $\displaystyle\int_{1/2}^1\frac{dx}{x-1}$ and $\displaystyle\int_1^2\frac{dx}{x-1}$ diverge near $x=1$ (substituting $u=x-1$ gives $\int u^{-1}\,du$), so by the Limit Comparison Test, $\displaystyle\int_{1/2}^1\frac{dx}{x\ln x}$ and $\displaystyle\int_1^2\frac{dx}{x\ln x}$ each diverge as well. Hence $\displaystyle\int_{1/2}^2\frac{dx}{x\ln x}$ diverges.

(2) $\tan\theta$ has a singularity at $\theta=\pi/2$. Writing $\varepsilon=\pi/2-\theta\to0^+$ as $\theta\to(\pi/2)^-$, we have $\tan\theta=\cot\varepsilon=\dfrac{\cos\varepsilon}{\sin\varepsilon}$, so $$\lim_{\theta\to(\pi/2)^-}\frac{\tan\theta}{1/(\pi/2-\theta)}=\lim_{\varepsilon\to0^+}\varepsilon\cdot\frac{\cos\varepsilon}{\sin\varepsilon}=\lim_{\varepsilon\to0^+}\cos\varepsilon\cdot\frac\varepsilon{\sin\varepsilon}=1.$$ Since $\displaystyle\int_0^{\pi/2}\frac{d\theta}{\pi/2-\theta}$ diverges near $\theta=\pi/2$, the Limit Comparison Test shows $\displaystyle\int_0^{\pi/2}\tan\theta\,d\theta$ diverges as well.

(3) As $t\to\infty$, $t^{3/2}-1\sim t^{3/2}$, so $\dfrac2{t^{3/2}-1}\sim\dfrac2{t^{3/2}}$. Since $\displaystyle\lim_{t\to\infty}\frac{2/(t^{3/2}-1)}{2/t^{3/2}}=\lim_{t\to\infty}\frac{t^{3/2}}{t^{3/2}-1}=1$ and $\displaystyle\int_4^\infty\frac{2\,dt}{t^{3/2}}$ is a convergent $p$-integral ($p=3/2\gt1$), the Limit Comparison Test shows $\displaystyle\int_4^\infty\frac{2\,dt}{t^{3/2}-1}$ converges.

(4) Since $e\gt2$, $(e/2)^x\gt1$ for $x\gt0$, i.e., $e^x\gt2^x$, so $e^x-2^x\gt0$ for $x\geq1$ and the integrand has no singularity. As $x\to\infty$, $$\lim_{x\to\infty}\frac{1/(e^x-2^x)}{e^{-x}}=\lim_{x\to\infty}\frac{e^x}{e^x-2^x}=\lim_{x\to\infty}\frac1{1-(2/e)^x}=1$$ (since $2/e\lt1$, $(2/e)^x\to0$). Since $\displaystyle\int_1^\infty e^{-x}\,dx$ converges, the Limit Comparison Test shows $\displaystyle\int_1^\infty\frac{dx}{e^x-2^x}$ converges.

Solution.

(1) As $t\to0^+$, $\dfrac{\sin t}t\to1$, so $t=0$ is not actually a singularity (a removable discontinuity), and $\displaystyle\int_0^1\frac{\sin t}t\,dt$ is an ordinary, finite definite integral. The issue is at $t\to\infty$. Integrating by parts with $u=1/t$, $dv=\sin t\,dt$, so $du=-t^{-2}\,dt$, $v=-\cos t$, $$\int_1^b\frac{\sin t}t\,dt=\left[-\frac{\cos t}t\right]_1^b-\int_1^b\frac{\cos t}{t^2}\,dt=\cos1-\frac{\cos b}b-\int_1^b\frac{\cos t}{t^2}\,dt.$$ As $b\to\infty$, $\left|\dfrac{\cos b}b\right|\leq\dfrac1b\to0$, and $\left|\dfrac{\cos t}{t^2}\right|\leq\dfrac1{t^2}$ with $\displaystyle\int_1^\infty\frac{dt}{t^2}$ convergent, so by the Comparison Test, $\displaystyle\int_1^\infty\frac{\cos t}{t^2}\,dt$ converges (absolutely) as well. Hence the right-hand side has a limit as $b\to\infty$, so $\displaystyle\int_1^\infty\frac{\sin t}t\,dt$ converges, and combined with the definite integral on $[0,1]$, $\displaystyle\int_0^\infty\frac{\sin t}t\,dt$ converges. (Since $\displaystyle\int_1^\infty\left|\frac{\sin t}t\right|dt$ is known to diverge, this convergence is conditional, not absolute.)

(2) Since $t^2\geq t$ for $t\geq1$, $e^{-t^2}\leq e^{-t}$, and $\displaystyle\int_1^\infty e^{-t}\,dt$ converges, so by the Comparison Test, $\displaystyle\int_1^\infty e^{-t^2}\,dt$ converges. Since $\displaystyle\int_0^1 e^{-t^2}\,dt$ is a finite definite integral, combining gives that $\displaystyle\int_0^\infty e^{-t^2}\,dt$ converges.