Improper integrals

Type 1: infinite intervals.

  • Let \(f(x)\) be a function, which has at most finite discontinuities on the interval \([a,b]\). Then the definite integral \[ \int_a^bf(x)\,\mathrm dx \] exists, and its value is the signed area between the graph \(y=f(x)\) and the \(x\)-axis \(y=0\).
    • Now we'll see how to define and calculate areas over infinite intervals.

Type 1: infinite intervals.

  • Definition.
    • If \(\int_a^tf(x)\,\mathrm dx\) exists for all \(t\ge a\), then \[ \int_a^\infty f(x)\,\mathrm dx=\lim_{t\to\infty}\int_a^tf(x)\,\mathrm dx, \] provided this limit exists (as a finite number).
    • If \(\int_t^bf(x)\,\mathrm dx\) exists for all \(t\le b\), then \[ \int_{-\infty}^b f(x)\,\mathrm dx=\lim_{t\to-\infty}\int_t^bf(x)\,\mathrm dx, \] provided this limit exists (as a finite number).
    • The integrals \(\int_a^\infty f(x)\,\mathrm dx\) and \(\int_{-\infty}^bf(x)\,\mathrm dx\) are called improper integrals. If the corresponding limit exists, then we say that the integral is convergent. Otherwise, we say it's divergent.
    • If both \(\int_a^\infty f(x)\,\mathrm dx\) and \(\int_{-\infty}^bf(x)\,\mathrm dx\) are convergent, then we let \[ \int_{-\infty}^\infty f(x)\,\mathrm dx=\int_{-\infty}^af(x)\,\mathrm dx+\int^\infty_af(x)\,\mathrm dx. \] It can be proven that the value is independent of the choice of \(a\) (7.8.76).

Type 1: infinite intervals.

  • \(\int_1^\infty\frac{1}{x^3}\,\mathrm dx\).
    • For \(t\ge1\), we have \[ \int_1^t\frac{1}{x^3}\,\mathrm dx=\left[-\frac{1}{2x^2}\right]_1^t=\frac12-\frac{1}{2t^2}. \]
    • Therefore, we get \[ \int_1^\infty\frac{1}{x^3}\,\mathrm dx=\lim_{t\to\infty}\frac{1}{x^3}\,\mathrm dx=\lim_{t\to\infty}\left(\frac12-\frac{1}{2t^2}\right)=\frac12, \] that is the improper integral is convergent.
  • \(\int_1^\infty\frac{1}{\sqrt x}\,\mathrm dx\).
    • For \(t\ge1\), we have \[ \int_1^\infty\frac{1}{\sqrt x}\,\mathrm dx=\left[2\sqrt x\right]_1^t=2t-2. \]
    • Therefore, we get \[ \int_1^\infty\frac{1}{\sqrt x}\,\mathrm dx=\lim_{t\to\infty}\int_1^t\frac{1}{\sqrt x}\,\mathrm dx=\lim_{t\to\infty}(2t-2)=\infty, \] that is the improper integral is divergent.

Type 1: infinite intervals.

  • For which \(p\) is the integral \(\int_1^\infty x^p\,\mathrm dx\) convergent?
    • If \(p\ne-1\), then we have \[ \int_1^\infty x^p\,\mathrm dx=\lim_{t\to\infty}\left[\frac{x^{p+1}}{p+1}\right]_1^t=\lim_{t\to\infty}\frac{t^{p+1}-1}{p+1}. \] This is convergent precisely when \(p<-1\).
    • If \(p=-1\), then we have \[ \int_1^\infty x^{-1}\,\mathrm dx=\lim_{t\to\infty}\left[\ln x\right]_1^t=\lim_{t\to\infty}\ln t=\infty. \]
    • Therefore, the improper integral \(\int_1^\infty x^p\,\mathrm dx\) is convergent precisely when \(p<1\).

Type 1: infinite intervals.

  • \(\int_{-\infty}^02^x\,\mathrm dx\).
    • We have \[ \int_{-\infty}^02^x\,\mathrm dx=\frac{1}{\ln2}\lim_{t\to-\infty}\left(\left[x2^x\right]_t^0-\int_t^02^x\,\mathrm dx\right)=\lim_{t\to-\infty}\left(\frac{t2^t}{\ln 2}-\frac{2^t-1}{(\ln2)^2}\right)=\frac{1}{(\ln2)^2}, \] therefore the integral is convergent.
  • \(\int_{-\infty}^\infty\frac{1}{\sqrt{1+x^2}}\,\mathrm dx\).
    • We have \[ \int\frac{1}{\sqrt{1+x^2}}\,\mathrm dx\stackrel{x=\tan\theta}{=}\int\frac{\sec^2\theta}{\sec\theta}\,\mathrm d\theta=\ln|\tan\theta+\sec\theta|+C=\ln|x+\sqrt{1+x^2}|+C, \] which gives \[ \int_0^\infty\frac{1}{\sqrt{1+x^2}}\,\mathrm dx=\lim_{t\to\infty}\left[\ln|x+\sqrt{1+x^2}|\right]_0^t=\lim_{t\to\infty}\left(\ln(t+\sqrt{1+t^2})\right)=\infty, \] therefore the improper integral is divergent.
  • Exercises. 7.8: 3, 5, 11, 15, 21, 25

Type 2: Discontinuous integrals

  • Another way to get improper integrals is to integrate functions with infinite discontinuities.
  • Definition.
    • If \(f\) is continuous on \([a,b)\), and it has an infinite discontinuity at \(b\), then \[ \int_a^bf(x)\,\mathrm dx=\lim_{t\to b}\int_a^tf(x)\,\mathrm dx, \] provided that the limit exists (as a finite number).
    • If \(f\) is continuous on \((a,b]\), and it has an infinite discontinuity at \(a\), then \[ \int_a^bf(x)|,\mathrm dx=\lim_{t\to a}\int_t^bf(x)\,\mathrm dx, \] provided that the limit exists (as a finite number).
    • In these cases, the integral \(\int_a^bf(x)\,\mathrm dx\) is called an improper integral. If the limits exists, then it is convergent, and if the limit does not exist, then it is divergent.
    • If \(f\) has an infinite discontinuity at \(c\) for \(a<c<b\), and the improper integrals \(\int_a^cf(x),\mathrm dx\) and \(\int_c^bf(x)\,\mathrm dx\) are convergent, then we let \[ \int_a^bf(x)\,\mathrm dx=\int_a^cf(x)\,\mathrm dx+\int_c^bf(x)\,\mathrm dx. \]

Type 2: Discontinuous integrals

  • \(\int_{-1}^3\frac{1}{\sqrt{x+1}}\,\mathrm dx\).
    • Note that \(f(x)=\frac{1}{\sqrt{x+1}}\) has an infinite discontinuity at \(x=-3\).
    • For \(-1<t\le 3\), we have \[ \int_t^3\frac{1}{\sqrt{x+1}}\,\mathrm dx=\left[2\sqrt{x+1}\right]_t^3=4-2\sqrt{t+1}. \]
    • Therefore, we get \[ \int_{-1}^3\frac{1}{\sqrt{x+1}}=\lim_{t\to-1}(4-2\sqrt{t+1})=4, \] which shows that the improper integral is convergent.
  • \(\int_0^{\pi/2}\tan x\,\mathrm dx\).
    • Note that \(f(x)=\tan x\) has an infinite discontinuity at \(x=\frac{\pi}{2}\).
    • We have \[ \int_0^{\pi/2}\tan x\,\mathrm dx=\lim_{t\to\pi/2}\left[-\ln|\cos x|\right]_0^t=\lim_{t\to\pi/2}(-\ln(\cos t))=\infty, \] therefore the integral is divergent.

Type 2: Discontinuous integrals

  • \(\int_{-2}^2\frac{1}{x^2-1}\,\mathrm dx\).
    • Note that \(f(x)=\frac{1}{x^2-1}\) has infinite discontinuities at \(x=-1,1\).
    • To get an antiderivative, we can use partial fractions. To solve \[ 1=A(x+1)+B(x-1), \] we need \(A=\frac12\) and \(B=-\frac12\).
    • Therefore, we get \[ \int\frac{1}{x^2-1}\,\mathrm dx=\frac{\ln|x-1|-\ln|x+1|}{2}+C. \]
    • This in turn gives \[ \int_{-2}^{-1}\frac{1}{x^2-1}\,\mathrm dx=\frac12\lim_{t\to-1}\left[\ln|x-1|-\ln|x+1|\right]_{-2}^t=\frac12\lim_{t\to-1}(\ln|t-1|-\ln|t+1|)=-\infty, \] therefore the integral is divergent.
    • WARNING. Note that if we had not noticed the infinite discontinuities, then we would have gotten the wrong answer \[ \int_{-2}^2\frac{1}{x^2-1}\,\mathrm dx=\frac12\left[\ln|x-1|-\ln|x+1|\right]_{-2}^2=-\ln3. \]
  • Exercises. 7.8: 29, 35, 39

A comparison test for improper integrals

  • Theorem. Suppose that \(f(x)\) and \(g(x)\) are continuous functions with \(f(x)\ge g(x)\ge0\) for \(x\ge a\). Then we have the following.
    • If \(\int_a^\infty f(x)\,\mathrm dx\) is convergent, then so is \(\int_a^\infty g(x)\,\mathrm dx\).
    • If \(\int_a^\infty g(x)\,\mathrm dx\) is divergent, then so is \(\int_a^\infty f(x)\,\mathrm dx\).
  • \(\int_0^\infty e^{-x^2}\,\mathrm dx\).
    • We have mentioned that the antiderivative of \(e^{-x^2}\) is not an elementary function.
    • We can write \[ \int_0^\infty e^{-x^2}\,\mathrm dx=\int_0^1e^{-x^2}\,\mathrm dx+\int_1^\infty e^{-x^2}\,\mathrm dx. \]
    • If \(x\ge1\), then \(-x^2\le-x\), and therefore \(e^{-x^2}\le e^{-x}\).
    • Then as we have \[ \int_1^\infty e^{-x}\,\mathrm dx=\lim_{t\to\infty}(e^{-1}-e^{-t})=e^{-1}, \] the comparison theorem shows that \(\int_1^\infty e^{-x^2}\,\mathrm dx\) is convergent, and thus so is \(\int_0^\infty e^{-x^2}\,\mathrm dx\).
  • Exercises. 7.8: 49, 51, 53.
  • More exercises. 7.8: 55, 63, 73, 79, 81