Calculus Improper integrals Improper integrals extend integration to infinite intervals and to functions that blow up at an endpoint, by taking limits.
Basic, Standard, Advanced: University Year 1 · 1st semester
Math problem generator
Please enable JavaScript to generate problems.
Guide Examples Practice Related
Definition Integrals over infinite intervals, or of functions that blow up at an endpoint, are defined as limits:
∫ a ∞ f ( x ) d x = lim R → ∞ ∫ a R f ( x ) d x , ∫ 0 1 f ( x ) d x = lim ε → + 0 ∫ ε 1 f ( x ) d x \int_a^{\infty} f(x)\,dx = \lim_{R\to\infty}\int_a^R f(x)\,dx,\qquad \int_0^1 f(x)\,dx = \lim_{\varepsilon\to+0}\int_\varepsilon^1 f(x)\,dx ∫ a ∞ f ( x ) d x = R → ∞ lim ∫ a R f ( x ) d x , ∫ 0 1 f ( x ) d x = ε → + 0 lim ∫ ε 1 f ( x ) d x If the limit is finite, the improper integral converges ; otherwise it diverges .
The basic cases
For p = 1 p = 1 p = 1 log \log log
Limits you will use These limits come up again and again (k > 0 k > 0 k > 0 n n n
lim R → ∞ R n e − R = 0 , lim R → ∞ log R R k = 0 , lim ε → + 0 ε k log ε = 0 , lim R → ∞ arctan R = π 2 \lim_{R\to\infty} R^n e^{-R} = 0,\qquad \lim_{R\to\infty} \frac{\log R}{R^k} = 0,\qquad \lim_{\varepsilon\to+0} \varepsilon^k\log\varepsilon = 0,\qquad \lim_{R\to\infty}\arctan R = \frac{\pi}{2} R → ∞ lim R n e − R = 0 , R → ∞ lim R k log R = 0 , ε → + 0 lim ε k log ε = 0 , R → ∞ lim arctan R = 2 π Famous improper integrals ∫ 0 ∞ x n e − x d x = n ! ( gamma function Γ ( n + 1 ) ) , ∫ − ∞ ∞ e − x 2 d x = π ( Gaussian integral ) \int_0^{\infty} x^n e^{-x}\,dx = n!\quad(\text{gamma function } \Gamma(n+1)),\qquad \int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}\quad(\text{Gaussian integral}) ∫ 0 ∞ x n e − x d x = n ! ( gamma function Γ ( n + 1 )) , ∫ − ∞ ∞ e − x 2 d x = π ( Gaussian integral ) The Gaussian integral is evaluated with a double integral and polar coordinates (see polar coordinates on the double integrals page).
Worked examples
Evaluate the improper integral.
∫ 0 1 1 x 3 d x \int_{0}^{1} \frac{1}{\sqrt[3]{x}}\,dx ∫ 0 1 3 x 1 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint The integrand blows up at x = 0 x = 0 x = 0 ∫ ε 1 \int_\varepsilon^1 ∫ ε 1 ε → + 0 \varepsilon \to +0 ε → + 0
Answer ∫ 0 1 1 x 3 d x = 3 2 \int_{0}^{1} \frac{1}{\sqrt[3]{x}}\,dx = \frac{3}{2} ∫ 0 1 3 x 1 d x = 2 3 Solution Replace the lower limit by ε \varepsilon ε ε → + 0 \varepsilon \to +0 ε → + 0
∫ 0 1 1 x 3 d x = lim ε → + 0 [ 3 2 x 2 3 ] ε 1 = lim ε → + 0 ( 3 2 − 3 2 ε 2 3 ) = 3 2 \begin{aligned}\int_{0}^{1} \frac{1}{\sqrt[3]{x}}\,dx &= \lim_{\varepsilon \to +0}\left[\frac{3}{2}\sqrt[3]{x^{2}}\right]_{\varepsilon}^{1} \\ &= \lim_{\varepsilon \to +0}\left(\frac{3}{2} - \frac{3}{2}\sqrt[3]{\varepsilon^{2}}\right) \\ &= \frac{3}{2}\end{aligned} ∫ 0 1 3 x 1 d x = ε → + 0 lim [ 2 3 3 x 2 ] ε 1 = ε → + 0 lim ( 2 3 − 2 3 3 ε 2 ) = 2 3 In general, ∫ 0 1 d x x p \int_0^1 \frac{dx}{x^p} ∫ 0 1 x p d x p < 1 p < 1 p < 1 p ≥ 1 p \ge 1 p ≥ 1
Evaluate the improper integral.
∫ 0 ∞ x e − 2 x d x \int_{0}^{\infty} xe^{-2x}\,dx ∫ 0 ∞ x e − 2 x d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Integrate by parts and use lim R → ∞ R e − a R = 0 \lim_{R \to \infty} Re^{-aR} = 0 lim R → ∞ R e − a R = 0
Answer ∫ 0 ∞ x e − 2 x d x = 1 4 \int_{0}^{\infty} xe^{-2x}\,dx = \frac{1}{4} ∫ 0 ∞ x e − 2 x d x = 4 1 Solution Integrating by parts,
∫ 0 R x e − 2 x d x = [ − 1 2 x e − 2 x ] 0 R + 1 2 ∫ 0 R e − 2 x d x \int_0^R xe^{-2x}\,dx = \left[-\frac{1}{2}xe^{-2x}\right]_0^R + \frac{1}{2}\int_0^R e^{-2x}\,dx ∫ 0 R x e − 2 x d x = [ − 2 1 x e − 2 x ] 0 R + 2 1 ∫ 0 R e − 2 x d x As R → ∞ R \to \infty R → ∞ R e − a R → 0 Re^{-aR} \to 0 R e − a R → 0 e − a R → 0 e^{-aR} \to 0 e − a R → 0
∫ 0 ∞ x e − 2 x d x = 0 + 1 2 ⋅ 1 2 = 1 4 \int_0^\infty xe^{-2x}\,dx = 0 + \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} ∫ 0 ∞ x e − 2 x d x = 0 + 2 1 ⋅ 2 1 = 4 1
Evaluate the improper integral.
∫ 0 ∞ e − 3 x sin x d x \int_{0}^{\infty} e^{-3x}\sin x\,dx ∫ 0 ∞ e − 3 x sin x d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Integrate by parts twice to find an antiderivative, then use e − a R → 0 e^{-aR} \to 0 e − a R → 0 R → ∞ R\to\infty R → ∞
Answer ∫ 0 ∞ e − 3 x sin x d x = 1 10 \int_{0}^{\infty} e^{-3x}\sin x\,dx = \frac{1}{10} ∫ 0 ∞ e − 3 x sin x d x = 10 1 Solution Integrating by parts twice (the original integral reappears and you solve for it), an antiderivative is
∫ e − 3 x sin x d x = − e − 3 x 10 ( 3 sin x + cos x ) + C \int e^{-3x}\sin x\,dx = -\frac{e^{-3x}}{10}\left(3\sin x + \cos x\right) + C ∫ e − 3 x sin x d x = − 10 e − 3 x ( 3 sin x + cos x ) + C As R → ∞ R \to \infty R → ∞ e − a R → 0 e^{-aR} \to 0 e − a R → 0 sin \sin sin cos \cos cos
∫ 0 ∞ e − 3 x sin x d x = 0 − ( − 1 10 ) = 1 10 \int_0^{\infty} e^{-3x}\sin x\,dx = 0 - \left(-\frac{1}{10}\right) = \frac{1}{10} ∫ 0 ∞ e − 3 x sin x d x = 0 − ( − 10 1 ) = 10 1 Practice problems
Evaluate the improper integral.
∫ 1 ∞ 1 x 2 x d x \int_{1}^{\infty} \frac{1}{x^{2}\sqrt{x}}\,dx ∫ 1 ∞ x 2 x 1 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Think of ∫ 1 ∞ \int_1^\infty ∫ 1 ∞ lim R → ∞ ∫ 1 R \lim_{R\to\infty}\int_1^R lim R → ∞ ∫ 1 R p > 1 p > 1 p > 1
Answer ∫ 1 ∞ 1 x 2 x d x = 2 3 \int_{1}^{\infty} \frac{1}{x^{2}\sqrt{x}}\,dx = \frac{2}{3} ∫ 1 ∞ x 2 x 1 d x = 3 2 Solution Replace the upper limit by R R R R → ∞ R \to \infty R → ∞
∫ 1 ∞ 1 x 2 x d x = lim R → ∞ [ − 2 3 x x ] 1 R = lim R → ∞ ( − 2 3 R R + 2 3 ) = 2 3 \begin{aligned}\int_{1}^{\infty} \frac{1}{x^{2}\sqrt{x}}\,dx &= \lim_{R \to \infty}\left[-\frac{2}{3x\sqrt{x}}\right]_{1}^{R} \\ &= \lim_{R \to \infty}\left(-\frac{2}{3R\sqrt{R}} + \frac{2}{3}\right) \\ &= \frac{2}{3}\end{aligned} ∫ 1 ∞ x 2 x 1 d x = R → ∞ lim [ − 3 x x 2 ] 1 R = R → ∞ lim ( − 3 R R 2 + 3 2 ) = 3 2 In general, ∫ 1 ∞ d x x p \int_1^\infty \frac{dx}{x^p} ∫ 1 ∞ x p d x p > 1 p > 1 p > 1 p ≤ 1 p \le 1 p ≤ 1
Evaluate the improper integral.
∫ 0 ∞ x 3 e − x d x \int_{0}^{\infty} x^{3}e^{-x}\,dx ∫ 0 ∞ x 3 e − x d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Let I n = ∫ 0 ∞ x n e − x d x I_n = \int_0^\infty x^n e^{-x}\,dx I n = ∫ 0 ∞ x n e − x d x I n = n I n − 1 I_n = nI_{n-1} I n = n I n − 1
Answer ∫ 0 ∞ x 3 e − x d x = 6 \int_{0}^{\infty} x^{3}e^{-x}\,dx = 6 ∫ 0 ∞ x 3 e − x d x = 6 Solution Let I n = ∫ 0 ∞ x n e − x d x I_n = \displaystyle\int_0^{\infty} x^n e^{-x}\,dx I n = ∫ 0 ∞ x n e − x d x lim R → ∞ R n e − R = 0 \displaystyle\lim_{R\to\infty} R^n e^{-R} = 0 R → ∞ lim R n e − R = 0
I n = [ − x n e − x ] 0 ∞ + n ∫ 0 ∞ x n − 1 e − x d x = n I n − 1 I_n = \left[-x^n e^{-x}\right]_0^{\infty} + n\int_0^{\infty} x^{n-1}e^{-x}\,dx = nI_{n-1} I n = [ − x n e − x ] 0 ∞ + n ∫ 0 ∞ x n − 1 e − x d x = n I n − 1 Also I 0 = ∫ 0 ∞ e − x d x = 1 I_0 = \displaystyle\int_0^{\infty} e^{-x}\,dx = 1 I 0 = ∫ 0 ∞ e − x d x = 1
I 3 = 3 ⋅ 2 ⋅ 1 ⋅ I 0 = 3 ! = 6 I_{3} = 3 \cdot 2 \cdot 1 \cdot I_0 = 3! = 6 I 3 = 3 ⋅ 2 ⋅ 1 ⋅ I 0 = 3 ! = 6 In general, ∫ 0 ∞ x n e − x d x = n ! \int_0^\infty x^n e^{-x}\,dx = n! ∫ 0 ∞ x n e − x d x = n ! Γ ( n + 1 ) = n ! \Gamma(n+1) = n! Γ ( n + 1 ) = n !
Evaluate the improper integral.
∫ 0 ∞ 1 ( x + 2 ) ( x + 4 ) d x \int_{0}^{\infty} \frac{1}{(x + 2)(x + 4)}\,dx ∫ 0 ∞ ( x + 2 ) ( x + 4 ) 1 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Use partial fractions, integrate, and use lim R → ∞ log R + a R + b = 0 \lim_{R\to\infty}\log\frac{R+a}{R+b} = 0 lim R → ∞ log R + b R + a = 0
Answer ∫ 0 ∞ 1 ( x + 2 ) ( x + 4 ) d x = 1 2 log 2 \int_{0}^{\infty} \frac{1}{(x + 2)(x + 4)}\,dx = \frac{1}{2}\log 2 ∫ 0 ∞ ( x + 2 ) ( x + 4 ) 1 d x = 2 1 log 2 Solution By partial fractions, 1 ( x + a ) ( x + b ) = 1 b − a ( 1 x + a − 1 x + b ) \dfrac{1}{(x+a)(x+b)} = \dfrac{1}{b-a}\left(\dfrac{1}{x+a} - \dfrac{1}{x+b}\right) ( x + a ) ( x + b ) 1 = b − a 1 ( x + a 1 − x + b 1 )
∫ 0 R 1 ( x + 2 ) ( x + 4 ) d x = 1 2 [ log x + 2 x + 4 ] 0 R = 1 2 ( log R + 2 R + 4 − log 2 4 ) \int_0^R \frac{1}{(x + 2)(x + 4)}\,dx = \frac{1}{2}\left[\log\frac{x + 2}{x + 4}\right]_0^R = \frac{1}{2}\left(\log\frac{R + 2}{R + 4} - \log\frac{2}{4}\right) ∫ 0 R ( x + 2 ) ( x + 4 ) 1 d x = 2 1 [ log x + 4 x + 2 ] 0 R = 2 1 ( log R + 4 R + 2 − log 4 2 ) As R → ∞ R \to \infty R → ∞ log R + a R + b → log 1 = 0 \log\dfrac{R + a}{R + b} \to \log 1 = 0 log R + b R + a → log 1 = 0
∫ 0 ∞ 1 ( x + 2 ) ( x + 4 ) d x = 1 2 log 2 \int_0^{\infty} \frac{1}{(x + 2)(x + 4)}\,dx = \frac{1}{2}\log 2 ∫ 0 ∞ ( x + 2 ) ( x + 4 ) 1 d x = 2 1 log 2