A definite integral needs a bounded interval and, for the Fundamental Theorem, a function continuous on it. An improper integral breaks one of those: an infinite limit of integration, or a function that is unbounded on the interval. Its value is defined by a limit of ordinary definite integrals.
Infinite limits of integration Example 1
Decide whether each integral converges, and find its value if it does.
(a) ∫ 0 ∞ 3 ( x + 1 ) ( x + 4 ) d x (b) ∫ 0 ∞ x 2 x 3 + 8 d x \begin{aligned}
&\text{(a)}\;\; \int_0^{\infty} \frac{3}{(x + 1)(x + 4)}\,dx \\[10pt]
&\text{(b)}\;\; \int_0^{\infty} \frac{x^2}{x^3 + 8}\,dx
\end{aligned} (a) ∫ 0 ∞ ( x + 1 ) ( x + 4 ) 3 d x (b) ∫ 0 ∞ x 3 + 8 x 2 d x (a) Partial fractions give
3 ( x + 1 ) ( x + 4 ) = 1 x + 1 − 1 x + 4 \frac{3}{(x + 1)(x + 4)} = \frac{1}{x + 1} - \frac{1}{x + 4} ( x + 1 ) ( x + 4 ) 3 = x + 1 1 − x + 4 1 , so an antiderivative is
ln x + 1 x + 4 \ln\frac{x + 1}{x + 4} ln x + 4 x + 1 for
x ≥ 0 x \ge 0 x ≥ 0 .
∫ 0 ∞ 3 ( x + 1 ) ( x + 4 ) d x = lim b → ∞ [ ln x + 1 x + 4 ] 0 b = lim b → ∞ ( ln b + 1 b + 4 − ln 1 4 ) = 0 + ln 4 = ln 4 \begin{aligned}
&\int_0^{\infty} \frac{3}{(x + 1)(x + 4)}\,dx \\[4pt]
&\quad = \lim_{b\to\infty} \left[\ln\frac{x + 1}{x + 4}\right]_0^b \\[4pt]
&\quad = \lim_{b\to\infty} \left(\ln\frac{b + 1}{b + 4} - \ln\frac{1}{4}\right) \\[4pt]
&\quad = 0 + \ln 4 = \ln 4
\end{aligned} ∫ 0 ∞ ( x + 1 ) ( x + 4 ) 3 d x = b → ∞ lim [ ln x + 4 x + 1 ] 0 b = b → ∞ lim ( ln b + 4 b + 1 − ln 4 1 ) = 0 + ln 4 = ln 4 The integral converges to
ln 4 ≈ 1.386 \ln 4 \approx 1.386 ln 4 ≈ 1.386 .
The region under y = 3/((x + 1)(x + 4)) from x = 0 on. It's infinitely long, and its area is ln 4. (b) With
u = x 3 + 8 u = x^3 + 8 u = x 3 + 8 ,
d u = 3 x 2 d x du = 3x^2\,dx d u = 3 x 2 d x , so an antiderivative is
1 3 ln ( x 3 + 8 ) \frac{1}{3}\ln\left(x^3 + 8\right) 3 1 ln ( x 3 + 8 ) .
∫ 0 ∞ x 2 x 3 + 8 d x = lim b → ∞ 1 3 [ ln ( x 3 + 8 ) ] 0 b = lim b → ∞ 1 3 ln b 3 + 8 8 = ∞ \begin{aligned}
&\int_0^{\infty} \frac{x^2}{x^3 + 8}\,dx \\[4pt]
&\quad = \lim_{b\to\infty} \frac{1}{3}\Big[\ln\left(x^3 + 8\right)\Big]_0^b \\[4pt]
&\quad = \lim_{b\to\infty} \frac{1}{3}\ln\frac{b^3 + 8}{8} = \infty
\end{aligned} ∫ 0 ∞ x 3 + 8 x 2 d x = b → ∞ lim 3 1 [ ln ( x 3 + 8 ) ] 0 b = b → ∞ lim 3 1 ln 8 b 3 + 8 = ∞ The limit doesn't exist, so the integral diverges.
Both integrands approach 0. What decides convergence is how fast: in (a) the integrand shrinks like
3 x 2 \frac{3}{x^2} x 2 3 , and in (b) only like
1 x \frac{1}{x} x 1 . For powers of
x x x the dividing line is exact.
The p-integral
∫ 1 ∞ d x x p = { 1 p − 1 if p > 1 , diverges if p ≤ 1. \int_1^{\infty} \frac{dx}{x^p} = \begin{cases} \dfrac{1}{p - 1} & \text{if } p > 1, \\[8pt] \text{diverges} & \text{if } p \le 1. \end{cases} ∫ 1 ∞ x p d x = ⎩ ⎨ ⎧ p − 1 1 diverges if p > 1 , if p ≤ 1. For
p ≠ 1 p \ne 1 p = 1 the integral up to
b b b is
b 1 − p − 1 1 − p \frac{b^{1-p} - 1}{1 - p} 1 − p b 1 − p − 1 , and
b 1 − p b^{1-p} b 1 − p approaches 0 only when
p > 1 p > 1 p > 1 . For
p = 1 p = 1 p = 1 it's
ln b \ln b ln b , which grows without bound.
When both limits are infinite, pick
c c c where the antiderivative is simple to evaluate.
Example 2
Evaluate the integral below.
∫ − ∞ ∞ d x x 2 + 8 x + 41 \int_{-\infty}^{\infty} \frac{dx}{x^2 + 8x + 41} ∫ − ∞ ∞ x 2 + 8 x + 41 d x Complete the square:
x 2 + 8 x + 41 = ( x + 4 ) 2 + 25 x^2 + 8x + 41 = (x + 4)^2 + 25 x 2 + 8 x + 41 = ( x + 4 ) 2 + 25 , so an antiderivative is
1 5 arctan x + 4 5 \frac{1}{5}\arctan\frac{x + 4}{5} 5 1 arctan 5 x + 4 . Split at
c = − 4 c = -4 c = − 4 , where the arctangent is 0.
∫ − ∞ ∞ d x x 2 + 8 x + 41 = lim a → − ∞ 1 5 [ arctan x + 4 5 ] a − 4 + lim b → ∞ 1 5 [ arctan x + 4 5 ] − 4 b = 1 5 ( 0 − ( − π 2 ) ) + 1 5 ( π 2 − 0 ) = π 5 \begin{aligned}
&\int_{-\infty}^{\infty} \frac{dx}{x^2 + 8x + 41} \\[4pt]
&\quad = \lim_{a\to-\infty} \frac{1}{5}\left[\arctan\frac{x + 4}{5}\right]_a^{-4} \\[4pt]
&\qquad + \lim_{b\to\infty} \frac{1}{5}\left[\arctan\frac{x + 4}{5}\right]_{-4}^b \\[4pt]
&\quad = \frac{1}{5}\left(0 - \left(-\frac{\pi}{2}\right)\right) + \frac{1}{5}\left(\frac{\pi}{2} - 0\right) \\[4pt]
&\quad = \frac{\pi}{5}
\end{aligned} ∫ − ∞ ∞ x 2 + 8 x + 41 d x = a → − ∞ lim 5 1 [ arctan 5 x + 4 ] a − 4 + b → ∞ lim 5 1 [ arctan 5 x + 4 ] − 4 b = 5 1 ( 0 − ( − 2 π ) ) + 5 1 ( 2 π − 0 ) = 5 π An antiderivative found by parts often leaves a limit like
b 2 e − 3 b b^2 e^{-3b} b 2 e − 3 b , of the form
∞ ⋅ 0 \infty \cdot 0 ∞ ⋅ 0 . Rewritten as
b 2 e 3 b \frac{b^2}{e^{3b}} e 3 b b 2 , it's of the form
∞ ∞ \frac{\infty}{\infty} ∞ ∞ , so L'Hospital's rule applies.
Example 3
Evaluate the integral below.
∫ 0 ∞ x 2 e − 3 x d x \int_0^{\infty} x^2 e^{-3x}\,dx ∫ 0 ∞ x 2 e − 3 x d x The tabular method, with
u = x 2 u = x^2 u = x 2 and
d v = e − 3 x d x dv = e^{-3x}\,dx d v = e − 3 x d x , gives the antiderivative
F ( x ) = − e − 3 x ( x 2 3 + 2 x 9 + 2 27 ) F(x) = -e^{-3x}\left(\frac{x^2}{3} + \frac{2x}{9} + \frac{2}{27}\right) F ( x ) = − e − 3 x ( 3 x 2 + 9 2 x + 27 2 ) .
∫ 0 ∞ x 2 e − 3 x d x = lim b → ∞ [ F ( x ) ] 0 b = lim b → ∞ ( F ( b ) − F ( 0 ) ) = 2 27 + lim b → ∞ F ( b ) \begin{aligned}
&\int_0^{\infty} x^2 e^{-3x}\,dx \\[4pt]
&\quad = \lim_{b\to\infty} \Big[F(x)\Big]_0^b \\[4pt]
&\quad = \lim_{b\to\infty} \big(F(b) - F(0)\big) \\[4pt]
&\quad = \frac{2}{27} + \lim_{b\to\infty} F(b)
\end{aligned} ∫ 0 ∞ x 2 e − 3 x d x = b → ∞ lim [ F ( x ) ] 0 b = b → ∞ lim ( F ( b ) − F ( 0 ) ) = 27 2 + b → ∞ lim F ( b ) Here
F ( b ) = − b 2 / 3 + 2 b / 9 + 2 / 27 e 3 b F(b) = -\frac{b^2/3 + 2b/9 + 2/27}{e^{3b}} F ( b ) = − e 3 b b 2 /3 + 2 b /9 + 2/27 . Applying L'Hospital's rule twice gives
lim b → ∞ F ( b ) = − lim b → ∞ 2 / 3 9 e 3 b = 0 \lim_{b\to\infty} F(b) = -\lim_{b\to\infty} \frac{2/3}{9e^{3b}} = 0 lim b → ∞ F ( b ) = − lim b → ∞ 9 e 3 b 2/3 = 0 , so the integral converges to
2 27 \frac{2}{27} 27 2 .
Unbounded integrands Example 4
Decide whether each integral converges, and find its value if it does.
(a) ∫ 0 2 x 4 − x 2 d x (b) ∫ 0 4 d x ( x − 2 ) 4 \begin{aligned}
&\text{(a)}\;\; \int_0^{2} \frac{x}{\sqrt{4 - x^2}}\,dx \\[10pt]
&\text{(b)}\;\; \int_0^{4} \frac{dx}{(x - 2)^4}
\end{aligned} (a) ∫ 0 2 4 − x 2 x d x (b) ∫ 0 4 ( x − 2 ) 4 d x (a) The integrand is unbounded as
x → 2 − x \to 2^- x → 2 − . With
u = 4 − x 2 u = 4 - x^2 u = 4 − x 2 ,
d u = − 2 x d x du = -2x\,dx d u = − 2 x d x , an antiderivative is
− 4 − x 2 -\sqrt{4 - x^2} − 4 − x 2 .
∫ 0 2 x 4 − x 2 d x = lim t → 2 − [ − 4 − x 2 ] 0 t = lim t → 2 − ( 2 − 4 − t 2 ) = 2 \begin{aligned}
&\int_0^{2} \frac{x}{\sqrt{4 - x^2}}\,dx \\[4pt]
&\quad = \lim_{t\to 2^-} \Big[-\sqrt{4 - x^2}\Big]_0^t \\[4pt]
&\quad = \lim_{t\to 2^-} \left(2 - \sqrt{4 - t^2}\right) = 2
\end{aligned} ∫ 0 2 4 − x 2 x d x = t → 2 − lim [ − 4 − x 2 ] 0 t = t → 2 − lim ( 2 − 4 − t 2 ) = 2 (b) The integrand is unbounded at
x = 2 x = 2 x = 2 , inside
[ 0 , 4 ] [0, 4] [ 0 , 4 ] . Applying the Fundamental Theorem straight across would give
[ − 1 3 ( x − 2 ) 3 ] 0 4 = − 1 12 \left[-\frac{1}{3(x - 2)^3}\right]_0^4 = -\frac{1}{12} [ − 3 ( x − 2 ) 3 1 ] 0 4 = − 12 1 , a negative number for a positive integrand, so something has gone wrong. Split at 2 and take the left piece first:
∫ 0 2 d x ( x − 2 ) 4 = lim t → 2 − [ − 1 3 ( x − 2 ) 3 ] 0 t = lim t → 2 − ( − 1 3 ( t − 2 ) 3 − 1 24 ) = ∞ \begin{aligned}
&\int_0^{2} \frac{dx}{(x - 2)^4} \\[4pt]
&\quad = \lim_{t\to 2^-} \left[-\frac{1}{3(x - 2)^3}\right]_0^t \\[4pt]
&\quad = \lim_{t\to 2^-} \left(-\frac{1}{3(t - 2)^3} - \frac{1}{24}\right) = \infty
\end{aligned} ∫ 0 2 ( x − 2 ) 4 d x = t → 2 − lim [ − 3 ( x − 2 ) 3 1 ] 0 t = t → 2 − lim ( − 3 ( t − 2 ) 3 1 − 24 1 ) = ∞ That piece diverges, so the whole integral diverges.
The region under y = 1/(x − 2)⁴ on [0, 4] runs up the asymptote x = 2 on both sides. Its area is infinite. Comparison This test goes beyond the AP course description, and it's often the quickest way to decide convergence when there's no antiderivative to work with.
The comparison test
Suppose
f f f and
g g g are continuous and
0 ≤ g ( x ) ≤ f ( x ) 0 \le g(x) \le f(x) 0 ≤ g ( x ) ≤ f ( x ) for
x ≥ a x \ge a x ≥ a . If
∫ a ∞ f ( x ) d x \int_a^{\infty} f(x)\,dx ∫ a ∞ f ( x ) d x converges, so does
∫ a ∞ g ( x ) d x \int_a^{\infty} g(x)\,dx ∫ a ∞ g ( x ) d x . If
∫ a ∞ g ( x ) d x \int_a^{\infty} g(x)\,dx ∫ a ∞ g ( x ) d x diverges, so does
∫ a ∞ f ( x ) d x \int_a^{\infty} f(x)\,dx ∫ a ∞ f ( x ) d x .
Example 5
Decide whether each integral converges.
(a) ∫ 1 ∞ d x x 3 + x (b) ∫ 1 ∞ d x x + sin 2 x \begin{aligned}
&\text{(a)}\;\; \int_1^{\infty} \frac{dx}{x^3 + \sqrt{x}} \\[10pt]
&\text{(b)}\;\; \int_1^{\infty} \frac{dx}{\sqrt{x} + \sin^2 x}
\end{aligned} (a) ∫ 1 ∞ x 3 + x d x (b) ∫ 1 ∞ x + sin 2 x d x (a) For
x ≥ 1 x \ge 1 x ≥ 1 ,
x 3 + x > x 3 x^3 + \sqrt{x} > x^3 x 3 + x > x 3 , so
0 < 1 x 3 + x < 1 x 3 0 < \frac{1}{x^3 + \sqrt{x}} < \frac{1}{x^3} 0 < x 3 + x 1 < x 3 1 . The p-integral with
p = 3 > 1 p = 3 > 1 p = 3 > 1 converges, so this integral converges.
(b) For
x ≥ 1 x \ge 1 x ≥ 1 ,
sin 2 x ≤ 1 ≤ x \sin^2 x \le 1 \le \sqrt{x} sin 2 x ≤ 1 ≤ x , so the denominator is at most
2 x 2\sqrt{x} 2 x and
1 x + sin 2 x ≥ 1 2 x > 0 \frac{1}{\sqrt{x} + \sin^2 x} \ge \frac{1}{2\sqrt{x}} > 0 x + s i n 2 x 1 ≥ 2 x 1 > 0 . The p-integral with
p = 1 2 ≤ 1 p = \frac{1}{2} \le 1 p = 2 1 ≤ 1 diverges, and so does half of it, so this integral diverges.