INTEGRATION 08
Improper Integrals
An improper integral is defined using a limit when the interval is infinite or the integrand becomes unbounded.
The main idea
Ordinary definite integrals use a finite interval and a function that remains finite on that interval.
Improper integrals extend the idea using limits.
1. Infinite intervals
Consider
\[ \int_1^\infty \frac{1}{x^2}\,dx. \]The upper limit is not a finite number, so we define the integral by introducing a finite endpoint \(b\):
2. Evaluate the finite integral first
Write
\[ \frac{1}{x^2}=x^{-2}. \]An antiderivative is
\[ -\frac{1}{x}. \]Therefore,
\[ \int_1^b \frac{1}{x^2}\,dx = \left[ -\frac1x \right]_1^b. \]So
\[ = -\frac1b+1. \]3. Now take the limit
As
\[ b\to\infty, \]we have
\[ \frac1b\to0. \]Therefore,
The interval is infinite, but the total accumulated value is finite.
4. Convergence
If the defining limit exists as a finite number, the improper integral is said to converge.
So
\[ \int_1^\infty \frac{1}{x^2}\,dx \]converges.
5. Divergence
Now consider
\[ \int_1^\infty \frac1x\,dx. \]Define
\[ \int_1^\infty \frac1x\,dx = \lim_{b\to\infty} \int_1^b \frac1x\,dx. \]Since
\[ \int\frac1x\,dx=\ln x, \]we obtain
\[ \lim_{b\to\infty}\ln b. \]But
\[ \ln b\to\infty. \]Therefore the integral does not approach a finite value.
6. Infinite interval to the left
An integral such as
\[ \int_{-\infty}^{a}f(x)\,dx \]is defined by
\[ \boxed{ \int_{-\infty}^{a}f(x)\,dx = \lim_{b\to-\infty} \int_b^a f(x)\,dx } \]7. Infinite in both directions
For
\[ \int_{-\infty}^{\infty}f(x)\,dx, \]we split at a finite point \(c\):
\[ \int_{-\infty}^{\infty}f(x)\,dx = \int_{-\infty}^{c}f(x)\,dx + \int_c^{\infty}f(x)\,dx. \]Both improper integrals must converge for the whole integral to converge in the usual sense.
8. Unbounded functions
Improper integrals also occur when the interval is finite but the function becomes unbounded.
Consider
\[ \int_0^1 \frac{1}{\sqrt{x}}\,dx. \]The function
\[ \frac1{\sqrt{x}} \]becomes unbounded as \(x\to0^+\).
So define
9. Evaluate the integral
Since
\[ \frac1{\sqrt{x}} = x^{-1/2}, \]an antiderivative is
\[ 2\sqrt{x}. \]Therefore,
\[ \int_a^1 x^{-1/2}\,dx = [2\sqrt{x}]_a^1. \]So
\[ = 2-2\sqrt{a}. \]As
\[ a\to0^+, \]we obtain
\[ 2-0=2. \]Therefore,
10. Unbounded does not automatically mean divergent
The previous function becomes infinitely large near \(x=0\), but the integral still converges.
So an infinite function value near one point does not by itself determine whether the total accumulation is infinite.
11. A comparison
Compare
\[ \int_0^1 x^{-1/2}\,dx \]with
\[ \int_0^1 x^{-1}\,dx. \]The first converges.
The second is
\[ \lim_{a\to0^+} \int_a^1 \frac1x\,dx. \]This gives
\[ -\ln a. \]As
\[ a\to0^+, \]we have
\[ -\ln a\to\infty. \]Therefore,
\[ \boxed{ \int_0^1 \frac1x\,dx \text{ diverges} } \]12. The \(p\)-integral
An important family is
\[ \int_1^\infty \frac{1}{x^p}\,dx. \]This converges exactly when
and diverges when
\[ p\le1. \]13. Near zero
For
\[ \int_0^1 \frac{1}{x^p}\,dx, \]the condition is reversed:
and diverges when
\[ p\ge1. \]14. Why limits are essential
We never substitute the symbol \(\infty\) directly into an antiderivative as though it were an ordinary number.
Instead, we introduce a finite variable and take a limit.
15. Visual idea
16. The practical method
17. The central idea
Improper integrals do not abandon the definition of a definite integral. They extend it through limits.