Calculus II: Improper integrals

1.Definition

The definite integral

	b
∫		f(x) dx
	a

is an improper integral when two cases occur:

1. The function f(x) is discontinue somewhere in the interval [a, b], (at least one point of discontinuity).

2. The interval [a,b] is unbounded,(at least one bound is infinite), that is
a → ± ∞ or
b → ± ∞

So how to apply the FTC?
Rules are available to overcome this kind of situations.

2.Integrals with discontimuous functions

If f is a continuous function in the interval [a, b]

1. execpt at the right on the bound a, then:

	b				b
∫		f(x) dx	= lim	∫		f(x) dx
	a		r→a+		r

2. execpt at the left on the bound b, then:

	b				r
∫		f(x) dx	= lim	∫		f(x) dx
	a		r→b-		a

3. execpt at a point c on the interval ]a, b[, then:

	b				c				b
∫		f(x) dx	=	∫		f(x) dx   +		∫		f(x) dx
	a				a				c

					r				b
			= lim	∫		f(x) dx	+  lim	∫		f(x) dx
			r→c-		a		r→c+		r

For each definition of the three above, if the limit exists, then the definition is satified and the integral converges, otherwise, the integral diverges.

3.Integrals with infinite bounds

We have three rules in this case:

1. f is a function continuous on the interval [a, +∞[:
2. f is a function continuous on the interval ]-∞, b]:
3. f is a function continuous on R = ]-∞, +∞[:

1. f is a function continuous on the interval [a, +∞[:

	+∞				r
∫		f(x) dx	= lim	∫		f(x) dx
	a		r→+∞		a

2. f is a function continuous on the interval ]-∞, b]:

	b				b
∫		f(x) dx	= lim	∫		f(x) dx
	-∞		r→-∞		r

3. f is a function continuous on R = ]-∞, +∞[:

	+∞				c				+∞
∫		f(x) dx	=	∫		f(x) dx   +		∫		f(x) dx
	-∞				-∞				c

					c				r
			= lim	∫		f(x) dx	+  lim	∫		f(x) dx
			r→-∞		r		r→+∞		c

For each definition of the three above, if the limit exists, then the definition is satified and the integral converges, otherwise, the integral diverges.

4. Examples

4.1. Example 1

	1
∫		dx /[1 - x2]1/2
	0

The function 1/[1 - x²]^1/2 is continuous in the interval [0, 1], except at the left of the point x = 1.

	1				r
∫		dx /[1 - x2]1/2	= lim	∫		dx /[1 - x2]1/2
	0		r→1-		0

= arcsin(1^-) - arcsin (0) = π/2

inverse function

	1
∫		dx /[1 - x2]1/2 = π/2
	0

4.2. Example 2

	1
∫		dx /x2
	-1

The function 1/x² is continuous in the interval [-1, 1], except at the point x = 0.

This function is even. It is equal to:

	1
2 ∫		dx /x2
	0

Therefore

The function 1/x² is continuous in the interval [0, 1], except at the right of the point x = 0.

∫		dx/x2 = - 1/x + const.

	1				1
∫		dx/x2	= lim	∫		dx/x2 = - 1/1 + 1/r = + ∞
	0		r→0+		r

	1
∫		dx /[1 - x2]1/2 diverges.
	0

4.3. Example 3

	+∞				0				+∞
∫		dx/(1 + x2)	=	∫		dx/(1 + x2)   +		∫		dx/(1 + x2)
	-∞				-∞				0

					0				r
			= lim	∫		dx/(1 + x2)	+  lim	∫		dx/(1 + x2)
			r→-∞		r		r→+∞		0

Since the integrand f(x) = 1/(1 + x²) is even:

									r
							= 2  lim	∫		dx/(1 + x2)
							r→+∞		0

= 2 [arctan(+∞) - arctan(0)] = 2 [π/2 - 0] = π

5. Exercises