Improper Integrals - ImproperIntegrals.nb

Finding Improper Integrals Symbolically

In "easy" cases, Mathematica can also compute improper integrals. You may use  "-Infinity" and/or "Infinity" as your limits of integration; here are some examples:

Integrate[1/(x^2 + 1), { x, 0, Infinity}]

Integrate[1/(y^2 - 1), {y, -Infinity,  -2}]

Integrate[1/x, {x, -1, 1}]

If you prefer a numerical answer (or if Mathematica cannot find the symbolic answer for you), put an "N" up front:

NIntegrate[1/(y^2 - 1), {y, -Infinity,  -2}]

Be careful: Mathematica does not always get it right!

Here is an example where Mathematica gets an incorrect answer, in fact the integral ("barely") converges!

Integrate[1/(x Log[x]^2), {x, 5, Infinity}]

The Limit Comparison Test

Recall the Integral Comparison Test: Suppose f (x) and g (x) are positive for all x. Then, if (lim_(x -> ∞))((f(x))/(g(x)))exists, is finite and does not equal 0, both integrals converge at ∞, or both integrals diverge at ∞.
Mathematica can help you to find the limit. In the example below, the Integral Comparison Test applies; since the integral (∫_10^∞)(1/(x^2)) dx converges, so does  (∫_10
              ^∞)((2 (x^2) + 4 x - 3)/(3 (x^4) + 2 (x^3) - 
                    2 (x^2) + 4 x - 7)) dx.

f[x_] = 
      (2  x^2 + 4  x - 3)/(3  x^4 + 2  x^3 - 2  x^2 + 4  x - 7)g[x_] = 1/x^2Limit[f[x]/g[x], x -> Infinity]

In the next example, the Integral Comparison Test does not apply, and indeed the first integral converges at ∞, while the second one diverges at ∞!

f[x_] = 1/(x Log[x]^2)g[x_] = 1/x\(Limit[f[x]/g[x], x -> Infinity]	\)

Etc.

Author: H. Knaust (helmut@math.utep.edu)
Created: August 2000
Last Edits: August 29, 2000