**Casio fx-9750GIII: Integrals with Infinite Limits**

**A Substitution to Get to Infinity... **

It is quite a challenge to calculate numerical integrals with infinite limits such as

∫( f(x) dx, a, ∞)

∫( f(x) dx, -∞, a)

∫( f(x) dx, -∞, ∞)

A trick is to substitute x = tan Θ. Then:

Θ = arctan x

dx = sec^2 Θ dΘ = cos^-2 Θ dΘ

Note that

lim t→∞ arctan t = π/2

lim t→-∞ arctan t = -π/2

In this blog, assume that radian angle mode is used in all calculations.

Let's go over some examples and see how it works. I used a Casio fx-9750GIII, however, this technique should work with all calculators with numerical integral calculations. Furthermore, changing the integral will allow for Simpson's Rule or Trapezoid Rule approximation.

For the upper limit, I approximate π/2.

A = 1.5708 (π/2 to 4 places)

B = 1.570796 (π/2 to 6 places)

C = 1.57079633 (π/2 to 8 places)

Example 1:

∫( e^(-x^2) dx, 0, ∞)

transforms to

∫( e^(-tan^2 Θ)/cos^2 Θ dΘ, 0, ≈π/2)

Calculations (fx-9750GIII):

Upper Limit: A (see above), Result: 0.8862269255

Upper Limit: B, Result: 0.8862269255

Upper Limit: C, Result: 0.8862269255

Example 2:

∫(x^0.5 * e^(-x) dx, 0, ∞)

transforms to

∫((tan Θ)^0.5 * e^(-tan Θ))/cos^2 Θ dΘ, 0, π/2)

Calculations:

Upper limits A, B, C: 0.862269255

Coincidently, ∫( e^(-x^2) dx, 0, ∞) = ∫(x^0.5 * e^(-x) dx, 0, ∞) = Γ(1.5) = √(π)/2

Example 3:

∫( e^x*(x^2 + 1) dx, -∞, 0)

transforms to

∫( e^(tan Θ) * (tan^2 Θ + 1)/cos^2 Θ dΘ, -π/2, 0)

= ∫( e^(tan Θ) * 1/cos^2 Θ * 1/cos^2 Θ dΘ, -π/2, 0)

∫( e^(tan Θ)/cos^4 Θ dΘ, -π/2, 0)

For upper limits A, B, C, the answer returned is 3, which is the exact answer.

For integrals like this all is needed is a four digit approximation of π/2 = 1.5708.

Let's keep going:

Example 4:

∫( (x^2 + 1)/(x^4 - 1) dx, 0, ∞)

transforms to

∫( 1/cos^Θ * (tan^2 Θ + 1)/(tan^4 Θ - 1) dΘ, 0, π/2)

= ∫( 1/(sin^4 Θ - cos^4 Θ) dΘ, 0, π/2)

On the fx-9750GIII get MA Error.

Example 5:

∫( 1/√(x^2 + 3*x + 1) dx, 0, ∞)

transforms to

∫( 1/(cos^2 Θ * √(tan^2 Θ + 3 * tan Θ + 1)) dΘ, 0, π/2)

like example 4, I get the MA Error.

Observation: the transformation works best if the integral involves some form of either of the following:

f(x) * e^(g(x))

or

f(x) * e^(-g(x))

Gamma

The Gamma Function is an excellent candidate for this transformation.

Γ(t) = ∫( x^(t-1) * e^-x dx, 0, ∞)

transforms to

∫( tan^(t-1) Θ * e^(-tan Θ)/cos^2 Θ dΘ, 0, π/2)

Source

Mier-Jedrzejuwicz, W.A.C. Ph.D. __Tips And Programs for the HP 32S__ Synthetix Publication. Berkeley, CA. September 1988. ISBN 0-937637-05-X

Download the book here: https://literature.hpcalc.org/items/1756

Happy calculating,

Eddie

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