Solar Scientific Calculators: Dealing with Integrals with Infinite Limits

Solar Scientific Calculators:  Dealing with Integrals with Infinite Limits

Integrals with Infinite Limits

Today's post deals with integrals with infinite limits in the forms:

∫( f(x) dx, x = a to x = ∞)

∫( f(x) dx, x = -∞ to x = ∞)

∫( f(x) dx, x = -∞ to x = a)

One method to deal with these integrals, as suggested by W.A.C. Mier-Jedrzejowicz Ph. D. (see the source), is to use the substitution

x = tan θ

Then:

dx = dθ/cos^2 θ

and θ = atan x.

Also, as x approaches π/2, tan x approaches +∞.

And, as x approaches -π/2, tan x approaches -∞.

With the substations, let's test four integrals on four solar-powered scientific calculators:

1.  Casio fx-991EX Classwiz
2.  Sharp EL-W516T
3.  Texas Instruments TI-36X Pro
4.  Casio fx-115ES Plus

Set the calculator to radians mode. 

Example 1:  ∫(1/x^2 dx, x = 1 to x = ∞) = 1

∫(1/x^2 dx, x = 1 to x = ∞)

with the substitutions x = tan θ and dx = dθ/(cos^2 θ):

∫( 1/tan^2 θ * dθ/cos^2 θ, θ = atan 1 to θ = π/2)

∫( 1/sin^2 θ * dθ, θ = atan 1 to θ = π/2)

We can evaulate the integral straight away.  Here are the results:

1.  Casio fx-991EX Classwiz
Time: 1.37 seconds
Answer: 1

2.  Sharp EL-W516T
Time: 38 seconds
Answer: 1

3.  Texas Instruments TI-36X Pro
Time: 4.5 seconds
Answer: 1

4.  Casio fx-115ES Plus
Time: 4.2 seconds
Answer: 1

A promising start.

Example 2:  ∫(e^(-0.5*x^2), x = 0 to x = ∞) ≈ 1.25331413732

∫(e^(-0.5*x^2), x = 0 to x = ∞)

with the substituions, this becomes:

∫(e^(-0.5 * tan^2 θ)/cos^2 θ dθ, θ = atan 0 to θ = π/2)

atan 0 = 0

But look at the denominator, we have cos^2 θ.  Since cos^2 π/2 = 0, there will be a problem.  Let's use an approximation of π/2 of 1.5708.

∫(e^(-0.5 * tan^2 θ)/cos^2 θ dθ, θ = 0 to θ = 1.5708)

Here are the results:

1.  Casio fx-991EX Classwiz
Time: 15.4 seconds
Answer: 1.253314137

2.  Sharp EL-W516T
Time: 1 minute, 8 seconds
Answer: errors out

3.  Texas Instruments TI-36X Pro
Time: 36 seconds
Answer: 1.253314138

4.  Casio fx-115ES Plus
Time: 1 minute, 6.8 seconds
Answer: 1.253314137

Example 3:  ∫(x^2*e^-x dx, x = 0 to x = ∞) = 2

∫(x^2*e^-x dx, x = 0 to x = ∞)

with the substitutions and simplification, we get:

∫( (sin^2 θ * e^(-tan θ))/cos^4 θ dθ, θ = 0 to θ = π/2)

Like the last situation, there is a potential problem with the denominator.  Let's see if we can use an approximation of π/2, this time using 1.57 in hopes to cut the calculation time down.

∫( (sin^2 θ * e^(-tan θ))/cos^4 θ dθ, θ = 0 to θ = 1.57)

Here are the results:

1.  Casio fx-991EX Classwiz
Time: 27 seconds
Answer: 2

2.  Sharp EL-W516T
Time: 1 minute, 34 seconds
Answer: 1.999999999

3.  Texas Instruments TI-36X Pro
Time: 1 minute, 9 seconds
Answer: 2

4.  Casio fx-115ES Plus
Time: 1 minute, 6.8 seconds
Answer: 1.253314137

Example 4:  ∫( e^-x/x^2 dx, x = 1 to x = ∞) ≈ 0.148495506776

∫( e^-x/x^2 dx, x = 1 to x = ∞)

with the substitutions and simplification, we get:

∫( e^(-tan θ)/sin^2 θ dθ, θ = 0 to θ = π/2)

I'm going to use the 1.57 approximation again and set the integral as:

∫( e^(-tan θ)/sin^2 θ dθ, θ = 0 to θ = 1.57)

Here are the results:

1.  Casio fx-991EX Classwiz
Time: errors out immediately
Answer: N/A

2.  Sharp EL-W516T
Time: 1 minute, 6 seconds
Answer: error

3.  Texas Instruments TI-36X Pro
Time: 7 seconds
Answer: 0.148495519

4.  Casio fx-115ES Plus
Time: errors out after 1 second
Answer: N/A

Some Observations

1.  Not all calculations of improper integrals will be successful.

2.  Out of the four calculators tested, from the four calculations:  the Casio fx-991ES is the fastest, but I found the most successful with the Texas Instruments TI-36X Pro.

3.  Be ready to spend a little for calculations by using this method.

Source:

Mier-Jedrzejowic, W.A.C. Ph.D.  Extend Your 41  London, UK 1985  ISBN 0-9510733-0-03

Eddie

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Calculus Revisited #15: Infinite Limits

We have arrived at the 15th blog entry of the 21 entry series of Calculus Revisited. It is all about infinity today, and some quick tips with deal with infinity and limits.

Infinite Limits

General Rules for Polynomials

Let p(x) be a polynomial with degree n. The lead term of p(x) is a_n*x^n.

If n is even and a_n is positive:
* p(x) →+∞ as x → +∞
* p(x) → +∞ as x → -∞

If n is even and a_n is negative:
* p(x) → -∞ as x → +∞
* p(x) → -∞ as x → -∞

If n is odd and a_n is positive:
* p(x) → +∞ as x → +∞
* p(x) → -∞ as x → -∞

If n is odd and a_n is negative:
* p(x) → -∞ as x → +∞
* p(x) → +∞ as x → -∞

Generally: 1 / p(x) → 0 as x → +∞ or x → -∞

If p(x)/q(x) have the same degree, then

p(x)/q(x) → (lead coefficient of p(x)) / (lead coefficient of q(x))

as x → +∞ or x → -∞

General Rules for Other Functions

ln x → -∞ as x → -∞
ln x → +∞ as x → +∞

e^x → 0 as x → -∞
e^x → +∞ as x → +∞

e^-x → +∞ as x → -∞
e^-x → 0 as x → +∞

sin x has no limit as x → +∞ or x → -∞

cos x has no limit as x → +∞ or x → -∞

tan x has no limit as x → +∞ or x → -∞

Note that sin x and cos x are both bounded between -1 and 1.

L'Hospital's Rule

For indeterminate forms 0/0, +∞/+∞, -∞/-∞ you can use the rule:

lim f(x)/g(x) =
x → a

lim f'(x)/g'(x)
x → a

Let's go over some problems.

Problems

1. Find: lim (1/x * sin) as x → ∞

We know that:

lim 1/x = 0 as x → ∞ and sin x is bounded.

Then the overall limit is 0 * (a number between -1 and 1) = 0

2. Find lim (e^(-x^2)) as x → ∞

In general, e^-x → 0 as x → ∞

Even though x^2 grows, e^-x shrinks faster and is dominant.

Hence lim (e^(-x^2)) = 0 as x → ∞.

The next three problems uses the p(x)/q(x) rules.

3. lim (x^2 + 1)/(x - 2) as x → ∞

Degree of numerator = 2
Degree of denominator = 1

Numerator will grow faster than the denominator.

Then lim (x^2 + 1)/(x - 2) → ∞ as x → ∞

4. lim (x^2 + 1)/(x^2 - 1) as x → ∞

Degree of numerator = 2
Degree of denominator = 2

Then lim (x^2 + 1)/(x^2 - 1) = 1/1 = 1 as x → ∞

5. lim (x^2 + 1)/(x^3 - 1) as x → ∞

Degree of numerator = 2
Degree of denominator = 3

Denominator will faster than numerator.

Then lim (x^2 + 1)/(x^3 - 1) = 0 as x → ∞

The next two problems uses L'Hospital's Rule.

6. lim (x^2 / e^x) as x → ∞

lim (x^2 / e^x) as → ∞ = ∞/∞ (Indeterminate form)

Applying the rule:

lim (2x / e^x ) as x → ∞ = ∞/∞ (Indeterminate form)

One more time:

lim (2 / e^x) as x → ∞ = 0

Therefore, lim (x^2 / e^x) = 0 as x → ∞

7. lim (sin x / x) as x → 0

lim (sin x / x) as x → 0 = 0/0 (Indeterminate form)

Applying the rule:

lim (cos x/1) as x→ 0 = 1/1 = 1

Therefore, lim (sin x / x) = 1 as x → 0

This concludes today's session. Next time we work with improper integrals (yes, there are such things!).

Have a great day,

Eddie

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