Saturday, November 8, 2025

RPN: Certain Integrals to Positive Infinity

RPN: Certain Integrals to Positive Infinity


Introduction


Today’s RPN session deals with improper integrals where the upper limit is positive infinity (∞).


∫( f(x) dx, a, ∞) → lim ∫( f(x) dx, a, t) as t → ∞


Some properties that will be use:

lim 1/t^n as t → ∞ approaches 0

lim e^(-t) as t → ∞ approaches 0

lim p(t) ÷ q(t) as t → ∞ approaches 0 where p(t) and q(t) are polynomials and

degree p(t) < degree q(t)

lim p(t) ÷ q(t) as t → ∞ approaches p_n ÷ q_n where p(t) and q(t) are polynomials and

degree p(t) = degree q(t). p_n and q_n are the leading coefficients of p(t) and q(t), respectively.


All integrals presented will have closed formulas because they have very friendly anti-derivatives.


( 1 ÷ (x^n) dx, a, ∞)



∫( 1 ÷ (x^n) dx, a, ∞) = 1 ÷ ((n - 1) * a^(n - 1))


(Abbreviated) Derivation:

∫( 1 ÷ (x^n) dx, a, ∞)

= lim (1 ÷ x^(n - 1) * -1 ÷ (n – 1) as x → ∞) - (1 ÷ a^(n – 1) * -1 ÷ (n – 1))

= ( 0 * -1 ÷ (n – 1) ) + 1 ÷ ((n – 1) * a^(n – 1))

= 1 ÷ ((n – 1) * a^(n – 1))


∫( 1 ÷ (x^n) dx, a, ∞) = 1 ÷ ((n - 1) * a^(n - 1))


HP 15C Code:

[42, 21, 11]: LBL A

[         1]: 1

[        30]: -

[    43, 36]: LSTx

[        20]: ×

[        15]: 1/x

[    43, 32]: RTN


Stack:

Y: a

X: n


Examples:

a = 2.75, n = 2: 4/11 ≈ 0.36364

a = 4.9, n = 3: ≈ 0.02082


( 1 ÷ ((x - r)*(x - s)) dx, a, ∞)


∫( 1 ÷ ((x - r)*(x - s)) dx, a, ∞)

= 1 ÷ (r - s) * ln( abs((a - s) ÷ (a – r)) )

For best results, a > max(s, r)


(Abbreviated) Derivation:

∫( 1 ÷ ((x - r)*(x - s)) dx, a, ∞)


Simply by partial fractions:

1 ÷ ((x - r)*(x – s)) = 1 ÷ ((r – s) * (x – r)) – 1 ÷ ((r – s) * (x – s))


Anti-derivative:

∫( 1 ÷ ((x - r)*(x - s)) dx)

= 1 ÷ (r – s) * ( ln(abs(x – r)) – ln(abs(x – s)) )

= 1 ÷ (r – s) * ln ( abs((x – r) ÷ (x – s)) )


Limit as x → ∞:

1 ÷ (r – s) * ln ( abs((x – r) ÷ (x – s)) )

= 1 ÷ (r – s) * ln ( abs((1 – r ÷ x) ÷ (1 – s ÷ x)) )

= 1 ÷ (r – s) * ln ( abs(1) )

= 1 ÷ (r – s) * ln(1)

= 0


When x = a

1 ÷ (r – s) * ln ( abs((a – r) ÷ (a – s)) )

= 1 ÷ (r – s) * ln ( abs(1 ÷ [(a – r) * (a -s)]) )

= 1 ÷ (r – s) * ln ( 1 ÷ [abs((a – s) ÷ (a – r))] )

= -1 ÷ (r – s) * ln ( abs((a – s) ÷ (a – r)) )


Then:

∫( 1 ÷ ((x - r)*(x - s)) dx, a, ∞) = 0 - (-1 ÷ (r – s) * ln ( abs((a – s) ÷ (a – r)) ))

= 1 ÷ (r - s) * ln ( abs((a - s) ÷ (a – r)) )


Code:

[42, 21, 12]: LBL B

[    44,  1]: STO 1

[        33]: R↓

[    44,  2]: STO 2

[        33]: R↓

[    44,  3]: STO 3

[45, 30,  1]: RCL- 1

[    45,  3]: RCL 3

[45, 30,  2]: RCL- 2

[        10]: ÷

[    43, 16]: ABS

[    43, 12]: LN

[    45,  2]: RCL 2

[45, 30,  1]: RCL- 1

[        15]: 1/x

[        20]: ×

[    43, 32]: RTN


Stack:

Z: a

Y: r

X: s


Examples:

a = 7.25, b = -3, s = 6: (ln 41 - ln 5) ÷ 9 ≈ 0.23379

a = 11, b = 4, s = 9: (ln 7 - ln 2) ÷ 5 ≈ 0.25055


( 1 ÷ e^x dx, a, ∞)


∫( 1 ÷ e^x dx, a, ∞) = 1 ÷ e^a


Code:

[42, 21, 13]: LBL C

[        12]: e^x

[        15]: 1/x

[    43, 32]: RTN


Stack:

X: a


Examples:

a = 4: e^(-4) ≈ 0.01832

a = 6: e^(-6) ≈ 0.00248


( x ÷ e^x dx, a, ∞)


∫( x ÷ e^x dx, a, ∞) = (a + 1) ÷ e^a


Code:

[42, 21, 14]: LBL D

[        12]: e^x

[    43, 36]: LSTx

[         1]: 1

[        40]: +

[        34]: x<>y

[        10]: ÷

[    43, 32]: RTN


Stack:

X: a


Examples:

a = 4: 5 ÷ e^(-4) ≈ 0.09158

a = 6: 7 ÷ e^(-6) ≈ 0.01735


Eddie


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RPN: Certain Integrals to Positive Infinity

RPN: Certain Integrals to Positive Infinity Introduction Today’s RPN session deals with improper integrals where the upper limit is...