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Integrating Reciprocals of Powers from First Principles

Integration from First Principles

Foreword: In this part of the paper, I explain how to integrate Reciprocals of Powers, from First Principles. (Use the latest Mozilla Firefox Browser, to see the equations properly.)

By: Chrysanthus Date Published: 4 Jun 2019

Introduction

In this part of the paper, I explain how to integrate Reciprocals of Powers, from First Principles. You should have read the previous parts of the series before reaching here, as this is the continuation.

Integration is area under the curve.
Consider the following diagram:

y = f(x)

You might have seen the following:

a b f ( x ) d x = lim δ x 0 x = a x = b f ( x ) δ x

This statement is not perfect. It should actually be:

a b f ( x ) d x = lim δ x 0 x = a x = b f ( x n ) δ x

where f(xn) is a particular height from the curve to the x-axis.

My Discovery

The widths of the vertical strips are equal.

As 𝛿x turns to 0, the number of strips increases.

n is the total number of strips.

δ x = b - a n

As 𝛿x → 0, n → ∞

The series and the formula for its summation, for all the heights of the strips from a to b, is obtained. If you do not know the formula for the series, develop one, and that may take time. Summing is done for the heights, from 1 to n.

The summation expression above in the statement, is replaced with the summation formula.

As n → ∞ the right-hand-side of the statement becomes the definite integral; and that concludes the scheme.

Let total area be represented by A. So, my statement for the Integration from First Principles is:

A = lim n x = a x = b f ( x n ) b - a n

In the following analysis, the binomial expansion to infinity, for a negative index, has been used.

Integration of the Term, 1/x

Let the definite integral be represented by A, and let the indefinite integral be represented by I. Let the summation expression (discrete summation) be Sn.

Here,

y = 1/x

i.e.

f(x) = 1/x1

S n = x = a x = b f ( x n ) δ x S n = [ 1 ( a + 1 δ x ) 1 + 1 ( a + 2 δ x ) 1 + 1 ( a + 3 δ x ) 1 + + 1 ( a + n δ x ) 1 ] δ x S n = [ 1 a ( 1 + 1 δ x a ) 1 + 1 a ( 1 + 2 δ x a ) 1 + 1 a ( 1 + 3 δ x a ) 1 + + 1 a ( 1 + n δ x a ) 1 ] δ x

δx is small, so nδx/a < 1.

S n = δ x a [ 1 + -1 1! ( δ x a ) 1 + (-1)(-2) 2! ( δ x a ) 2 + (-1)(-2)(-3) 3! ( δ x a ) 3 + ] S n = δ x a + 1 + -1 1! ( 2 δ x a ) 1 + (-1)(-2) 2! ( 2 δ x a ) 2 + (-1)(-2)(-3) 3! ( 2 δ x a ) 3 + ] S n = δ x a + 1 + -1 1! ( 3 δ x a ) 1 + (-1)(-2) 2! ( 3 δ x a ) 2 + (-1)(-2)(-3) 3! ( 3 δ x a ) 3 + ] S n = δ x a + S n = δ x a + 1 + -1 1! ( n δ x a ) 1 + (-1)(-2) 2! ( n δ x a ) 2 + (-1)(-2)(-3) 3! ( n δ x a ) 3 + ]

Rearranging:

S n = δ x a [ 1 + 1 + 1 + + 1 n ] S n = δ x a + -1 1! ( 1 δ x a ) 1 + -1 1! ( 2 δ x a ) 1 + -1 1! ( 3 δ x a ) 1 + + -1 1! ( n δ x a ) 1 ] S n = δ x a + (-1)(-2) 2! ( 1 δ x a ) 2 + (-1)(-2) 2! ( 2 δ x a ) 2 + (-1)(-2) 2! ( 3 δ x a ) 2 + + (-1)(-2) 2! ( n δ x a ) 2 ] S n = δ x a + (-1)(-2)(-3) 3! ( 1 δ x a ) 3 + (-1)(-2)(-3) 3! ( 2 δ x a ) 3 + (-1)(-2)(-3) 3! ( 3 δ x a ) 3 + + (-1)(-2)(-3) 3! ( n δ x a ) 3 ] S n = δ x a + S n = δ x a + ]
S n = ( δ x a ) 1 r=1 n r 0 S n = δ x a + (-1) 1! ( δ x a ) 2 r=1 n r 1 S n = δ x a + (-1)(-2) 2! ( δ x a ) 3 r=1 n r 2 S n = δ x a + (-1)(-2)(-3) 3! ( δ x a ) 4 r=1 n r 3 S n = δ x a + S n = δ x a + ]

For the general term, rm, you need my general formula pattern for the power series, which is:

r = 1 n r m = 1 m+1 n m+1 + 1 2 n m + lower powers of n ...

Let the definite integral be represented by A, and let the indefinite integral be represented by I.

δ x = b - a n
S n = ( b-a an ) 1 [ n ] S n = δ x a - ( b-a an ) 2 [ n 2 2 + n 1 2 ] S n = δ x a + ( b-a an ) 3 [ n 3 3 + n 2 2 + n 1 6 ] S n = δ x a - ( b-a an ) 4 [ n 4 4 + n 3 2 + n 2 4 ] S n = δ x a + S n = δ x a - ]
S n = ( b-a a ) 1 S n = δ x a - ( b-a a ) 2 1 2 + ( b-a an ) 2 1 2 S n = δ x a + ( b-a a ) 3 1 3 + ( b-a an ) 3 1 3 + lower terms in n S n = δ x a - ( b-a a ) 4 1 4 + ( b-a an ) 4 1 4 + lower terms in n S n = δ x a + S n = δ x a - ]

As n → ∞, the fractions with denominators in n turn to 0, and Sn → A

A = 1 1 a (b-a) 1 - 1 2 a 2 (b-a) 2 + 1 3 a 3 (b-a) 3 - 1 4 a 4 (b-a) 4 + .the definite integral

If b = x and a = 1,

I = (x-1) 1 - 1 2 (x-1) 2 + 1 3 (x-1) 3 - 1 4 (x-1) 4 + .the indefinite integral

This is the indefinite integral according to the Taylor series expansion of ln(x) about x = 1.

I = In(x) = (x-1) 1 - 1 2 (x-1) 2 + 1 3 (x-1) 3 - 1 4 (x-1) 4 + .still the indefinite integral

Integration of the Term, 1/x2

Let the definite integral be represented by A, and let the indefinite integral be represented by I. Let the summation expression (discrete summation) be Sn.

Here,

y = 1/x2

i.e.

f(x) = 1/x2

S n = x = a x = b f ( x n ) δ x S n = [ 1 ( a + 1 δ x ) 2 + 1 ( a + 2 δ x ) 2 + 1 ( a + 3 δ x ) 2 + + 1 ( a + n δ x ) 2 ] δ x S n = [ 1 a2 ( 1 + 1 δ x a ) 2 + 1 a2 ( 1 + 2 δ x a ) 2 + 1 a2 ( 1 + 3 δ x a ) 2 + + 1 a2 ( 1 + n δ x a ) 2 ] δ x

δx is small, so nδx/a < 1.

S n = δ x a 2 [ 1 + -2 1! ( δ x a ) 1 + (-2)(-3) 2! ( δ x a ) 2 + (-2)(-3)(-4) 3! ( δ x a ) 3 + ] S n = δ x a + 1 + -2 1! ( 2 δ x a ) 1 + (-2)(-3) 2! ( 2 δ x a ) 2 + (-2)(-3)(-4) 3! ( 2 δ x a ) 3 + ] S n = δ x a + 1 + -2 1! ( 3 δ x a ) 1 + (-2)(-3) 2! ( 3 δ x a ) 2 + (-2)(-3)(-4) 3! ( 3 δ x a ) 3 + ] S n = δ x a + S n = δ x a + 1 + -2 1! ( n δ x a ) 1 + (-2)(-3) 2! ( n δ x a ) 2 + (-2)(-3)(-4) 3! ( n δ x a ) 3 + ]

Rearranging:

S n = δ x a 2 [ 1 + 1 + 1 + + 1 n ] S n = δ x a + -2 1! ( 1 δ x a ) 1 + -2 1! ( 2 δ x a ) 1 + -2 1! ( 3 δ x a ) 1 + + -2 1! ( n δ x a ) 1 ] S n = δ x a + (-2)(-3) 2! ( 1 δ x a ) 2 + (-2)(-3) 2! ( 2 δ x a ) 2 + (-2)(-3) 2! ( 3 δ x a ) 2 + + (-2)(-3) 2! ( n δ x a ) 2 ] S n = δ x a + (-2)(-2)(-4) 3! ( 1 δ x a ) 3 + (-2)(-3)(-4) 3! ( 2 δ x a ) 3 + (-2)(-3)(-4) 3! ( 3 δ x a ) 3 + + (-2)(-3)(-4) 3! ( n δ x a ) 3 ] S n = δ x a + S n = δ x a + ]
S n = ( δ x a 2 ) 1 r=1 n r 0 S n = δ x a + (-2) a11! ( δ x a ) 2 r=1 n r 1 S n = δ x a + (-2)(-3) a12! ( δ x a ) 3 r=1 n r 2 S n = δ x a + (-2)(-3)(-4) a13! ( δ x a ) 4 r=1 n r 3 S n = δ x a + S n = δ x a + ]

For the general term, rm, you need my general formula pattern for the power series, which is:

r = 1 n r m = 1 m+1 n m+1 + 1 2 n m + lower powers of n ...

Let the definite integral be represented by A, and let the indefinite integral be represented by I.

δ x = b - a n
S n = 1 a ( b-a an ) 1 [ n ] S n = δ x a - 2 a ( b-a an ) 2 [ n 2 2 + n 1 2 ] S n = δ x a + 3 a ( b-a an ) 3 [ n 3 3 + n 2 2 + n 1 6 ] S n = δ x a - 4 a ( b-a an ) 4 [ n 4 4 + n 3 2 + n 2 4 ] S n = δ x a + S n = δ x a - ]
S n = 1 a ( b-a a ) 1 S n = δ x a - 2 a ( b-a a ) 2 1 2 + 2 a ( b-a an ) 2 1 2 S n = δ x a + 3 a ( b-a a ) 3 1 3 + 3 a ( b-a an ) 3 1 3 + lower terms in n S n = δ x a - 4 a ( b-a a ) 4 1 4 + 4 a ( b-a an ) 4 1 4 + lower terms in n S n = δ x a + S n = δ x a - ]

As n → ∞, the fractions with denominators in n turn to 0, and Sn → A

A = 1 1 a2 (b-a) 1 - 1 1 a 3 (b-a) 2 + 1 1 a 4 (b-a) 3 - 1 1 a 5 (b-a) 4 +

.looks like the definite integral, but it is not, because the summation formula still needs to be obtained, and the limit as n → ∞ determined.


A = 1 a [ 1 1 a1 (b-a) 1 - 1 1 a 2 (b-a) 2 + 1 1 a 3 (b-a) 3 - 1 1 a 4 (b-a) 4 + ] a A = [ + 1 1 a1 (b-a) 1 - 1 1 a 2 (b-a) 2 + 1 1 a 3 (b-a) 3 - 1 1 a 4 (b-a) 4 + ]

Multiplying both sides by -1 gives:

- a A = [ - 1 1 a1 (b-a) 1 + 1 1 a 2 (b-a) 2 - 1 1 a 3 (b-a) 3 + 1 1 a 4 (b-a) 4 - ]

Add 1 to both sides:

- a A + 1 = 1 - 1 1 a1 (b-a) 1 + 1 1 a 2 (b-a) 2 - 1 1 a 3 (b-a) 3 + 1 1 a 4 (b-a) 4 -

If (b-a)/a is less than 1, then the right-hand-side is the reciprocal as follows:

- a A + 1 = 1 ( 1 + ( b-a a ) ) 1

Multiplying numerator and denominator of right-hand-side by a, gives:

- a A + 1 = a ( a + ( b-a ) ) 1 - a A = a b 1 - 1 A = - a ab 1 - - 1 a 1 A = - 1 b 1 - - 1 a 1 .the definite integral

Now, there is no value of a, for which we can have the indefinite integral, of the type resulting from the reversal of differentiation, which is 1/x. So, by analogy with the natural result of the integration of 1/x to give ln(x), I have chosen 1 for a, and x for b, to have the indefinite integral, from first principles. Remember, that b > a. So as a turns to infinity, b also turns to infinity faster. So you cannot end up with the result of -1/x for the indefinite integral; it would not make sense.

If b = x and a = 1,

I = - 1 x 1 - - 1 1 . the indefinite integral from First Principles

The definite integral above, has been proven for (b-a)/a < 1. When (b-a)/a >= 1, for example, b=7 and a=3, divide the x-axis interval into portions, such that for each portion, the corresponding (b-a)/a is less than 1. If we divide the interval from 3 to 7 into portions of 1 unit each, that is, b1=4, a1=3; b2=5, a2=4; b3=6, a3=5; and b4=7, a4=6; then

A = ( - 1 4 1 - - 1 3 1 ) + ( - 1 5 1 - - 1 4 1 ) + ( - 1 6 1 - - 1 5 1 ) + ( - 1 7 1 - - 1 6 1 ) A = - 1 7 1 - - 1 3 1 . still a definite integral

Integration of the Term, 1/x3

Let the definite integral be represented by A, and let the indefinite integral be represented by I. Let the summation expression (discrete summation) be Sn.

Here,

y = 1/x3

i.e.

f(x) = 1/x3

S n = x = a x = b f ( x n ) δ x S n = [ 1 ( a + 1 δ x ) 3 + 1 ( a + 2 δ x ) 3 + 1 ( a + 3 δ x ) 3 + + 1 ( a + n δ x ) 3 ] δ x S n = [ 1 a3 ( 1 + 1 δ x a ) 3 + 1 a3 ( 1 + 2 δ x a ) 3 + 1 a3 ( 1 + 3 δ x a ) 3 + + 1 a3 ( 1 + n δ x a ) 3 ] δ x

δx is small, so nδx/a < 1.

S n = δ x a 3 [ 1 + -3 1! ( δ x a ) 1 + (-3)(-4) 2! ( δ x a ) 2 + (-3)(-4)(-5) 3! ( δ x a ) 3 + ] S n = δ x a + 1 + -3 1! ( 2 δ x a ) 1 + (-3)(-4) 2! ( 2 δ x a ) 2 + (-3)(-4)(-5) 3! ( 2 δ x a ) 3 + ] S n = δ x a + 1 + -3 1! ( 3 δ x a ) 1 + (-3)(-4) 2! ( 3 δ x a ) 2 + (-3)(-4)(-5) 3! ( 3 δ x a ) 3 + ] S n = δ x a + S n = δ x a + 1 + -3 1! ( n δ x a ) 1 + (-3)(-4) 2! ( n δ x a ) 2 + (-3)(-4)(-5) 3! ( n δ x a ) 3 + ]

Rearranging:

S n = δ x a 3 [ 1 + 1 + 1 + + 1 n ] S n = δ x a + -3 1! ( 1 δ x a ) 1 + -3 1! ( 2 δ x a ) 1 + -3 1! ( 3 δ x a ) 1 + + -3 1! ( n δ x a ) 1 ] S n = δ x a + (-3)(-4) 2! ( 1 δ x a ) 2 + (-3)(-4) 2! ( 2 δ x a ) 2 + (-3)(-4) 2! ( 3 δ x a ) 2 + + (-3)(-4) 2! ( n δ x a ) 2 ] S n = δ x a + (-3)(-4)(-5) 3! ( 1 δ x a ) 3 + (-3)(-4)(-5) 3! ( 2 δ x a ) 3 + (-3)(-4)(-5) 3! ( 3 δ x a ) 3 + + (-3)(-4)(-5) 3! ( n δ x a ) 3 ] S n = δ x a + S n = δ x a + ]
S n = ( δ x a 3 ) 1 r=1 n r 0 S n = δ x a + (-3) a21! ( δ x a ) 2 r=1 n r 1 S n = δ x a + (-3)(-4) a22! ( δ x a ) 3 r=1 n r 2 S n = δ x a + (-3)(-4)(-5) a23! ( δ x a ) 4 r=1 n r 3 S n = δ x a + S n = δ x a + ]

For the general term, rm, you need my general formula pattern for the power series, which is:

r = 1 n r m = 1 m+1 n m+1 + 1 2 n m + lower powers of n ...

Let the definite integral be represented by A, and let the indefinite integral be represented by I.

δ x = b - a n
S n = 1 a 2 ( b-a an ) 1 [ n ] S n = δ x a + (-3) 1! a 2 ( b-a an ) 2 [ n 2 2 + n 1 2 ] S n = δ x a + (-3)(-4) 2! a 2 ( b-a an ) 3 [ n 3 3 + n 2 2 + n 1 6 ] S n = δ x a + (-3)(-4)(-5) 3! a 2 ( b-a an ) 4 [ n 4 4 + n 3 2 + n 2 4 ] S n = δ x a + S n = δ x a + ]
S n = 1 a 2 ( b-a a ) 1 S n = δ x a + (-3) 1! a 2 ( b-a a ) 2 1 2 + (-3) 1! a 2 ( b-a an ) 2 1 2 S n = δ x a + (-3)(-4) 2! a 2 ( b-a a ) 3 1 3 + (-3)(-4) 2! a 2 ( b-a an ) 3 1 3 + lower terms in n S n = δ x a + (-3)(-4)(-5) 3! a 2 ( b-a a ) 4 1 4 + (-3)(-4)(-5) 3! a 2 ( b-a an ) 4 1 4 + lower terms in n S n = δ x a + S n = δ x a + ]

As n → ∞, the fractions with denominators in n turn to 0, and Sn → A

A = 1 1 !a3 (b-a) 1 + (-3) 2! a 4 (b-a) 2 + (-3)(-4) 3! a 5 (b-a) 3 + (-3)(-4)(-5) 4! a 6 (b-a) 4 +

.looks like the definite integral, but it is not, because the summation formula still needs to be obtained, and the limit as n → ∞ determined.


A = 1 a 2 [ 1 1!a1 (b-a) 1 + (-3) 2! a 2 (b-a) 2 + (-3)(-4) 3! a 3 (b-a) 3 + (-3)(-4)(-5) 4! a 4 (b-a) 4 + ] a 2 A = [ + 1 1!a1 (b-a) 1 + (-3) 2! a 2 (b-a) 2 + (-3)(-4) 3! a 3 (b-a) 3 + (-3)(-4)(-5) 4! a 4 (b-a) 4 + ]

Multiplying both sides by -2 gives:

- 2 a 2 A = [ + (-2) 1!a1 (b-a) 1 + (-2)(-3) 2! a 2 (b-a) 2 + (-2)(-3)(-4) 3! a 3 (b-a) 3 + (-2)(-3)(-4)(-5) 4! a 4 (b-a) 4 + ]

Add 1 to both sides:

- 2 a 2 A + 1 = 1 + (-2) 1!a1 (b-a) 1 + (-2)(-3) 2! a 2 (b-a) 2 + (-2)(-3)(-4) 3! a 3 (b-a) 3 + (-2)(-3)(-4)(-5) 4! a 4 (b-a) 4 +

If (b-a)/a is less than 1, then the right-hand-side is the reciprocal as follows:

- 2 a 2 A + 1 = 1 ( 1 + ( b-a a ) ) 2

Multiplying numerator and denominator of right-hand-side by a2, gives:

- 2 a 2 A + 1 = a 2 ( a + ( b-a ) ) 2 - 2 a 2 A = a 2 b 2 - 1 A = - a 2 2 a2 b2 - - 1 2 a2 A = - 1 2 b2 - - 1 2 a2 .the definite integral

Now, there is no value of a, for which we can have the indefinite integral, of the type resulting from the reversal of differentiation, which is 1/2x2. So, by analogy with the natural result of the integration of 1/x to give ln(x), I have chosen 1 for a, and x for b, to have the indefinite integral, from first principles. Remember, that b > a. So as a turns to infinity, b also turns to infinity faster. So you cannot end up with the result of -1/2x2 for the indefinite integral; it would not make sense.

If b = x and a = 1,

I = - 1 2 x2 - - 1 2 . the indefinite integral from First Principles

The definite integral above, has been proven for (b-a)/a < 1. When (b-a)/a >= 1, for example, b=7 and a=3, divide the x-axis interval into portions, such that for each portion, the corresponding (b-a)/a is less than 1. If we divide the interval from 3 to 7 into portions of 1 unit each, that is, b1=4, a1=3; b2=5, a2=4; b3=6, a3=5; and b4=7, a4=6; then

A = ( - 1 2 (42) - - 1 2 (32) ) + ( - 1 2 (52) - - 1 2 (42) ) + ( - 1 2 (62) - - 1 2 (52) ) + ( - 1 2 (72) - - 1 2 (62) ) A = - 1 2 (72) - - 1 2 (32) . still a definite integral

Integration of the Term, 1/x4

Let the definite integral be represented by A, and let the indefinite integral be represented by I. Let the summation expression (discrete summation) be Sn.

Here,

y = 1/x4

i.e.

f(x) = 1/x4

S n = x = a x = b f ( x n ) δ x S n = [ 1 ( a + 1 δ x ) 4 + 1 ( a + 2 δ x ) 4 + 1 ( a + 3 δ x ) 4 + + 1 ( a + n δ x ) 4 ] δ x S n = [ 1 a4 ( 1 + 1 δ x a ) 4 + 1 a4 ( 1 + 2 δ x a ) 4 + 1 a4 ( 1 + 3 δ x a ) 4 + + 1 a4 ( 1 + n δ x a ) 4 ] δ x

δx is small, so nδx/a < 1.

S n = δ x a 4 [ 1 + -4 1! ( δ x a ) 1 + (-4)(-5) 2! ( δ x a ) 2 + (-4)(-5)(-6) 3! ( δ x a ) 3 + ] S n = δ x a + 1 + -4 1! ( 2 δ x a ) 1 + (-4)(-5) 2! ( 2 δ x a ) 2 + (-4)(-5)(-6) 3! ( 2 δ x a ) 3 + ] S n = δ x a + 1 + -4 1! ( 3 δ x a ) 1 + (-4)(-5) 2! ( 3 δ x a ) 2 + (-4)(-5)(-6) 3! ( 3 δ x a ) 3 + ] S n = δ x a + S n = δ x a + 1 + -4 1! ( n δ x a ) 1 + (-4)(-5) 2! ( n δ x a ) 2 + (-4)(-5)(-6) 3! ( n δ x a ) 3 + ]

Rearranging:

S n = δ x a 4 [ 1 + 1 + 1 + + 1 n ] S n = δ x a + -4 1! ( 1 δ x a ) 1 + -4 1! ( 2 δ x a ) 1 + -4 1! ( 3 δ x a ) 1 + + -4 1! ( n δ x a ) 1 ] S n = δ x a + (-4)(-5) 2! ( 1 δ x a ) 2 + (-4)(-5) 2! ( 2 δ x a ) 2 + (-4)(-5) 2! ( 3 δ x a ) 2 + + (-4)(-5) 2! ( n δ x a ) 2 ] S n = δ x a + (-4)(-5)(-6) 3! ( 1 δ x a ) 3 + (-4)(-5)(-6) 3! ( 2 δ x a ) 3 + (-4)(-5)(-6) 3! ( 3 δ x a ) 3 + + (-4)(-5)(-6) 3! ( n δ x a ) 3 ] S n = δ x a + S n = δ x a + ]
S n = ( δ x a 4 ) 1 r=1 n r 0 S n = δ x a + (-4) 1! ( δ x a 3 ) 2 r=1 n r 1 S n = δ x a + (-4)(-5) 2! ( δ x a 3 ) 3 r=1 n r 2 S n = δ x a + (-4)(-5)(-6) 3! ( δ x a 3 ) 4 r=1 n r 3 S n = δ x a + S n = δ x a + ]

For the general term, rm, you need my general formula pattern for the power series, which is:

r = 1 n r m = 1 m+1 n m+1 + 1 2 n m + lower powers of n ...

Let the definite integral be represented by A, and let the indefinite integral be represented by I.

δ x = b - a n
S n = 1 a 3 ( b-a an ) 1 [ n ] S n = δ x a + (-3) 1! a 3 ( b-a an ) 2 [ n 2 2 + n 1 2 ] S n = δ x a + (-3)(-4) 2! a 3 ( b-a an ) 3 [ n 3 3 + n 2 2 + n 1 6 ] S n = δ x a + (-3)(-4)(-5) 3! a 3 ( b-a an ) 4 [ n 4 4 + n 3 2 + n 2 4 ] S n = δ x a + S n = δ x a + ]
S n = 1 a 3 ( b-a a ) 1 S n = δ x a + (-3) 1! a 3 ( b-a a ) 2 1 2 + (-3) 1! a 3 ( b-a an ) 2 1 2 S n = δ x a + (-3)(-4) 2! a 3 ( b-a a ) 3 1 3 + (-3)(-4) 2! a 3 ( b-a an ) 3 1 3 + lower terms in n S n = δ x a + (-3)(-4)(-5) 3! a 3 ( b-a a ) 4 1 4 + (-3)(-4)(-5) 3! a 3 ( b-a an ) 4 1 4 + lower terms in n S n = δ x a + S n = δ x a + ]

As n → ∞, the fractions with denominators in n turn to 0, and Sn → A

A = 1 1 !a4 (b-a) 1 + (-3) 2! a 5 (b-a) 2 + (-3)(-4) 3! a 6 (b-a) 3 + (-3)(-4)(-5) 4! a 7 (b-a) 4 +

.looks like the definite integral, but it is not, because the summation formula still needs to be obtained, and the limit as n → ∞ determined.


A = 1 a 3 [ 1 1!a1 (b-a) 1 + (-4) 2! a 2 (b-a) 2 + (-4)(-5) 3! a 3 (b-a) 3 + (-4)(-5)(-6) 4! a 4 (b-a) 4 + ] a 3 A = [ + 1 1!a1 (b-a) 1 + (-4) 2! a 2 (b-a) 2 + (-4)(-5) 3! a 3 (b-a) 3 + (-4)(-5)(-6) 4! a 4 (b-a) 4 + ]

Multiplying both sides by -3 gives:

- 3 a 3 A = [ + (-3) 1!a1 (b-a) 1 + (-3)(-4) 2! a 2 (b-a) 2 + (-3)(-4)(-5) 3! a 3 (b-a) 3 + (-3)(-4)(-5)(-6) 4! a 4 (b-a) 4 + ]

Add 1 to both sides:

- 3 a 3 A + 1 = 1 + (-3) 1!a1 (b-a) 1 + (-3)(-4) 2! a 2 (b-a) 2 + (-3)(-4)(-5) 3! a 3 (b-a) 3 + (-3)(-4)(-5)(-6) 4! a 4 (b-a) 4 +

If (b-a)/a is less than 1, then the right-hand-side is the reciprocal as follows:

- 3 a 3 A + 1 = 1 ( 1 + ( b-a a ) ) 3

Multiplying numerator and denominator of right-hand-side by a3, gives:

- 3 a 3 A + 1 = a 3 ( a + ( b-a ) ) 3 - 3 a 3 A = a 3 b 3 - 1 A = - a 3 3 a3 b3 - - 1 3 a3 A = - 1 3 b3 - - 1 3 a3 .the definite integral

Now, there is no value of a, for which we can have the indefinite integral, of the type resulting from the reversal of differentiation, which is 1/3x3. So, by analogy with the natural result of the integration of 1/x to give ln(x), I have chosen 1 for a, and x for b, to have the indefinite integral, from first principles. Remember, that b > a. So as a turns to infinity, b also turns to infinity faster. So you cannot end up with the result of -1/3x3 for the indefinite integral; it would not make sense.

If b = x and a = 1,

I = - 1 3 x3 - - 1 3 . the indefinite integral from First Principles

The definite integral above, has been proven for (b-a)/a < 1. When (b-a)/a >= 1, for example, b=7 and a=3, divide the x-axis interval into portions, such that for each portion, the corresponding (b-a)/a is less than 1. If we divide the interval from 3 to 7 into portions of 1 unit each, that is, b1=4, a1=3; b2=5, a2=4; b3=6, a3=5; and b4=7, a4=6; then

A = ( - 1 3 (43) - - 1 3 (33) ) + ( - 1 3 (53) - - 1 3 (43) ) + ( - 1 3 (63) - - 1 3 (53) ) + ( - 1 3 (73) - - 1 3 (63) ) A = - 1 3 (73) - - 1 3 (33) . still a definite integral

Integration of the General Term, 1/xm

1/x1
A = 1 1 a (b-a) 1 - 1 2 a 2 (b-a) 2 + 1 3 a 3 (b-a) 3 - 1 4 a 4 (b-a) 4 + .the definite integral

If b = x and a = 1,

I = In|x| = (x-1) 1 - 1 2 (x-1) 2 + 1 3 (x-1) 3 - 1 4 (x-1) 4 + . the indefinite integral
1/xm, where m > 1

A = - 1 (m-1) b(m-1) - - 1 (m-1) a(m-1) .the definite integral

Now, there is no value of a, for which we can have the indefinite integral, of the type resulting from the reversal of differentiation, which is 1/[(m-1)x(m-1)]. So, by analogy with the natural result of the integration of 1/x to give ln(x), I have chosen 1 for a, and x for b, to have the indefinite integral, from first principles. Remember, that b > a. So as a turns to infinity, b also turns to infinity faster. So you cannot end up with the result of 1/[(m-1)x(m-1)] for the indefinite integral; it would not make sense.

If b = x and a = 1,

I = - 1 (m-1) x(m-1) - - 1 (m-1) . the indefinite integral from First Principles

This is the end of this part of the series. Continue in the next part.

Chrys


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