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EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1

Part I, Section I,Chapter 8. Translated and annotated by Ian Bruce. page 340

CHAPTER VIII

CONCERNING THE EVALUATION OF INTEGRALS ON

TAKING CERTAIN CASES ONLY

PROBLEM 38 330. The value of the integral

( )1mx dx

xx assigned, that it receives on putting 1x = , clearly with the integral thus determined so that it vanishes on putting 0x = .

SOLUTION For the simplest cases, in which m = 0 or m = 1, we have after integration on putting x = 1

( ) ( )21 1and 1dx xdx

xx xx.

= =

Then from above 120 we have seen in general that

( ) ( )( )1 1 11 11 1 1 ;

m m mx dx m x dxm mxx xx

x xx+

+ + =

hence in the case x = 1 there will be

( ) ( )1 1

11 1

m mx dx m x dxmxx xx

+

+ =

for which from the simplest to the greater values of the exponents m we obtain by progressing :

( ) 21mx dx

xx

= ( )1 1

xdxxx

=

( )12 21

xxdxxx

= ( )3 2

31x dx

xx=

( )4 31

2 4 21x dx

xx

= ( )

5 2 43 51

x dxxx

=

( )6 3 51

2 4 6 21x dx

xx

= ( )

7 2 4 63 5 71

x dxxx

=

( )8 3 5 71

2 4 6 8 21x dx

xx

= ( )

9 2 4 6 83 5 7 91

x dxxx

= . . . . . .

( )( )2 13 5 2 1

2 4 6 2 21

n nx dxnxx

= ( ) ( )2 1 2 4 6 2

3 5 7 2 11

nx dx nnxx

+ +

=

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1

Part I, Section I,Chapter 8. Translated and annotated by Ian Bruce. page 341

COROLLARY 1

331. Hence the integral( )1

mx dxxx on putting x = 1 is expressed algebraically in the cases in

which m is an odd whole number, but in the cases in which it is even, it involves the quadrature of the circle ; for designates the periphery of the circle, of which the diameter is equal to 1.

COROLLARY 2

332. If we should multiply the two last formulas together, there is produced

( ) ( ) ( )2 2 1 1

22 11 1

n nx dx x dxnxx xx

++

=

evidently on putting x = 1, since that is known to be true, even if n is not a whole number. [This result can be shown generally from betta and gamma functions. See the O.O. edition for more details about this.]

COROLLARY 3 333. Hence this equality stands, if we should put vx z= , with the same conditions, because on taking x = 0 or z = 1. Then there becomes :

( ) ( ) ( )2 1 2 2 1

2 21

22 11 1

nv v nv v

v vz dz z dz

nz zvv

+ +

+ = ,

and on putting 2 1nv v + = it becomes on putting z = 1

( ) ( ) ( )2 21

211 1

v

v vz dz z dz

vz z

+

+ = .

SCHOLIUM 1

334. Since such a product of two integrals is able to be shown, consequently it is more worthy to be noted, as this equality stands, even if neither formula can be shown, either algebraically or through . Just as if v = 2 and 0 = , there becomes

( ) ( )4 412 2 41 1

dz zzdzz z

= = and in a similar manner,

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1

Part I, Section I,Chapter 8. Translated and annotated by Ian Bruce. page 342

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( )

3

6 6

4

6 6

4

8 8

6

8 8

5

10

13 2 61 1

16 2 121 1

14 2 81 1

112 2 241 1

1

3 0 becomes

3 1 becomes

4 0 becomes

4 2 becomes

5 0 becomes

dz z dzz z

zdz z dzz z

dz z dzz z

zzdz z dzz z

dz zz

v , , ,

v , , ,

v , , ,

v , , ,

v , ,

= = = =

= = = =

= = = =

= = = =

= =

( )

( ) ( )

( ) ( )

( ) ( )

10

6

10 10

7

10 10

3 8

10 10

15 2 101

110 2 201 1

115 2 301 1

120 2 401 1

5 1 becomes

5 2 becomes

5 3 becomes

dzz

zdz z dzz z

zzdz z dzz z

z dz z dzz z

,

v , , ,

v , , ,

v , , ,

= =

= = = =

= = = =

= = = =

which theorems without doubt are all worthy of attention.

SCHOLIUM 2

335. Hence also the value of the integral( )

mx dxx xx can be deduced easily on putting x = 1;

for if we write x zz= , the integral is made into this : ( )2

12

mz dzzz , concerning which for

the case x = 1 we arrive at the following values :

( ) ( )

( ) ( )

( )

( )

4

5

3

13 5 72 4 6 8

13 5 7 912 2 4 6 810

132 4

13 52 4 6

:

dx x dxx xx x xx

xdx x dxx xx x xx

xxdxx xx

x dxx xx

, ,

, ,

,

,

= =

= =

=

=

( )( )13 5 2 1

2 4 6 2 m mx dx

mx xx. =

Hence the formulas of this kind of integrals, involving more complicated values which they take on putting x = 1, can be succinctly expressed by series, the use of which we indicate with several examples.

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1

Part I, Section I,Chapter 8. Translated and annotated by Ian Bruce. page 343

EXAMPLE 1

336. To show the value of the integral ( )41

dxx on putting x = 1 by a series.

This form can be given to the integral [See E605 for this integral arising from an elastic curve]:

( )( )

12

11dx

xxxx

+ ,

so that we have :

( ) ( ) ( )44 6 813 13 5 13 5 71

2 2 4 2 4 6 2 4 6 8111 etc ;dx dx

xxxxx x x x .

= + + hence from the individual terms of the integral for the case x = 1 there arises :

( ) ( )419 19 25 19 25 491

2 4 416 416 36 416 36 6411 etc .dx

x.

= + + COROLLARY

337. In a like manner in the same case x = 1 there is found :

( )

( ) ( )

( )

4

2 2 2 2 2 2

2 2 2 2 2 24

3

4

1 1 1 1 13 5 7 9 11 41

1 3 1 3 5 1 3 5 712 2 2 4 2 4 6 2 4 6 81

6 8 102 43 3 5 5 7 7 9 9111

1 + etc ,

etc ,

etc ;

xdxx

xxdxx

x dxx

.

.

.

= + + =

= + +

= + +

but

( ) ( )3

4

41 12 21

1x dxx

x

= and thus 12= on putting x = 1, from which this final series is

12= .

[Note that the first is Gregory's formula for 4 .]

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1

Part I, Section I,Chapter 8. Translated and annotated by Ian Bruce. page 344

EXAMPLE 2

338. To show the value of the integral 11

axxxxdx

+ in the case x = 1 by a series.

Since there shall be ( ) 2 4 3 61131 112 2 4 2 4 61 1 etcaxx axx a x a x ., + = + +

it becomes on multiplying by

( )1dx

xx and on integrating :

( )2 31 1113 1113 5111 2 2 2 2 2 4 4 2 2 4 4 6 61 etcaxxxxdx a a a . ,+ = + + from which the periphery of the ellipse can be recognised [See E154].

EXAMPLE 3 339. To show the value of the integral

( )1dx

x xx in the case x = 1 by a series.

This formula may be represented thus ( )( )

121dx x

x xx

+

, so that it becomes :

( ) ( )3 313 13 512 2 4 2 4 61 etcdxx xx x x x . , + + from which this series may be obtained :

( ) ( )19 19 251

4 416 416 361 etcdx

x x xx. , = + +

which does not differ from the first series; which is not surprising, since on putting x = zz this formula is reduced to that.

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1

Part I, Section I,Chapter 8. Translated and annotated by Ian Bruce. page 345

PROBLEM 39

340. To define the value of the integral ( )121 1

nmx dx xx which vanishes on putting

x = 0. SOLUTION

The reductions given above in 118 produce in this case :

( ) ( ) ( )12

2 211 21 12 21 1 ;

mx xxm mm mx dx xx x dx xx

++ + + + + + = +

hence on taking 2 1n = there will be

( ) ( )1211 12 2

2 11 1n nm mn

m nx dx xx x dx xx+ +

+ + = on putting x = 1. There since in the preceding problem the value

( )1

1

mx dxxx

can be assigned, as for the sake of brevity we put = M, hence we can progress to the following :

( )

( )

( ) ( )( )( ) ( )( )( )

1

12

32

52

1

1 11

1 131 3

1 13 51 3 5

1

1

1

mx dxxx

mm

mm m

mm m m

M ,

x dx xx M ,

x dx xx M ,

x dx xx M

+

+ +

+ + +

=

=

=

=

and in general

( ) ( )( )( )( ) ( )12 13 5 2 11

1 3 5 2 11nnm

m m m m nx dx xx M .

+ + + + =

Now there are two cases to be considered carefully, since 1m is either an even or odd number ; for if 1m is even, then

( )( )

13 5 222 4 6 1 ;

mmM

=

but if 1m is odd, then ( )( )

2 4 6 23 5 7 1 .

mmM

=

Hence the following values are deduced :

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1

Part I, Section I,Chapter 8. Translated and annotated by Ian Bruce.