Integrals of Wallis

In Analysis, the integral of Wallis constitute a family of Intégrale S introduced by John Wallis.

Definition, first properties

One calls usually integral of Wallis the terms of the real continuation $(W_n) _ {\, N \, \ in \, \ mathbb {NR} \,}$ defined by:

W_n = \ int_0^ {\ frac {\ pi} {2}} \ sin^n (X) \, dx

, or in an equivalent way (by the change of variable

x = \ frac {\ pi} {2} - t

W_n = \ int_0^ {\ frac {\ pi} {2}} \ cos^n (X) \, dx

In particular, the first two terms of this continuation are:

\ quad W_0= \ frac {\ pi} {2} \ qquad \,

and

\ quad W_1=1 \,

The continuation $\ (W_n)$ is decreasing, with strictly positive terms. Indeed, for all $n \ in \, \ mathbb {NR}$ :

$\ W_n > 0$ : it is the integral of a function continuous, positive, and not identically null on the interval of integration
$W_ {N} - W_ {N + 1} = \ int_0^ {\ frac {\ pi} {2}} \ sin^ {N} (X) \, dx - \ int_0^ {\ frac {\ pi} {2}} \ sin^ {N + 1} (X) \, dx = \ int_0^ {\ frac {\ pi} {2}} \ sin^ {N} (X) \, - \ sin (X) \, dx \ geqslant 0$

(by linearity of the integral and because the last integral is that of a positive function on the interval of integration)

Nota: decreasing and undervalued (by 0), the continuation

\ (W_n)

converges, and its limit is positive or null; in fact, it is null, as that results from the equivalent obtained further.

Relation of recurrence, calculation of the integrals of Wallis

A Intégration by parts will make it possible to establish a relation of interesting Récurrence:

By noticing that for any reality $x$ , $\ quad \ sin^2 (X) = 1 \ cos^2 (X)$ , one has for entire naturalness N :

$\ int_0^ {\ frac {\ pi} {2}} \ sin^ {n+2} (X) \, dx = \ int_0^ {\ frac {\ pi} {2}} \ sin^n (X) \ left \, dx$

$\ int_0^ {\ frac {\ pi} {2}} \ sin^ {n+2} (X) \, dx = \ int_0^ {\ frac {\ pi} {2}} \ sin^n (X) \, dx - \ int_0^ {\ frac {\ pi} {2}} \ sin^n (X) \ cos^2 (X) \, dx$ (relation $\ mathbf {(1)}$ )

One then integrates by parts the second integral of the second member:

$\ int_0^ {\ frac {\ pi} {2}} \ sin^n (X) \ cos^2 (X) \, dx = \ left \ frac {1} {n+1} \ sin^ {n+1} (X) \ cos (X) \ right_0^ {\ frac {\ pi} {2}} + \ int_0^ {\ frac {\ pi} {2}} \ \ frac {1} {n+1} \ sin^ {n+1} (X) \ sin (X) \, dx$

While deferring in $\ mathbf {(1)}$ , one obtains then:

$W_ {n+2} =W_n - {1 \ over {n+1}} \, W_ {n+2}$

from where

\ quad (n+2) \, W_ {n+2} = (n+1) \, W_n

(relation

\ mathbf {(2)}

)

This results in the well-known relation:

$n \, W_n = (n-1) \, W_ {N2} \ qquad \,$ valid for $n \ geqslant 2 \ qquad \,$ . There is there a relation of recurrence giving $W_n$ according to $W_ {N2}$ , IE the nth term of the continuation according to that which precedes it by two rows.

From this relation and values of

W_0

and

W_1

, one draws an expression from the terms of the continuation, according to the parity of their row. As follows:

for $\ quad n=2 \, p$ , $\ quad W_ {2 \, p} = \ frac {2 \, p-1} {2 \, p} \, \ frac {2 \, p-3} {2 \, p-2} \ cdots \ frac {1} {2} \, W_0= \ frac {(2 \, p)!}{2^ {2 \, p} \, (p!)^2} \ frac {\ pi} {2}$
for $\ quad n=2 \, p+1$ , $\ quad W_ {2 \, p+1} = \ frac {2 \, p} {2 \, p+1} \, \ frac {2 \, p-2} {2 \, p-1} \ cdots \ frac {2} {3} \, W_1= \ frac {2^ {2 \, p} \, (p!)^2} {(2 \, p +1)!}$

It is noticed that the terms of even row are irrational, while those of odd row are rational.

An equivalent of the continuation of the integrals of Wallis

Of the formula of preceding recurrence $\ mathbf {(2)}$ , one deduces initially that:

\ W_ {N + 1} \ sim W_n

(equivalence of two continuations).

Indeed, for all $n \ in \, \ mathbb {NR}$ :

\ W_ {N + 2} \ leqslant W_ {N + 1} \ leqslant W_n

(the continuation being decreasing) thus:

\ frac {W_ {N + 2}} {W_n} \ leqslant \ frac {W_ {N + 1}} {W_n} \ leqslant 1

(since

\ W_n > 0

), which is written:

\ frac {N + 1} {N + 2} \ leqslant \ frac {W_ {N + 1}} {W_n} \ leqslant 1

(according to the relation

\ mathbf {(2)}

By framing, one concludes that

\ frac {W_ {N + 1}} {W_n} \ to 1

, is

\ W_ {N + 1} \ sim W_n

Then one establishes following equivalence:

$W_n \ sim \ sqrt {\ frac {\ pi} {2 \, N}} \ quad$ (either still $\ quad \ lim_ {N \ rightarrow \ infty} \ sqrt N \, W_n= \ sqrt {\ pi /2} \ quad$ ).

Application to the formula of Stirling

Known following equivalence is supposed (established in the article on the Formule of Stirling):

\ N \! \ sim C \, \ sqrt {N} \ left (\ frac {N} {\ mathrm {E}} \ right) ^n

, where

\ C \ in \ R^*

One now proposes to determine the constant

\ C

using equivalents of

W_ {2 \, p}

Of the preceding paragraph results equivalence:

W_ {2 \, p} \ sim \ sqrt {\ frac {\ pi} {4 \, p}} = \ frac {\ sqrt {\ pi}} {2} \, \ frac {1} {\ sqrt {p}}

(relation

\ mathbf {(3)}

)

In addition, by using the equivalent of the factorial given supra :

W_ {2 \, p} = \ frac {(2 \, p)!}{2^ {2 \, p} \, (p \!)^2} \, \ frac {\ pi} {2} \ sim \ frac {C \, \ left (\ frac {2 \, p} {\ mathrm {E}} \ right) ^ {2p} \, \ sqrt {2 \, p}} {2^ {2p} \, C^2 \, \ left (\ frac {p} {\ mathrm {E}} \ right) ^ {2p} \, \ left (\ sqrt {p} \ right) ^2} \, \ frac {\ pi} {2}

, is:

W_ {2 \, p} \ sim \ frac {\ pi} {C \, \ sqrt {2}} \, \ frac {1} {\ sqrt {p}}

(relation

\ mathbf {(4)}

)

Of equivalences

\ mathbf {(3)}

and

\ mathbf {(4)}

, one deduces by transitivity:

\ frac {\ pi} {C \, \ sqrt {2}} \, \ frac {1} {\ sqrt {p}} \ sim \ frac {\ sqrt {\ pi}} {2} \, \ frac {1} {\ sqrt {p}}

, from where:

\ frac {\ pi} {C \, \ sqrt {2}} = \ frac {\ sqrt {\ pi}} {2}

, and finally

C = \ sqrt {2 \, \ pi}

One thus established the formula of Stirling in its final version:

\ N \! \ sim \ sqrt {2 \, \ pi \, N} \, \ left (\ frac {N} {\ mathrm {E}} \ right) ^n

Application to the calculation of the integral of Gauss

One can easily use the integrals of Wallis to calculate the Intégrale Gauss.

Let us check initially the following inequalities:

$\ forall N \ in \ mathbb N^* \ quad \ forall U \ in \ mathbb R_+ \ quad U \ leqslant N \ quad \ Rightarrow \ quad (1-u/n) ^n \ leqslant e^ {- U}$
$\ forall N \ in \ mathbb N^* \ quad \ forall U \ in \ mathbb R_+ \ qquad e^ {- U} \ leqslant (1+u/n) ^ {- N}$

Indeed by posing

\ quad u/n=t

the first inequality (for which

t \ in

) is equivalent to

1-t \ leqslant e^ {- T}

. As for the second she is written

e^ {- T} \ leqslant (1+t) ^ {- 1}

, which returns to

e^t \ geqslant 1+t

. These 2 inequalities are immediate consequences of the convexity of the exponential function (or if one prefers study of the function

t \ mapsto e^t-1 - t

Posing $u=x^2$ then and using the property elementary of the integrals (" impropres") (the convergence of the integrals is immediate) the framing is obtained:

$\ int_0^ {\ sqrt N} (1-x^2/n) ^n dx \ leqslant \ int_0^ {\ sqrt N} e^ {- x^2} dx \ leqslant \ int_0^ {+ \ infty} e^ {- x^2} dx \ leqslant \ int_0^ {+ \ infty} (1+x^2/n) ^ {- N} dx$ .

However the integrals of framing are brought back easily to integrals of Wallis. For that of left it is enough to pose $x= \ sqrt N \, \ sin \, T$ (T varying of 0 with $\ pi /2$ ) and she is written $\ sqrt N \, W_ {2n+1}$ . As for that of right-hand side, one can pose $x= \ sqrt N \, \ tan \, t$ (T varying of $0$ with $\ pi /2$ ) which gives $\ sqrt N \, W_ {2n-2}$ .

As it was seen that $\ lim_ {N \ rightarrow + \ infty} \ sqrt N \; W_n= \ sqrt {\ pi /2}$ , one from of deduced that $\ int_0^ {+ \ infty} e^ {- x^2} dx = \ sqrt {\ pi} /2$ .

Note: There exists many other methods of calculating of the integral of Gauss, including one method much more direct.

Foot-note

The same properties lead to the produced of Wallis, which expresses

\ frac {\ pi} {2} \,

(see '' π '') in the form of an infinite Produit.

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