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5.11: Integral Definitions of Some Functions

This page is a draft and is under active development. 

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By Theorem 2 in §10, f exists on I whenever the function f:E1E is regulated on I, and E is complete. Hence whenever such an f is given, we can define a new function F by setting

F=xaf

on I for some aI. This is a convenient method of obtaining new continuous functions, differentiable on IQ (Q countable). We shall now apply it to obtain new definitions of some functions previously defined in a rather strenuous step-by-step manner.

I. Logarithmic and Exponential Functions. From our former definitions, we proved that

lnx=x11tdt,x>0.

Now we want to treat this as a definition of logarithms. We start by setting

f(t)=1t,tE1,t0,

and f(0)=0.

Then f is continuous on I=(0,+) and J=(,0), so it has an exact primitive on I and J separately (not on E1). Thus we can now define the log function on I by

x11tdt=logx (also written lnx) for x>0.

By the very definition of an exact primitive, the log function is continuous and differentiable on I=(0,+); its derivative on I is f. Thus we again have the symbolic formula

(logx)=1x,x>0.

If x<0, we can consider log(x). Then the chain rule (Theorem 3 of §1) yields

(log(x))=1x. (Verify!)

Hence

(log|x|)=1x for x0.

Other properties of logarithms easily follow from (1). We summarize them now.

Theorem 5.11.1

(i) log1=111tdt=0.

(ii) logx<logy whenever 0<x<y.

(iii) limx+logx=+ and limx0+logx=.

(iv) The range of log is all of E1.

(v) For any positive x,yE1,

log(xy)=logx+logy and log(xy)=logxlogy.

(vi) logar=rloga,a>0,rN.

(vii) loge=1, where e=limn(1+1n)n.

Proof

(ii) By (2), (logx)>0 on I=(0,+), so logx is increasing on I.

(iii) By Theorem 5 in §10,

limx+logx=11tdt=+

since

n=11n=+(Chapter 4, §13, Example (b)).

Hence, substituting y=1/x, we obtain

limy0+logy=limx+log1x.

However, by Theorem 2 in §5 (substituting s=1/t),

log1x=1/x11tdt=x11sds=logx.

Thus

limy0+logy=limx+log1x=limx+logx=

as claimed. (We also proved that log1x=logx.)

(iv) Assertion (iv) now follows by the Darboux property (as in Chapter 4, §9, Example (b)).

(v) With x,y fixed, we substitute t=xs in

xy11tdt=logxy

and obtain

logxy=xy11tdt=y1/x1sds=11/x1sds+y11sds=log1x+logy=logx+logy.

Replacing y by 1/y here, we have

logxy=logx+log1y=logxlogy.

Thus (v) is proved, and (vi) follows by induction over r.

(vii) By continuity,

loge=limxelogx=limnlog(1+1n)n=limnlog(1+1/n)1/n,

where the last equality follows by (vi). Now, L'Hôpital's rule yields

limx0log(1+x)x=limx011+x=1.

Letting x run over 1n0, we get (vii).

Note 1. Actually, (vi) holds for any rE1, with ar as in Chapter 2, §§11-12. One uses the techniques from that section to prove it first for rational r, and then it follows for all real r by continuity. However, we prefer not to use this now.

Next, we define the exponential function ("exp") to be the inverse of the log function. This inverse function exists; it is continuous (even differentiable) and strictly increasing on its domain (by Theorem 3 of Chapter 4, §9 and Theorem 3 of Chapter 5, §2) since the log function has these properties. From (logx)=1/x we get, as in 2,

(expx)=expx(cf. §2, Example (B)).

The domain of the exponential is the range of its inverse, i.e., E1 (cf. Theorem 1(iv)). Thus expx is defined for all xE1. The range of exp is the domain of log, i.e., (0,+). Hence exp x>0 for all xE1. Also, by definition,

exp(logx)=x for x>0exp0=1 (cf. Theorem 1(i)), and expr=er for rN.

Indeed, by Theorem 1(vi) and (vii), log er=rloge=r. Hence (6) follows. If the definitions and rules of Chapter 2, §§11-12 are used, this proof even works for any r by Note 1. Thus our new definition of exp agrees with the old one.

Our next step is to give a new definition of ar, for any a,rE1(a>0). We set

ar=exp(rloga) or logar=rloga(rE1).

In case rN, (8) becomes Theorem 1(vi). Thus for natural r, our new definition of ar is consistent with the previous one. We also obtain, for a,b>0,

(ab)r=arbr;ars=(ar)s;ar+s=aras;(r,sE1).

The proof is by taking logarithms. For example,

log(ab)r=rlogab=r(loga+logb)=rloga+rlogb=logar+logbr=log(arbr).

Thus (ab)r=arbr. Similar arguments can be given for the rest of (9) and other laws stated in Chapter 2, §§11-12.

We can now define the exponential to the base a(a>0) and its inverse, log a, as before (see the example in Chapter 4, §5 and Example (b) in Chapter 4, §9). The differentiability of the former is now immediate from (7), and the rest follows as before.

II. Trigonometric Functions. These shall now be defined in a precise analytic manner (not based on geometry).

We start with an integral definition of what is usually called the principal value of the arcsine function,

arcsinx=x011t2dt.

We shall denote it by F(x) and set

f(x)=11x2 on I=(1,1).

(F=f=0 on E1I.) Thus by definition, F=f on I.

Note that f exists and is exact on I since f is continuous on I. Thus

F(x)=f(x)=11x2>0for xI,

and so F is strictly increasing on I. Also, F(0)=00f=0.

We also define the number π by setting

π2=2arcsin12=2F(c)=2c0f,c=12.

Then we obtain the following theorem.

Theorem 5.11.2

F has the limits

F(1)=π2 and F(1+)=π2.

Thus F becomes relatively continuous on ¯I=[1,1] if one sets

F(1)=π2 and F(1)=π2,

i.e.,

arcsin1=π2 and arcsin(1)=π2.

Proof

We have

F(x)=x0f=c0f+xcf,c=12.

By substituting s=1t2 in the last integral and setting, for brevity, y= 1x2, we obtain

xcf=xc11t2dt=cy11s2ds=F(c)F(y).(Verify!)

Now as x1, we have y=1x20, and hence F(y)F(0)=0 (for F is continuous at 0). Thus

F(1)=limx1F(x)=limy0(c0f+cyf)=c0f+F(c)=2c0f=π2.

Similarly, one gets F(1+)=π/2.

The function F as redefined in Theorem 2 will be denoted by F0. It is a primitive of f on the closed interval ¯I (exact on I). Thus F0(x)=x0f, 1x1, and we may now write

π2=10f and π=01f+10f=11f.

Note 2. In classical analysis, the last integrals are regarded as so-called improper integrals, i.e., limits of integrals rather than integrals proper. In our theory, this is unnecessary since F0 is a genuine primitive of f on ¯I.

For each integer n (negatives included), we now define Fn:E1E1 by

Fn(x)=nπ+(1)nF0(x) for x¯I=[1,1]Fn=0 on ¯I.

Fn is called the n th branch of the arcsine. Figure 26 shows the graphs of F0 and F1 ( that of F1 is dotted). We now obtain the following theorem.

Theorem 5.11.3

(i) Each Fn is differentiable on I=(1,1) and relatively continuous on ¯I=[1,1].

(ii) Fn is increasing on ¯I if n is even, and decreasing if n is odd.

(iii) Fn(x)=(1)n1x2 on I.

(iv) Fn(1)=Fn1(1)=nπ(1)nπ2;Fn(1)=Fn1(1)=nπ+(1)nπ2.

Proof

Screen Shot 2019-06-26 at 2.04.32 PM.png

The proof is obvious from (12) and the properties of F0. Assertion (iv) ensures that the graphs of the Fn add up to one curve. By (ii), each Fn is one to one (strictly monotone) on ¯I. Thus it has a strictly monotone inverse on the interval ¯Jn=Fn [[1,1]], i.e., on the Fn -image of ¯I. For simplicity, we consider only

¯J0=[π2,π2] and J1=[π2,3π2],

as shown on the Y -axis in Figure 26. On these, we define for x¯J0

sinx=F10(x)

and

cosx=1sin2x,

and for x¯J1

sinx=F11(x)

and

cosx=1sin2x.

On the rest of E1, we define sinx and cosx periodically by setting

sin(x+2nπ)=sinx and cos(x+2nπ)=cosx,n=0,±1,±2,.

Note that by Theorem 3(iv),

F10(π2)=F11(π2)=1.

Thus (13) and (14) both yield sinπ/2=1 for the common endpoint π/2 of ¯J0 and ¯J1, so the two formulas are consistent. We also have

sin(π2)=sin(3π2)=1,

in agreement with (15). Thus the sine and cosine functions (briefly, s and c) are well defined on E1.

Theorem 5.11.4

The sine and cosine functions (s and c) are differentiable, hence continuous, on all of E1, with derivatives s=c and c=s; that is,

(sinx)=cosx and (cosx)=sinx.

Proof

It suffices to consider the intervals ¯J0 and ¯J1, for by (15), all properties of s and c repeat themselves, with period 2π, on the rest of E1.

By (13),

s=F10 on ¯J0=[π2,π2],

where F0 is differentiable on I=(1,1). Thus Theorem 3 of §2 shows that s is differentiable on J0=(π/2,π/2) and that

s(q)=1F0(p) whenever pI and q=F0(p);

i.e., qJ and p=s(q). However, by Theorem 3(iii),

F0(p)=11p2.

Hence,

s(q)=1sin2q=cosq=c(q),qJ.

This proves the theorem for interior points of ¯J0 as far as s is concerned.

As

c=1s2=(1s2)12 on J0 (by (13)),

we can use the chain rule (Theorem 3 in §1) to obtain

c=12(1s2)12(2s)s=s

on noting that s=c=(1s2)12 on J0. Similarly, using (14), one proves that s=c and c=s on J1 (interior of ¯J1).

Next, let q be an endpoint, say, q=π/2. We take the left derivative

s(q)=limxqs(x)s(q)xq,xJ0.

By L'Hôpital's rule, we get

s(q)=limxqs(x)1=limxqc(x)

since s=c on J0. However, s=F10 is left continuous at q (why?); hence so is c=1s2. (Why?) Therefore,

s(q)=limxqc(x)=c(q),as required.

Similarly, one shows that s+(q)=c(q). Hence s(q)=c(q) and c(q)=s(q) as before.

The other trigonometric functions reduce to s and c by their defining formulas

tanx=sinxcosx,cotx=cosxsinx,secx=1cosx, and cscx=1sinx,

so we shall not dwell on them in detail. The various trigonometric laws easily follow from our present definitions; for hints, see the problems below.


This page titled 5.11: Integral Definitions of Some Functions is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform.

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