       Fwd: RE: Antiderivatives and Definite Integrals

• To: mathgroup at smc.vnet.net
• Subject: [mg39841] Fwd: [mg39713] RE: [mg39670] Antiderivatives and Definite Integrals
• From: Garry Helzer <gah at math.umd.edu>
• Date: Sat, 8 Mar 2003 02:50:48 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```
Begin forwarded message:

> From: "Wolf, Hartmut" <Hartmut.Wolf at t-systems.com>
To: mathgroup at smc.vnet.net
> Date: Mon Mar 3, 2003  11:48:37  PM US/Eastern
> To: mathgroup at smc.vnet.net
> Subject: [mg39841] [mg39713] RE: [mg39670] Antiderivatives and Definite Integrals
>
>
>> -----Original Message-----
>> From: Garry Helzer [mailto:gah at math.umd.edu]
To: mathgroup at smc.vnet.net
> To: mathgroup at smc.vnet.net
>> Sent: Saturday, March 01, 2003 8:48 AM
>> To: mathgroup at smc.vnet.net
>> Subject: [mg39841] [mg39713] [mg39670] Antiderivatives and Definite Integrals
>>
>>
>> The antiderivative of Sqrt[1+Cos[x]] discussed here recently (sorry, I
>> lost the thread) provides an amusing illustration of the fact that
>> Mathematica does not always evaluate definite integrals by first
>> finding an antiderivative and then substituting in the upper and lower
>> limits. (See the Mathematica book A.9.5) Make the definitions
>>
>> f[x_] = Integrate[Sqrt[1 + Cos[t]], {t, 0, x}]
>> g[x_] := Integrate[Sqrt[1 + Cos[t]], {t, 0, x}]
>>
>> Then f[2Pi] is 0 (wrong) and g[2Pi] if 4Sqrt (correct). Of course
>> f[x]==g[x] returns True.
>>
>> Garry Helzer
>> Department of  Mathematics
>> University of Maryland
>> 1303 Math Bldg
>> College Park, MD 20742-4015
>>
>>
>
> You might be interested to observe:
>
>
> In:= f[x_] = Integrate[Sqrt[1 + Cos[t]], {t, 0, x}]
> Out= 2*Sqrt[1 + Cos[x]]*Tan[x/2]
>
> In:= g[x_] := Integrate[Sqrt[1 + Cos[t]], {t, 0, x}]
>
> In:= ?g
> Global`g
> g[x_] := Integrate[Sqrt[1 + Cos[t]], {t, 0, x}]
>
> In:= Plot[f[x], {x, 0, 2Pi}] // Timing
> Out= {0.03 Second, = Graphics =}
>
> ...what you called "wrong".

Well, it is wrong. From calculus we know that
1. A function defined as an integral with a variable upper limit is an
antiderivative.
2. An antiderivative is, by definition, differentiable.
3. A differentiable function is continuous.

So, by this plot, f[x] is not the antiderivative of any function.

But this may be too rigid a view. After all, D[f[x],x]//FullSimplify
brings you back to the integrand. This is a formal computation that
misses the points of discontinuity. Perhaps the algorithms used by
Mathematica are guaranteed only to produce such formal antiderivatives.
A  superficial look a the Risch algorithm suggests that this might be
the case since it proceeds by formal manipulation in extension fields
of fields of rational functions.

As I was typing this a stranger wandered in looking for an arc
length(it happens). His problem provides another example. Let
r=2-Sin[t] and try Integrate[Sqrt[r^2 +D[r,t]^2],{t,0,Pi/2}]//N. The
result is about -15, but arc length should not be negative. Set

h[x_]= Integrate[Sqrt[r^2 +D[r,t]^2],{t,0,x}]

and again we get a formal antiderivative , the plot of which reveals
discontinuities. It cannot be carelessly used to evaluate definite
integrals. (If you simplify Sqrt[r^2 +D[r,t]^2] before integrating you
get an honest antiderivative.)

In:=
> Plot[g[x], {x, 0, 2Pi}, PlotRange -> {All, {0, 6}}] // Timing
> Out= {4.096 Second, = Graphics =}
>
> ...what you called "right".

It is right. But the corresponding function for the arc length problem
is wrong.
>
>
> Redefine:
>
> In:= f[x_] /; -Pi < x < Pi = f[x]
> Out= 2*Sqrt[1 + Cos[x]]*Tan[x/2]
>
> In:= f[x_] /; Pi < x < 3 Pi = f[x] + 4*Sqrt
> Out= 4*Sqrt + 2*Sqrt[1 + Cos[x]]*Tan[x/2]

Or  4SqrtRound[x/(2Pi)]+ 2*Sqrt[1 + Cos[x]]*Tan[x/2] . But these
formulas are less than perfect since they are indeterminate at odd
multiples of Pi.

>
>
> In:=
> Plot[f[x], {x, 0, 2Pi}, PlotRange -> {All, {0, 6}}] // Timing
> Out= {0.03 Second, = Graphics =}
>
> Don't be confused by that double use of f (as a function, and as a
> shortcut
> for an algebraic expression).
>
> --
> Hartmut Wolf
>
>
>
Garry Helzer
Department of  Mathematics
University of Maryland
1303 Math Bldg
College Park, MD 20742-4015

```

• Prev by Date: Re: exporting an expression which is readable in fortran 77
• Next by Date: What is the function doing
• Previous by thread: RE: Antiderivatives and Definite Integrals
• Next by thread: Re: Fwd: RE: Antiderivatives and Definite Integrals