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Derivatives and Antiderivatives

Date: 10/31/2001 at 19:41:31
From: Robert
Subject: integral sqrt(cos(x^2)) dx


When I run the formula int(sqrt(cos(x^2)),x=0..1) in Maple it just 
returns the formula. I know it's because there is no antiderivative, 
but why isn't there one?

Date: 11/02/2001 at 14:43:17
From: Doctor Rob
Subject: Re: integral sqrt(cos(x^2)) dx

Thanks for writing to Ask Dr. Math, Robert.

There are many, many functions that behave similarly. For any of them, 
its antiderivative exists, but there is no expression for it in closed 
form in terms of familiar functions and constants. Examples are 
x^x, e^(x^2), e^(e^x)), e^(-x)/x, 1/ln(x), ln(1-x)/x, sin(x)/x, 
sin(x^2), sin(sin(x)), sqrt(sin(x)), e^sin(x), ln(sin(x)), 
sqrt(x^3+a*x^2+b*x+c), and so on.

The nonexistence of such an expression is a consequence of the fact 
that we only have a fairly small restricted set of functions to use to 
try to express antiderivatives: constant, rational, irrational 
algebraic, exponential, logarithmic, trigonometric, inverse 
trigonometric, hyperbolic, and inverse hyperbolic functions. This may 
seem like a big list to you, but it is not big enough for those 

One way to think of this is that the antiderivative is the solution of 
a certain differential equation:

   dy/dx = sqrt(cos(x^2)).

You probably are aware that not all such solutions are expressible 
that way. Bessel functions and many other so-called "special" 
functions that crop up in engineering and physics arise as such

A similar situation exists with trying to solve the general quintic
equation. The roots cannot be expressed in terms of the equation's
coefficients using the above functions. The roots exist, and are
continuous functions of the coefficients, but those functions just
don't have simple expressions.


- Doctor Rob, The Math Forum   
Associated Topics:
College Calculus
High School Calculus

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