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Liouville's theorem tells us that not all indefinite integrals of elementary functions (such as $\sin(x)/x$, $x^x)$ exist. However, this does not always seem to be the case when we have a definite integral.

For example, while the definite integral in terms of elementary functions for sinc(x) does not exist, the definite integral of sinc(x) from 0 through infinity is equal to $\pi/2$.

My question is: Suppose we have a function f(x) composed of elementary functions. When do we know the definite integral of f(x) over some interval has a closed form in terms of elementary functions?

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What do you mean, an elementary function? A definite integral is just a number (if it exists) –  mixedmath Jan 22 '12 at 14:05
    
The integrands I am considering have parameters. There is information on what I mean by elementary further down the page... –  pbs Jan 22 '12 at 15:03
    
I don't mean to talk down to you or anything. I really don't understand, and I don't see where you are referring to - "further down the page." Can you clarify? –  mixedmath Jan 22 '12 at 15:09
    
Hi mixedmath, thanks for your reply.I was thinking of "elementary functions" as those which can be expressed as a Taylor series whose coefficients are not exotic in the sense that they involve only "finite" terms such as k!, Pi^k, the Bernoulli numbers B_k, etc., but not things like zeta(k) where it may turn out we are dealing with zeta(odd) for which there is no known closed form, etc. Especially elementary as in trigonometric, exponential, logarithmic. –  pbs Jan 22 '12 at 16:20
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4 Answers 4

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If a closed form is known for the indefinite integral then we may find the definite integral (if the limits at the bounds are defined!) else it is much more difficult.

We have to try different methods like convert to more standard forms (say Laplace or Fourier transform) using change of variable, consult tables (yes computer algebra searches in tables too!), try to recognize very general and elaborate functions like Meijer G, seek symmetries, try changes of variable, integration or differentiation under the integral sign, try to obtain a solvable differential equation and so on...

We should note that the concept of 'closed form' is itself extensible. Some people will accept only algebraic, logarithmic, trigonometric and their inverses but you may add some special functions like $\Gamma$, $\zeta$, $\rm LamberW$, hypergeometric or their extension Meijer G...

To appreciate the difficulty of the task you may look at this paper of Davenport 'The Difficulties of Definite Integration' or discussions about undecidability 'An undecidable property of definite integrals'.

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When you have a Taylor series then integration is straightforward since $\displaystyle \int f(n)x^n dx=\frac{f(n)x^{n+1}}{n+1}$ and you get a simple series but the integral will often not be elementary! Example $\int \frac{e^{-x}}x dx$ is a new function the 'Exponential integral'. Remember that even the simple $1/x$ becomes a logarithm. Integration and differential equations create new functions as opposed to differentiation. Sometimes the definite integral may be simpler because the 'special function' has simple values at certain points. –  Raymond Manzoni Jan 22 '12 at 15:09
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Any definite integral(without parameters) is a constant number. Any constant is an elementary function, i.e. a polynomial of zero order.

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Thanks for your post. Sorry I should have been clearer - I am thinking of definite integrals with parameters. –  pbs Jan 22 '12 at 14:50
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An "indefinite integral" can be written as "definite integral with parameters" by a change of variables. For example, indefinite integral $$\int_0^x e^{-t^2}\,dt $$ is equal to definite integral (with parameter $x$) $$\int_0^1 x e^{-x^2 u^2}\,du$$ –  GEdgar Jan 22 '12 at 18:14
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It depends what you mean by expressing it as an elementary function: an elementary function of what? In a sense, any definite integral is itself an elementary function simply because it is constant.

But if you're asking if you can express any definite integral as an elementary function of the limits, then the answer is no. The reason for this is because when you do indefinite integration, what you're really doing is integrating between a constant and a variable limit. So when you write $$\int f(x)\, dx$$ what you really mean is $$\int_a^x f(t)\, dt$$ where $a$ is a (possibly infinite) constant, $x$ is the variable and $t$ is the dummy variable that you're integrating over. Then the $a$ determines what is commonly called the 'constant of integration'.

Definite integration is then just the indefinite integral with $x$ constant, say $x=b$, or with $x \to \infty$. As such, there's really no difference as regards expressing the integral as an elementary function of the limits; it just so happens that sometimes you get nicely-expressible constants for some values of the limits.

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Also, note that $\int_0^x f(t)\;dt = x\int_0^1 f(xt)\;dt$, so restricting to a situation with fixed limits but some kind of parameter in the integrand is not going to help you. –  Henning Makholm Jan 22 '12 at 15:48
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If the indefinite integral exists then the definite integral automatically exists (the definite integral is found by merely substituting numbers into the indefinite integral subtracting in the usual manner), but it might diverge if integrated to infinity. Is that what you mean?

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