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If a function's integral can't be written, then how can we find exact values for it over areas? Can we only ever estimate it? Why can't we make new functions to define these strange unwritable anti-derivatives?

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My favorite example is $e^{-x^2}$. Its definite integral over the real plane is $\sqrt {\pi}$, yet there is no real solution to the antiderivative. – cygorx May 3 '14 at 22:30
It is not so much that we make new functions. Rather, we give a name to one of these antiderivatives. One of the early ones to be given a name in exactly this manner is – Will Jagy May 3 '14 at 22:31
Ive seen tons of examples, and they perplex me as to why something so definable is so unreachable. Math, as beautiful as she is, is such a frustrating tease sometimes. – Asimov May 3 '14 at 22:31
Out here in the wilderness beyond calculus homework problems, the vast majority of functions can not be written as combinations of elementary ones. We can "cheat" by inventing new elementary functions, like the ones mentioned in Bennett Gardiner's answer, but that's only a tiny step forward. – bubba May 3 '14 at 23:37
You could just as well ask how the sine function can have values that can't be written explicitly, like $\sin 1$. There's no "closed form" expression for this number, which may feel familiar since it has a geometric meaning as the $y$-coordinate of the point 1 radian along the unit circle from $(1,0)$ going counterclockwise. But the geometric meaning of an integral of $e^{-x^2}$ doesn't seem to satisfy you, so can a geometric meaning for $\sin 1$? The number $\sin 1$ can be written as an infinite series, and that can be used to estimate it well; do you have a problem with the number $\sin 1$? – KCd May 4 '14 at 19:36
up vote 31 down vote accepted

We can, and do all the time! For example,

The Gamma function $$ \Gamma (z) = \int_0^{\infty} t^{z-1}\mathrm{e}^{-t} \ \mathrm{d}t. $$

The Beta function $$ \mathrm{B}(z,y) = \int_0^1 t^{z-1}(1-t)^{y-1}\,\mathrm{d}t. $$

The Exponential integral function $$ \mathrm{E}_1(z) = \int_z^\infty \frac{e^{-t}}{t}\, \mathrm{d}t. $$

The Error function $$ \operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_{0}^x e^{-t^2}\,\mathrm dt.$$

The Elliptic integral of the second kind $$ E(\phi,k)=\int_0^{\phi} \sqrt{1-k^2\sin^2\theta} \ \mathrm{d}\theta. $$

The Logarithmic integral function $$ {\rm Li} (x) = \int_2^x \frac{\mathrm{d}t}{\ln t}, $$

and many, many more very important "special functions' are defined by definite integrals. If you go back even to elementary functions, you can define the logarithm via the following integral -

The Logarithm $$ \ln (t) = \int_1^t \frac{1}{x} \, dx. $$

As to your first question, how do we find areas under these curves if we don't have an elementary antiderivative? Well, how do you find the area under the curve $1/t$ from $1$ to $5$? The above integral tells you the value is $\ln 5$, but what is that value, exactly? We can only approximate it, given the best methods we have!

The same is true of all the above functions. At some special values they have exact values, given perhaps by integers, rational or irrational numbers, or a combination of common mathematical constants such as $\pi,\mathrm{e},\gamma,$ Catalan's constant, etc. (another interesting question is - why are these constants special enough to have names? Because they come up all the time! The same is true for the above functions) .

But for almost all values we must approximate the value of the function by computing the definite integral numerically (Trapezoid rule, Simpson's rule, more advanced techniques), or using some other representation of the integral such as an infinite sum etc.

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+1 for mentioning the natural logarithm in the context of the others. It's a good reminder that "elementary" is a somewhat subjective notion. – heropup May 4 '14 at 0:28

Just because we can't write down a closed form for an antiderivative (that we know exists), that does not mean we are forced to only estimate its definite integral.

For example, there does not exist a closed form for the antiderivative $\int e^{-x^2}\,dx$, but that does not mean we are limited to merely estimating $\int_0^1 e^{-x^2}\,dx$. In fact, this is a main thrust of the theory of Taylor series!

Using $\displaystyle e^x=\sum_{n=0}^\infty {x^n\over n!}$, we see $$ e^{-x^2}=\sum_{n=0}^\infty {(-x^2)^n\over n!}=\sum_{n=0}^\infty {(-1)^n x^{2n}\over n!}=1-x^2+{x^4\over 2!}-{x^6\over 3!}+\cdots $$ Then, by term-by-term integration of Taylor series, \begin{align*} \int_0^1 e^{-x^2}\,dx&=\int_0^1 \left(1-x^2+{x^4\over 2!}-{x^6\over 3!}+\cdots\right)\,dx\\ &=\left[x-{x^3\over 3}+{x^5\over 5\cdot2!}-{x^7\over 7\cdot3!}+\cdots\right]_0^1\\ &=1-{1\over 3}+{1\over 5\cdot2!}-{1\over 7\cdot3!}+\cdots \end{align*} Thus, we have found the exact value---not an estimate---for the desired area under the curve even though no closed form antiderivative is available. It is exact in the sense that we can find the desired area to any prescribed accuracy you specify by summing sufficiently many terms in the series.

Objecting to such a situation is akin, IMO, to objecting to the fact that $\sqrt 2$ is not rational. Ok, it's not rational, but we can still compute $\sqrt 2$ exactly in the same sense that we computed the area above exactly. The point is to be able to compute the quantity of interest to a prescribed level of accuracy; I am indifferent as to the form the particular computation takes (apart from mere subjective personal preference).

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by the way, does this mean we can write the antiderivative of $e^{-x^2}$ as an polynomial with infinite terms? – Ant May 3 '14 at 22:52
I think his answer implies that @JohnD considers an infinite sum to be a more fundamental result then the definite integral. Computing the actual numerical result still requires you to at some point truncate the infinite series, just as we would have had to accept some error in computing the definite integral numerically. Sometimes we prefer a sum representation, other times an integral representation can be nicer and easier to deal with. If the terms in the sum are pleasant enough, and the sum converges quickly enough, perhaps the sum is objectively better. However this is not always the case – Bennett Gardiner May 3 '14 at 23:03
@BennettGardiner To clarify, I find $\int e^{-x^2}\,dx$ as well as $\int_0^1 e^{-x^2}\,dx$ to be a perfectly acceptable, and well-defined result. The reason I segued into infinite numerical series was to address the implicit accusation that "lack of a closed form implies approximation/inexactness". – JohnD May 4 '14 at 0:00
@Ant Yes, we can! – JohnD May 4 '14 at 0:09
@JohnD Your example would be more convincing if you picked an definite integral which turns out to have a rational value. Then you could really say that you found an exact value, instead of replacing one approximation by another. – fgp May 4 '14 at 15:30

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