The Gaussian Integrand and Its Antiderivative

The Gaussian Function

The function \(e^{-x^2}\) is called the Gaussian function. Its graph is smooth, positive, symmetric about the vertical axis, and approaches zero as \(|x|\) grows. It appears throughout mathematics and applications, including probability and models of measurement error.

Its formula looks simple, but its antiderivative does not have an expression in elementary functions. That fact does not mean the function cannot be integrated. It means we must distinguish two questions: whether an elementary formula exists for an antiderivative, and whether a definite integral has a value that can be represented or approximated.

Why the Familiar Substitution Does Not Finish the Job

For an exponential composite \(e^{g(x)}\), the chain rule gives \(\frac{d}{dx}e^{g(x)}=g'(x)e^{g(x)}\). This readily produces an antiderivative when the integrand includes the factor \(g'(x)\). For the Gaussian function, the inner function is \(-x^2\), whose derivative is \(-2x\). That factor is absent.

Proving There Is No Elementary Antiderivative

The distinction between “this substitution does not work” and “no elementary antiderivative exists” matters. The latter is a stronger claim and requires a result about elementary functions. Here, an elementary function means one built from rational functions using finitely many algebraic operations, exponentials, logarithms, and compositions of these functions.

We use a specialized form of Liouville’s theorem for an exponential of a polynomial. For a polynomial \(g(x)\), an elementary antiderivative of \(e^{g(x)}\) can exist only if there is a rational function \(R(x)\) satisfying \(R'(x)+g'(x)R(x)=1\). In fact, under the usual setting of elementary functions over the complex numbers, this condition is also sufficient: if it holds, then \((R(x)e^{g(x)})'=e^{g(x)}\).

To verify the last claim, first suppose \(R\) has a pole at a finite complex number \(a\). If the pole has order \(m\geq 1\), then \(R'\) has a pole of order \(m+1\). The term \(-2xR(x)\) has a pole of order at most \(m\), including when \(a=0\), where multiplication by \(x\) lowers the pole order by one. Thus the highest-order pole in \(R'-2xR\) cannot cancel. But the right side, \(1\), has no pole. This is a contradiction. A rational function without finite poles must be a polynomial.

Now suppose \(R\) is a nonzero polynomial of degree \(n\), with leading term \(c x^n\), where \(c\neq 0\). Then \(-2xR(x)\) has leading term \(-2c x^{n+1}\), while \(R'(x)\) has degree at most \(n-1\). The degree-\((n+1)\) term cannot cancel, so \(R'-2xR\) cannot equal the constant \(1\). The zero polynomial also fails, since it makes the left side zero. There is no rational solution, and the criterion therefore rules out an elementary antiderivative.

A Definite Integral Still Defines an Antiderivative

The Fundamental Theorem of Calculus gives a direct way to define a primitive even when no elementary formula is available. Define

Because \(e^{-t^2}\) is continuous, the first part of the Fundamental Theorem of Calculus gives \(F'(x)=e^{-x^2}\). So \(F\) is an antiderivative, but the result above shows it is not an elementary one. The integral definition is a precise representation, not an elementary closed form.

For any real numbers \(a\) and \(b\), the same theorem gives

This identity holds for either orientation of the endpoints; when \(b<a\), the definite integral has the usual reversed orientation. Since \(e^{-x^2}\) is even, \(F\) is odd: for \(a\geq 0\), changing variables \(t=-u\) shows \(F(-a)=-F(a)\). Consequently, integrals over symmetric intervals satisfy \(\int_{-a}^{a}e^{-x^2}\,dx=2F(a)\).

A Series Representation for a Definite Integral

A definite integral can also be represented by a convergent series. The exponential series gives \(e^{-x^2}=\sum_{n=0}^{\infty}(-1)^n x^{2n}/n!\). On \([0,1]\), this series converges uniformly: the absolute value of its \(n\)th term is at most \(1/n!\), and \(\sum_{n=0}^{\infty}1/n!\) converges. Uniform convergence permits integration term by term on this interval. Since \(\int_0^1 x^{2n}\,dx=1/(2n+1)\), we obtain

This is an exact value in series form. It is not an elementary antiderivative, and it does not turn the integral into a finite combination of elementary functions. It does, however, give a practical approximation with a rigorous error bound.

A Common Mistake

A frequent error is to infer that the definite integral cannot be evaluated because no elementary antiderivative exists. That conclusion confuses two different tasks. An elementary antiderivative is one possible tool for evaluating a definite integral through the Fundamental Theorem of Calculus; its absence does not prevent other exact representations, bounds, or numerical approximations.

The Gaussian function is a useful example precisely because both statements are true: its antiderivative is non-elementary, while its definite integrals remain well-defined and amenable to exact or approximate study. These are not competing conclusions; they address different meanings of integration.