Why this integrand is a useful example
The integrand \(\sin(x^2)\) is built from familiar functions, yet its antiderivative cannot be expressed using elementary functions. It is a useful example because it separates two ideas that are easy to confuse: a technique failing to find an antiderivative does not prove that no elementary antiderivative exists, but a separate theorem can establish that stronger conclusion.
We will see why the obvious substitution leaves a difficult integral, establish non-elementarity using a criterion from Liouville's theorem, and construct an exact antiderivative as an infinite series. The series gives a precise representation; it does not turn the answer into an elementary formula.
The tempting substitution
For an integral containing a composite function, substitution is especially useful when the derivative of the inner function is also present. Here the inner function is \(x^2\), whose derivative is \(2x\), but the integrand contains no factor of \(x\).
Worked Example: What the substitution reveals
Consider \(\int \sin(x^2)\,dx\). Set \(u=x^2\). Then \(du=2x\,dx\), so \(dx=du/(2x)\). On an interval where \(x>0\), \(x=\sqrt{u}\), and the transformed integral is
The new integrand is \(\sin(u)/(2\sqrt{u})\), not a basic sine antiderivative: the extra factor \(1/\sqrt{u}\) remains. On an interval where \(x<0\), one instead has \(x=-\sqrt{u}\), with the corresponding sign accounted for. This substitution changes the form of the problem, but does not evaluate it.
For comparison, if the original integrand were \(2x\sin(x^2)\), the same substitution would give \(\int\sin u\,du=-\cos u+C=-\cos(x^2)+C\). Differentiation checks that result: \(\frac{d}{dx}[-\cos(x^2)]=2x\sin(x^2)\). The missing factor in our original integral is exactly what prevents this direct rule from applying.
This distinction is important. “The substitution does not finish the integral” is a statement about one method. It is not, by itself, a proof that no elementary formula exists.
What non-elementary means here
An elementary function is one formed from constants and the variable using a finite number of algebraic operations, roots, exponentials, logarithms, and trigonometric or inverse trigonometric functions, with compositions allowed. An elementary antiderivative of \(f\) is an elementary function \(F\) satisfying \(F'(x)=f(x)\) throughout the interval under consideration.
Some familiar integrands have elementary antiderivatives: for example, \(\sin x\) has antiderivative \(-\cos x\). For \(\sin(x^2)\), the chain rule would produce a factor \(2x\) when differentiating \(-\cos(x^2)\), so that candidate is not an antiderivative of the integrand as given.
A criterion for sums of exponentials
To establish that no elementary antiderivative exists, we use a specialized consequence of Liouville's theorem on elementary integration. We state the criterion in the form needed here. Let \(p_1,\ldots,p_m\) be complex polynomials such that \(p_j-p_k\) is nonconstant whenever \(j\ne k\). For a finite sum \(\sum_{j=1}^{m} c_j e^{p_j(x)}\), with constant coefficients \(c_j\), an elementary antiderivative can exist only if, for each term, there is a rational function \(R_j(x)\) satisfying
Here “rational” means a quotient of polynomials. The criterion gives a necessary condition: showing that even one required rational function cannot exist rules out an elementary antiderivative. In the application below, both required equations fail.
To see why the equation is relevant, differentiate the product \(R_j(x)e^{p_j(x)}\). The product rule gives
Thus a rational solution of the stated equation would provide an antiderivative for that exponential term in product form. Liouville's theorem supplies the stronger necessary-condition claim: allowing the full class of elementary functions cannot avoid the rational-function test in this setting.
Why the criterion rules out an elementary antiderivative
Write the sine in terms of complex exponentials:
The two exponents are \(p_1(x)=ix^2\) and \(p_2(x)=-ix^2\). Their difference is \(2ix^2\), which is nonconstant, so the criterion applies. It requires rational functions satisfying
Neither equation has a rational solution. The argument is the same for each, so consider the form \(R'+\lambda xR=c\), where \(\lambda\ne0\) and \(c\ne0\). If \(R\) had a finite pole at \(x=a\) of order \(m\ge1\), its derivative would have a pole of order \(m+1\). The term \(\lambda xR\) has pole order at most \(m\), including when \(a=0\), where multiplication by \(x\) lowers the pole order. It cannot cancel the higher-order pole in \(R'\). The right side is constant, so this is impossible. Consequently, \(R\) has no finite poles; a rational function with no finite poles is a polynomial.
If \(R\) is a nonzero polynomial of degree \(n\), then \(\lambda xR\) has degree \(n+1\), while \(R'\) has degree at most \(n-1\). The highest-degree term cannot cancel, so \(R'+\lambda xR\) cannot be the nonzero constant \(c\). The zero polynomial also fails because it gives \(0\), not \(c\). Thus no required rational function exists.
An exact antiderivative as a power series
Non-elementary does not mean that an antiderivative cannot be defined or calculated to useful accuracy. The Taylor series for sine converges for every real input. Substituting \(x^2\) for its input gives
Integrating the series term by term on any bounded interval gives an antiderivative. Term-by-term integration is valid here because the Taylor series converges uniformly on every bounded interval. Choose the constant so that the antiderivative equals zero at \(x=0\), and define
Worked Example: Verify the series antiderivative
Differentiate the general term of \(F\). The power rule gives
Because the power series and its differentiated series converge on every bounded interval, we may differentiate term by term. Therefore
Also, every term of \(F(0)\) is zero, so \(F(0)=0\). This verifies both the derivative and the chosen normalization. Every other antiderivative on an interval differs from \(F\) by a constant.
The series is an exact representation, not an elementary antiderivative. It can nonetheless be used to compute values. Its terms also show the local behavior: the first term is \(x^3/3\), so near zero the antiderivative beginning at zero is approximately \(x^3/3\).
Worked Example: Bound the value at \(x=1\)
At \(x=1\), the first four terms of the series for \(F\) are
For the absolute value \(a_n=1/((4n+3)(2n+1)!)\) of the \(n\)-th term at \(x=1\), the ratio of consecutive terms is
The terms therefore decrease to zero, and the series alternates. The alternating-series remainder bound says that after these four terms, the error has the sign of the next term and magnitude at most that term. The next term, for \(n=4\), is \(1/(19\cdot9!)\). Hence
This gives an exact error bound without treating the power series as an elementary formula. The interval is checked directly from the alternating-series remainder rule: the next term is positive, and all later terms decrease in magnitude.
Common pitfall: treating an incomplete substitution as proof
A common mistake is to stop after the substitution \(u=x^2\) and claim that the transformed integral proves non-elementarity. It does not. A change of variable can make an integral less familiar without proving anything about whether an elementary antiderivative exists.
Another incorrect calculation is
which silently applies the chain rule in reverse while omitting the derivative of \(x^2\). Differentiating the proposed answer exposes the error:
in general. The correction is to distinguish the integrand \(2x\sin(x^2)\), which has the elementary antiderivative \(-\cos(x^2)\), from \(\sin(x^2)\), whose antiderivative is non-elementary. The latter still has an exact power-series representation.
What to take from the example
The integrand \(\sin(x^2)\) illustrates three complementary points. First, substitution works directly when the required inner derivative is present; without it, the substitution may leave a new integral rather than solve the problem. Second, failure of familiar rules is not proof of non-elementarity; a theorem is needed to rule out all elementary candidates. Third, non-elementary antiderivatives remain mathematically precise objects: here the convergent power series gives an exact antiderivative and a way to control approximations.
Check Your Understanding
Answer these questions using the definitions and calculations above.
- When \(u=x^2\), what factor is missing from \(\sin(x^2)\) for the direct substitution to give a basic sine integral?
- Why does failure of that substitution not prove that no elementary antiderivative exists?
- What rational-function equation does the Liouville-theorem criterion require for the term \(e^{ix^2}/(2i)\)?
- Why can a rational solution of \(R'+\lambda xR=c\), with \(\lambda\ne0\) and \(c\ne0\), have no finite poles?
- Differentiate the general term of the series for \(F(x)\) and identify the resulting series.
