Non-Elementary Integrals

Some integrals have no formula at all, and Joseph Liouville proved it. Run his test on a polynomial times e^−x², see why the answer hangs on one number, then train on integrands where the right answer may be None exists.

For a polynomial g	$\int g (x) e^{- x^{2}} d x = R (x) e^{- x^{2}} + C ⟺ \int_{- \infty}^{\infty} g (x) e^{- x^{2}} d x = 0$
The name for the missing formula	$\int e^{- x^{2}} d x = \frac{π}{2} erf x + C$

Worked example — Non-Elementary Integrals

Whole-line area of g(x)·e⁻ˣ²	1.7725
Elementary antiderivative of g(x)·e⁻ˣ²	none
Share with a formula, coefficients −3 to 3	6.12%
The trainer's integrand, integrated by Simpson	2.2405

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Non-Elementary Integrals

A function is elementary if it can be built from x and constants by finitely many additions, multiplications, divisions, roots of polynomial equations, exponentials and logarithms; complex values are allowed, which brings in the trigonometric functions and their inverses. An integrand has an elementary antiderivative if some elementary F has F′ = f. Liouville's theorem restricts the form such an F can take, and for f·e^g with f and g rational, g not constant, it becomes one test: an elementary antiderivative exists exactly when R′ + g′R = f has a rational solution R, and then R·e^g is one.

What each symbol means

R: the rational function Liouville's test looks for; for g(x)e^−x² it must be a polynomial
K: g₀ + g₂/2 + 3g₄/4 + 15g₆/8 + …, the multiple of ∫e^−x² dx left over after integrating by parts: the whole-line area divided by √π

Where the formula comes from

The form any answer must take. Liouville’s theorem narrows the search to one shape: if ∫g(x)e^−x² dx has a formula, it is R(x)e^−x² for some polynomial R. Differentiate that and ask it to equal g.
$\int g (x) e^{- x^{2}} d x = R (x) e^{- x^{2}} \Rightarrow g = R^{'} - 2 x R$
Match the powers. For a cubic g, R is a quadratic. The x³ and x terms can always be matched. The x² term and the constant term both fix r₁, once as −g₂/2 and once as g₀, and they agree only when g₀ + g₂/2 = 0.
$R = r_{2} x^{2} + r_{1} x + r_{0} : R^{'} - 2 x R = - 2 r_{2} x^{3} - 2 r_{1} x^{2} + (2 r_{2} - 2 r_{0}) x + r_{1} \Rightarrow r_{1} = g_{0} = - \frac{g _{2}}{2}$

Breaks when: The test on this page covers f·e^g with f and g rational, and through x = e^u the integrands x^m/ln x. Square roots of polynomials need a different argument: Conrad's Example 4.3 explains why ∫dx/√P(x) is not elementary for P of degree 3 or more without repeated roots, and Hardy credits Liouville with the standard elliptic integrals, an ellipse's arc length among them.

Problems solved in full

The elementary neighbour 2 steps
Find ∫x e^−x² dx.
1. The x is −½ times the derivative of −x², so substitute:
  
  $u = - x^{2}, d u = - 2 x d x, x d x = - \frac{1}{2} d u$
2. One exponential:
  
  $\int x e^{- x^{2}} d x = - \frac{1}{2} \int e^{u} d u = - \frac{1}{2} e^{- x^{2}} + C$
Answer

−½e^−x² + C. One factor of x is the whole difference between this integral and ∫e^−x² dx, which has no formula at all.

$\int x e^{- x^{2}} d x = - \frac{1}{2} e^{- x^{2}} + C$
Four decimals from a series 3 steps
Evaluate ∫₀¹ e^−x² dx to four decimal places.
Show solution
1. No antiderivative to use, but e^−x² has a power series, and it can be integrated term by term:
  
  $e^{- x^{2}} = k \geq 0 \sum \frac{( - 1 ) ^{k} x ^{2 k}}{k !} \Rightarrow \int_{0}^{1} e^{- x^{2}} d x = k \geq 0 \sum \frac{( - 1 ) ^{k}}{k ! ( 2 k + 1 )}$
2. The terms alternate and shrink, so add them until the next one is too small to matter:
  
  $1 - \frac{1}{3} + \frac{1}{10} - \frac{1}{42} + \frac{1}{216} - \frac{1}{1320} + \frac{1}{9360} = 0.74684$
3. For an alternating series the error is smaller than the first term left out:
  
  $∣ R ∣ < \frac{1}{7 ! \cdot 15} = \frac{1}{75600} \approx 1.3 \times 1 0^{- 5}$
Answer

0.7468. Stopping one term earlier, at 1/1320, leaves an error bound of 1/9360 ≈ 1.1×10⁻⁴, which cannot vouch for the fourth decimal.

$\int_{0}^{1} e^{- x^{2}} d x \approx 0.7468$
The same method on sin(x²) 2 steps
Evaluate ∫₀¹ sin(x²) dx to four decimal places.
Show solution
1. sin(x²) has no elementary antiderivative either. Its series, integrated term by term:
  
  $sin x^{2} = k \geq 0 \sum \frac{( - 1 ) ^{k} x ^{4 k + 2}}{( 2 k + 1 )!} \Rightarrow \int_{0}^{1} sin x^{2} d x = k \geq 0 \sum \frac{( - 1 ) ^{k}}{( 2 k + 1 )! ( 4 k + 3 )}$
2. Three terms, and the fourth bounds the error:
  
  $\frac{1}{3} - \frac{1}{42} + \frac{1}{1320} = 0.31028, ∣ R ∣ < \frac{1}{75600}$
Answer

0.3103. This is a Fresnel integral: Augustin Fresnel needed it in 1818 for his theory of the diffraction of light.

$\int_{0}^{1} sin x^{2} d x \approx 0.3103$
Parts, down to the integral that has no formula 3 steps
Find ∫x² e^−x² dx, and its value from 0 to ∞, given ∫₀^∞ e^−x² dx = √π/2.
Show solution
1. Split x² e^−x² as x · x e^−x²; the second factor has the antiderivative of the first problem:
  
  $u = x, d v = x e^{- x^{2}} d x, v = - \frac{1}{2} e^{- x^{2}}$
2. One round of parts leaves ∫e^−x² dx, which has no formula, so it stays as it is:
  
  $\int x^{2} e^{- x^{2}} d x = - \frac{1}{2} x e^{- x^{2}} + \frac{1}{2} \int e^{- x^{2}} d x$
3. Over [0, ∞) the boundary term is 0 at both ends and the integral left is known:
  
  $\int_{0}^{\infty} x^{2} e^{- x^{2}} d x = [- \frac{1}{2} x e^{- x^{2}}]_{0}^{\infty} + \frac{1}{2} \cdot \frac{π}{2} = \frac{π}{4}$
Answer

−½x e^−x² + (√π/4) erf x + C, and √π/4 ≈ 0.4431 over [0, ∞). erf is the name given to the integral that has no formula, scaled so that it tends to 1.

$\int x^{2} e^{- x^{2}} d x = - \frac{1}{2} x e^{- x^{2}} + \frac{π}{4} erf x + C, \int_{0}^{\infty} x^{2} e^{- x^{2}} d x = \frac{π}{4} \approx 0.4431$
Two pieces without formulas that add up to one 1 steps
Find ∫e^x(1/x − 1/x²) dx for x > 0.
Show solution
1. Neither ∫eˣ/x dx nor ∫eˣ/x² dx is elementary. But differentiate eˣ/x by the quotient rule:
  
  $\frac{d}{d x} \frac{e ^{x}}{x} = \frac{e ^{x} x - e ^{x}}{x ^{2}} = e^{x} (\frac{1}{x} - \frac{1}{x ^{2}})$
Answer

eˣ/x + C. Liouville’s theorem is about the whole integrand: a sum can have a formula when neither of its terms does.

$\int e^{x} (\frac{1}{x} - \frac{1}{x ^{2}}) d x = \frac{e ^{x}}{x} + C$
Which c makes (4x⁴ + c)e^(−x²) integrable? 4 steps
For which c does (4x⁴ + c)e^−x² have an elementary antiderivative, and what is it? The page's boxes stop at x³. Press Balanced for the same question one degree lower.

Open this problem in the tool
Show solution
1. Integrating by parts, each x^2m leaves (2m − 1)!!/2^m times ∫e^−x² dx behind; for x⁴ that weight is 3/4. Add up the leftovers:
  
  $K = c + 4 \cdot \frac{3 \cdot 1}{2 ^{2}} = c + 3$
2. A formula needs the leftover to vanish:
  
  $K = 0 \Rightarrow c = - 3$
3. Then R has degree 3. Matching powers of R′ − 2xR = 4x⁴ − 3 from the top down:
  
  $- 2 r_{3} = 4, - 2 r_{2} = 0, 3 r_{3} - 2 r_{1} = 0, 2 r_{2} - 2 r_{0} = 0 \Rightarrow R = - 2 x^{3} - 3 x; r_{1} = - 3 = g_{0}$
4. Check by differentiating:
  
  $\frac{d}{d x} (- (2 x^{3} + 3 x) e^{- x^{2}}) = (- 6 x^{2} - 3 + 4 x^{4} + 6 x^{2}) e^{- x^{2}} = (4 x^{4} - 3) e^{- x^{2}}$
Answer

c = −3, and ∫(4x⁴ − 3)e^−x² dx = −(2x³ + 3x)e^−x² + C. Any other c leaves (c + 3)∫e^−x² dx behind, which has no formula, and the whole-line area is then (c + 3)√π instead of zero.

$c = - 3 : \int (4 x^{4} - 3) e^{- x^{2}} d x = - (2 x^{3} + 3 x) e^{- x^{2}} + C$

“No formula” for e⁻ˣ² is a theorem 🖖

Press The bell curve. The trainer asks for ∫3e^−x² dx, and the right answer is None exists. Nobody is waiting for a cleverer substitution: Joseph Liouville proved in the 1830s that no combination of powers, roots, exponentials, logarithms and trigonometric functions, however long, has e^−x² as its derivative. Given Liouville's theorem, the last step fits in a paragraph, and Brian Conrad's notes give it: if an elementary antiderivative existed, one would have the form R(x)e^−x² with R a rational function, so R′ − 2xR = 1. A polynomial R makes the left side's degree too high; a pole of R gives R′ a pole of one order more, which nothing on either side cancels. And yet the antiderivative exists. The lower panel draws it from Simpson's rule, a smooth rising curve, and the results print ∫₀¹ 3e^−x² dx = 2.2405. What is missing is a formula, not a function, and mathematicians named a scaled copy of it: the error function erf(x) = (2/√π)∫₀ˣ e^−t² dt, so the trainer’s antiderivative is (3√π/2)·erf(x) + C. Press 1/ln x: the same theorem, through x = e^u, rules out a formula for the logarithmic integral ∫dx/ln x, and the page draws that antiderivative just the same.

Times a polynomial, e⁻ˣ² has a formula exactly when its area over the whole line is zero 🖖

Liouville's test can be run, not just quoted. Type a cubic g into the boxes and step through it: any formula for ∫g(x)e^−x² dx would have to be R(x)e^−x² with R a polynomial, so R′ − 2xR must equal g, and matching powers from the top down either finds R or ends in a contradiction at the constant term. Press x²·e⁻ˣ². The steps force R = −½x and then ask −½ to equal 0, so there is no formula, and the area over the whole line is √π/2 = 0.8862. Press Balanced, (2x² − 1)e^−x²: the steps find R = −x, so the integral is −xe^−x² + C, and the area is 0.0000. The two facts are one fact. A formula R(x)e^−x² dies away at both ends, so the area between them, F(∞) − F(−∞), is zero; and integrating by parts reduces every x^2me^−x² to a formula plus a fixed multiple of ∫e^−x² dx, a multiple that adds up to the area divided by √π. In numbers: a formula exists exactly when g₀ + g₂/2 = 0, and the odd powers never matter. The green and red shading make it visible: they must cancel.

Count the integrands in a family and those with a formula all but vanish 🖖

Liouville's theorem is about the integrand in front of you. It does not say what fraction of integrals have no formula, and cannot, because “most” needs a family and a way of counting. So count a family. Take every cubic g with integer coefficients from −3 to 3, 2,401 of them. Times e^−x², 147 have an elementary antiderivative, 6.12%, and they are exactly the ones with g₂ = −2g₀: three lit cells of the 49 in the grid, with g₁ and g₃ free. Widen the coefficients to −10 … 10 and the share drops to 2.49%; to −100 … 100, 0.25%; each tenfold widening divides it by about ten, as the table shows. A formula needs one exact coincidence among the coefficients, and an exact coincidence is a line drawn through a square that keeps growing. The trainer gives you a formula about half the time only because it was built to; drawn at random from this family, almost no integrand would have one.

Learning path

Techniques of integration

References (2)

The criterion the page runs: “Choose f, g ∈ ℂ(X) with f ≠ 0 and g nonconstant. The function f(x)e^g(x) can be integrated in elementary terms if and only if there exists a rational function R ∈ ℂ(X) such that R′(X) + g′(X)R(X) = f(X)” (Theorem 4.4, p. 8), deduced from Liouville's theorem (Theorem 4.1) and proved in §5; applied to e^(−x²) in Example 4.6 and to eˣ/x, equivalently dt/log t, in Example 4.7 (pp. 9–10). Example 4.3 gives ∫dx/√P(x) for P of degree 3 or more without repeated roots: B. Conrad, “Impossibility theorems for elementary integration”, Clay Mathematics Institute Academy Colloquium lecture notes (2005), Theorems 4.1 and 4.4, Examples 4.3, 4.6 and 4.7.
Liouville's results and their dates: the theorem that an elementary ∫eˣy dx (y algebraic) must be eˣ times a rational expression in y, from his 1835 memoir in Crelle's Journal (vol. 13, pp. 93–118), with “the ‘logarithm-integral’ … is really a new transcendent” (ch. VI §4, pp. 58–59); the reduction of every ∫eˣR(x) dx to ∫eˣ/x dx (p. 58, the eˣ(1 − 1/x)³ and eˣ(1 − 2/x)² examples); a footnote crediting the general method with “the ‘error function’ integral ∫e^(−x²) dx is not an elementary function” (§5, p. 59); Liouville's proof for the standard elliptic integrals (ch. V §19, p. 47); and the bibliography of his memoirs, 1833–1840: G. H. Hardy, The Integration of Functions of a Single Variable, 2nd ed., Cambridge Tracts in Mathematics and Mathematical Physics no. 2, Cambridge University Press (1916), Chapter V §19, Chapter VI §§4–5, and Appendix I.