# Is the integral of $e^{-x^2}$ from $0$ to $1$ known to be irrational?

Is it known whether $$\int_0^1 e^{-x^2} \, dx$$ is irrational? It is well-known that $$\int_0^\infty e^{-x^2} \, dx=\frac{\sqrt{\pi}}{2}$$ which is irrational, but what about the prior integral? Also, I know there is probably no nice "closed-form" number as its solution (ala Tim Chow).

This question arises from looking at Catalan's constant, denoted by $$G$$ on Wikipedia. One integral representation of $$G$$ is $$\int_0^1 \frac{\tan^{-1}(x)}{x} \, dx$$ and it looks like the integrand function on this interval, $$[0,1]$$, can be approximated by a bell-type curve:$$f(x)=e^{-ax^{2n}}$$ where say $$a,n \in \mathbb{Q}.$$

P.S. Since $$\int_0^1 e^{-x^2} \, dx+\int_1^\infty e^{-x^2} \, dx = \int_0^\infty e^{-x^2} \, dx,$$ at least one of the two integrals on the LHS must be irrational.

Edit: Using Desmos, I found that if $$a=\frac{1}{4.1}$$ and $$n=0.8486609$$, then $$H(n) = \left|\int_0^1 \frac{\tan^{-1}(x)}{x} \, dx-\int_0^1 e^{-ax^{2n}} \, dx \right| = 3.975\times10^{-9}.$$

And this function has a zero, but it seems unless the zero $$n$$ is a familiar algebraic number, it might not be too helpful. But perhaps it can be used to generate some type of sequence.

• Just to note: $$\int_0^1 e^{-x^2}dx = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)n!}.$$ Dec 26, 2023 at 0:48
• @BrendanMcKay are you sure? I believe these $2n+1$ ruin it, because the denominator grows faster than the error term. Dec 26, 2023 at 3:30
• Write $X=\int_0^1 e^{-x^2}\,dx$. Replacing $x$ by $\frac1x$ yields $X=\int_1^\infty x^2 e^{-x^2}\,dx$. Integrating by parts, we obtain $X=\frac12\left(\frac1e+\int_1^\infty e^{-x^2}\,dx\right)$. Since $\int_1^\infty=\frac{\sqrt\pi}2-X$ we end up with $X=\frac13\left(\frac1e+\frac{\sqrt\pi}2\right).$ If that is rational, then $1,e$ and $e\sqrt\pi$ are linearly dependent over $\mathbb{Q}$, which I doubt.
– user473423
Dec 26, 2023 at 7:47
• @Echo That substitution doesn't seem correct, numerically it's a completely different number Dec 26, 2023 at 8:17
• By continued fraction computation, neither $X=\int_0^1 e^{-x^2}$ nor $\mathrm{erf}(1)=2X/\sqrt\pi$ is a rational number with a denominator less than $10^{10000}$. Dec 26, 2023 at 12:10

Following @te4's comment, we can look at the function $$f(x) = \frac{\sqrt{\pi}}{2\sqrt x} \operatorname{erf}(\sqrt{x}) = \sum_{n=0}^\infty \frac{(-1)^n x^n}{(2n+1) n!}.$$ Note that it's an E-function, because $$\log\bigl(\operatorname{lcm}(1, 3, \ldots , 2n+1)\bigr) = O(n),$$ and thus so is its derivative, $$f'(x)$$. We will apply the Siegel–Shidlovsky theorem to $$f(x)$$ and $$g(x) = f'(x)$$. We have the linear differential equations $$f'(x) = g(x), \qquad g'(x) = -\frac1{2x}f(x) - \left(1 + \frac{3}{2x}\right) g(x).$$

Since $$f'(x) + \frac{f(x)}{2x} = \frac{e^{-x}}{2x}$$ is elementary, and since $$f(x)$$ is non-elementary, we can conclude that $$f(x)$$ and $$f'(x)$$ are algebraically independent, so the Siegel–Shidlovsky theorem tells us that $$f(1)$$ and $$f'(1)$$ are algebraically independent. Hence $$f(1) = \int_0^1 {e^{-x^2} dx}$$ is transcendental, so in particular it is irrational.

• Thank you. This helps a lot. This might seem trivial for some, but I just figured out that if ${a,b}$ are algebraically independent over say $\mathbb{Q}$, then each of $a$ and $b$ individually are also algebraically independent which implies transcendental. Suppose $a$ and $b$ are not transcendental. Then there exist polynomials $P$ and $Q$ over $\mathbb{Q}$ such that $Q(a)=0$ and $R(b)=0$. But this means we can take $P(x,y) = Q(x) + R(y)$. Then $P(a,b) = Q(a) + R(b) = 0$. Contradiction. Dec 27, 2023 at 0:17
• I just don't understand the second to last point about $2xf'(x) + f(x) = e^{-x}$ being elementary, and $f(x)$ being non-elementary implies $f(x)$ and $f′(x)$ are algebraically independent. From direct computation we also see that $f'(x)$ is non-elementary. Does it also have to do with the fact that $e^{-x}$ is a transcendental function? Dec 27, 2023 at 13:45
• @MatthewAlbano From $2xf'(x) + f(x) = e^{-x}$, if $P(f(x), f'(x)) = 0$ for some polynomial $P$, then there is a polynomial $Q$ such that $Q(f(x), e^{-x}) = 0$, which contradicts $f(x)$ being non-elementary, because the roots of a polynomial with elementary coefficients are elementary. Dec 28, 2023 at 4:28
• Got it. Thanks. Dec 28, 2023 at 14:55
• @MatthewAlbano your argument that $a$ and $b$ are both transcendental when $\{a,b\}$ is algebraically independent over $\mathbf Q$ is incorrect. The negation of them both being transcendental is not that they are both algebraic, but only that at least one is algebraic. Dec 29, 2023 at 15:02

$$\newcommand{\Q}{\mathbb Q}\newcommand{\erf}{\operatorname{erf}}$$(This answer had been posted before I saw Command Master's answer. I am leaving it here, since it contains more and/or different details; in particular, here the argument that the function $$\erf$$ is non-elementary is not used.)

The answer is yes. Moreover, the value of the integral $$\begin{equation*} I:=\int_0^1 e^{-t^2}\,dt \end{equation*}$$ is transcendental.

To show this, let us use some of the comments, with added details:

Step 0: For real $$x>0$$, let $$\begin{equation*} y(x):=\frac{\sqrt\pi}{2\sqrt x}\,\erf\sqrt x=\sum_{n=0}^\infty c_n\frac{x^n}{n!}, \end{equation*}$$ with $$\begin{equation*} c_n:=\frac{(-1)^n}{2n+1} \end{equation*}$$ and $$y(0):=1$$, so that $$\begin{equation*} I=y(1). \end{equation*}$$

Step 1: The function $$y$$ is an E-function over $$\Q$$.

Indeed, of the three bulleted conditions listed in the definition of an E-function in the Wikipedia article, only the last, third one is nontrivial. To verify it, let $$q_n$$ be the least common multiple of $$1,\dots,2n+1$$. Then, in view of the prime number theorem,
$$\begin{multline*} q_n=\prod_{p\le2n+1}p^{\lfloor\log_p(2n+1)\rfloor} \le\prod_{p\le2n+1}p^{\log_p(2n+1)} \\ =\prod_{p\le2n+1}(2n+1)=(2n+1)^{\pi_{2n+1}} \\ =(2n+1)^{(1+o(1))(2n+1)/\ln(2n+1)} \\ =e^{(1+o(1))(2n+1)}=O(n^{n\varepsilon}) \end{multline*}$$ for each real $$\varepsilon>0$$, where the product $$\prod_{p\le2n+1}$$ is over all primes $$p\le2n+1$$ and $$\pi_k$$ is the number of primes $$\le k$$.

Step 2: Letting $$y_1(x):=y(x)$$ and $$y_2(x):=y'(x)$$, and using Command Master's comment, we get the system of ODEs $$\begin{equation*} y'_1(x)=y_2(x),\quad y'_2(x)=-\frac1{2x}\,y_1(x)-\Big(1+\frac3{2x}\Big)y_2(x). \tag{1}\label{1} \end{equation*}$$ We now want to use the Siegel–Shidlovsky theorem. The only possibly nontrivial condition of this theorem to check here is that the "functions" $$y_1(x)$$ and $$y_2(x)$$ are algebraically independent over $$\Q(x)$$ or, equivalently, that the "functions" $$\begin{equation*} v_1(t):=y_1(t^2)=\frac{\sqrt\pi}{2t}\,\erf t\quad\text{and}\quad v_2(t):=y_2(t^2)=\sqrt\pi\,\frac{t\erf' t-\erf t}{2t^3} \end{equation*}$$ are algebraically independent over $$\Q(t^2)$$. So, it is enough to show that $$\erf t$$ and $$\erf' t$$ are algebraically independent over $$\Q(t)$$.

To do this, suppose that $$\begin{equation*} \sum_{(j,k)\in F}P_{j,k}\,\erf^{\,j} (\erf')^k=0 \end{equation*}$$ for some nonempty finite subset $$F$$ of the set $$\{0,1,\dots\}^2$$ and some polynomial functions $$P_{j,k}$$. The latter identity can be rewritten as $$\begin{equation*} \sum_{j=0}^m Q_j \erf^{\,j}=0 \tag{2}\label{2} \end{equation*}$$ for some integer $$m\ge0$$, where each $$Q_j(t)$$ is rational in $$(t,\erf'(t))$$ and $$Q_m=1$$. Differentiating \eqref{2}, we get an identity of the form \eqref{2}, but with $$m-1$$ in place of $$m$$ (if $$m\ge1$$). So, without loss of generality $$m=0$$. So, $$Q_0=0$$. Similarly repeatedly rewriting and differentiating, we conclude that all the coefficients of the $$(\erf')^k$$'s in $$Q_0$$ are $$0$$. This confirms that $$\erf t$$ and $$\erf' t$$ are algebraically independent over $$\Q(t)$$, and hence $$y_1(x)$$ and $$y_2(x)$$ are algebraically independent over $$\Q(x)$$.

Step 3: By the Siegel–Shidlovsky theorem, $$I=y(1)$$ and $$\frac1e=y'(1)+\frac12\,y(1)$$ are algebraically independent. In particular, it follows that $$I$$ and $$e$$ are transcendental and hence irrational. $$\quad\Box$$

• Why does the derivative of $y_2$ involve $y_1$? $y_2$ is elementary but $y_1$ is not! Dec 26, 2023 at 18:44
• Your formula for $y$ is not quite correct as the power series in $x$ has half the coefficients zero - it is a power series in $x^2$ Dec 26, 2023 at 20:21
• @Zerox : Thank you for your comment. This has now been fixed. Dec 26, 2023 at 21:46
• @Conrad : Thank you for your comment. This has now been fixed. Dec 26, 2023 at 21:47
• that's a very nice detailed answer Dec 26, 2023 at 22:36