TITLE: Translate a German paper by Hilbert
QUESTION [6 upvotes]: Please translate Hilbert's paper Ein Beitrag zur Theorie des Legendre'schen Polynoms. I cannot find an existing translation for it. Thanks:)

REPLY [91 votes]: Acta mathematica 18, printed April 3rd, 1894
A CONTRIBUTION TO THE THEORY OF THE LEGENDRE POLYNOMIAL
BY 
DAVID HILBERT
Königsberg, Prussia
This communication at hand is about the question of the smallest value different from $0$ which the integral
$$
I = \int\limits_\alpha^\beta \left[ f(x) \right]^2 \, dx \quad (\beta \gt \alpha)
$$
is able to attain, if one chooses for $f(x)$ a polynomial (lit.: "whole rational function") of degree $n - 1$ with integer coefficients and if one understands $\alpha$ and $\beta$ as given constants.
If one sets
$$
f(x) = a_1 x^{n - 1} + a_2 x^{n - 2} + \cdots + a_n
$$
the integral turns into a positive quadratic form of the $n$ variables $a_1$, $a_2, \ldots, a_n$:
$$
I = \sum^{i,k} \alpha_{ik} a_i a_k, \quad (i, k = 1, 2, \ldots, n)
$$
whose coefficients are given by the formula
$$
\alpha_{ik} = 
\int\limits_\alpha^\beta x^{2n-i-k} dx =
\frac{\beta^{2n-i-k+1}-\alpha^{2n-i-k+1}}{2n-i-k+1} 
$$
To get an upper bound for the minimum of this quadratic form $I$ it is needed to calculate its discriminant
$$
D_{\alpha \beta} =
\left|
\begin{matrix} 
\alpha_{11} & \alpha_{12} & \cdots & \alpha_{1n} \\
\alpha_{21} & \alpha_{22} & \cdots & \alpha_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
\alpha_{n1} & \alpha_{n2} & \cdots & \alpha_{nn}
\end{matrix}
\right| .
$$
(end of page 1)
If we replace in this determinant each element by its integral expression and thereby use for all elements of the first horizontal row the integration variable $x = x_1$, for all elements of the $2^{\tiny\mbox{nd}}, \ldots, n^{\tiny\mbox{th}}$ horizontal row regarding the integration variables $x = x_2, \ldots, x_n$, the discriminant of the quadratic form $I$ is represented as $n$-fold integral of the following form:
$$
D_{\alpha \beta} = 
\int\limits_\alpha^\beta \ldots \int\limits_\alpha^\beta
x_1^{n-1} x_2^{n-2} \ldots x_n^0
\prod^{i,k}(x_i - x_k) dx_1 \ldots dx_n .
$$
The interchange of the $n$ integration variables $x_1,\ldots,x_n$ and
the addition of the resulting equations gives
$$
D_{\alpha \beta} = 
\frac{1}{\left|\underline{n}\right.}
\int\limits_\alpha^\beta
\ldots
\int\limits_\alpha^\beta
\prod^{i,k} (x_i - x_k)^2 dx_1 \ldots dx_n
$$
and if we -- using
$$
x_i = \frac{1}{2}(\beta - \alpha) y_i + \frac{1}{2}(\beta + \alpha)
$$
introduce the new integration variables $y_1,\ldots,y_n$, 
we obtain the formula
$$
D_{\alpha \beta} =
\left(
\frac{\beta - \alpha}{2}
\right)^{n^2} D, 
$$
where for abbreviation $D = D_{-1, +1}$ has been set.
For example it follows for $\alpha = 0$, $\beta = 1$
$$
D = 2^n
\left|
\begin{matrix} 
1 & 0 & \frac{1}{3} & 0 & \frac{1}{5} & \cdots \\
0 & \frac{1}{3} & 0 & \frac{1}{5} & 0 & \cdots \\
\frac{1}{3} & 0 & \frac{1}{5} & 0 & \frac{1}{7} & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\end{matrix}
\right|
= 2^{n^2}
\left|
\begin{matrix} 
1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & 
\cdots & \frac{1}{n} \\
\frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} &
\cdots & \frac{1}{n + 1} \\ 
\frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & 
\cdots & \frac{1}{n + 2} \\ 
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
\frac{1}{n} & \frac{1}{n+1} & \frac{1}{n+2} & \frac{1}{n+3} &
\cdots & \frac{1}{2n - 1} 
\end{matrix}
\right| .
$$
(end of page 2)
To calculate $D$ we develop the polynomial (lit.: "whole rational function") $f(x)$ into a progressing series of Legendre polynomials $X_m$.
Because of
$$
x^m = c_m X_m + c_m' X_{m-1} + \cdots + c_m^{(m)} X_0,
$$
where
$$
c_m = \frac{\left| \underline{m} \right.}{1 \cdot 3 \cdot 5 \cdots (2m-1)},
$$
we get
$$
\begin{align}
f(x) &=
a_1 (c_{n-1} X_{n-1} + c_{n-1}' X_{n-2} + \cdots) +
a_2 (c_{n-2} X_{n-2} + c_{n-2}' X_{n-3} + \cdots) + \cdots \\
&= c_{n-1} a_1 X_{n-1} +
(c_{n-1}' a_1 + c_{n-2} a_2) X_{n-2} +
(c_{n-1}'' a_1 + c_{n-2}' a_2 + c_{n-3} a_3 X_{n-3}) + \cdots
\end{align}
$$
and with the help of the formulas
$$
\int\limits_{-1}^{+1} X_m^2 dx = \frac{2}{2m+1}, \quad
\int\limits_{-1}^{+1} X_m X_k dx = 0 \quad
(m \ne k)
$$
it therefore follows
$$
\left[ I \right]_{\alpha = -1 \atop \beta = +1} =
\int\limits_{-1}^{+1} f^2(x) \, dx =
\frac{2}{2n-1} b_1^2 +
\frac{2}{2n-3} b_2^2 +
\frac{2}{2n-5} b_3^2 + \cdots ,
$$
where
$$
\begin{align}
b_1 &= c_{n-1} a_1, \\
b_2 &= c_{n-1}' a_1 + c_{n-2} a_2, \\
b_3 &= c_{n-1}'' a_1 + c_{n-2}' a_2 + c_{n-3} a_3, \\
. \,\,\, & . \,\,\, . \,\,\, . \,\,\, . \,\,\, . \,\,\, . \,\,\, . 
\,\,\, . \,\,\, . \,\,\, . \,\,\, . \,\,\, . \,\,\, . \,\,\, .
\end{align}
$$
has been set. Because of this representation as sum of squares of linear forms one wins for the disciminant $D$ the value
$$
\begin{align}
D &=
\frac{2^n}{1 \cdot 3 \cdot 5 \ldots (2n-1)}
(c_0 c_1 \cdots c_{n-1})^2 \\
&= 2^{n^2} 
\frac{\left\{
1^{n-1} 2^{n-2} \ldots (n-2)^2 (n-1)^1 
\right\}^4}{1^{2n-1} 2^{2n-2} \ldots (2n-2)^2 (2n-1)^1}
\end{align}
$$
and herein is the right side exactly identical with
$\frac{2^{n^2}}{\Delta}$ where $\Delta$ means that value, which I, in my treatise "About the Discriminant of the in the Finite terminating Hypergeometric Series" ("Über die Discriminante der im Endlichen abbrechenden hypergeometrischen Reihe") [1],
[1] Crelle's Journal, Vol. 103, p. 342.
(end of page 3)
got for the discriminant of a $n$-th degree Legendre polynomial under a  certain linear transformation. 
Under consideration of the formula for $D$ stated above, from this follows the result

The discriminant of the quadratic form $\int\limits_0^1 f^2(x)\,dx$ has the value 
  $$
\left|
\begin{matrix} 
1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & 
\cdots & \frac{1}{n} \\
\frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} &
\cdots & \frac{1}{n + 1} \\ 
\frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & 
\cdots & \frac{1}{n + 2} \\ 
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
\frac{1}{n} & \frac{1}{n+1} & \frac{1}{n+2} & \frac{1}{n+3} &
\cdots & \frac{1}{2n - 1} 
\end{matrix}
\right| 
=
\frac{\left\{
1^{n-1} 2^{n-2} \ldots (n-2)^2 (n-1)^1 
\right\}^4}{1^{2n-1} 2^{2n-2} \ldots (2n-2)^2 (2n-1)^1}
$$
  and is equal to the reciprocal value of the disciminant of 
  the $n^{\tiny\mbox{th}}$ degree equation
  $$
\xi^n + 
\binom{n}{1}^2 \xi^{n-1} +
\binom{n}{2}^2 \xi^{n-2} + \ldots + 1 = 0, 
$$
  whose left side, by a linear transformation of the variable $\xi$, can be turned into the Legendre polynomial $X_n$.

We now return to the original posed question.
Application of the Stirling forumla, if $N$ is a positive number,
delivers the equation
$$
N l 1 + (N-1) l 2 + \ldots + 
2 l (N-1) + 1 l N =
\frac{1}{2}N^2 l N - \frac{3}{4} N^2 (1 + \varepsilon_N),
$$
where $\varepsilon_N$ means a with growing $N$ vanishing number.
With the help of this formula one easily finds
$$
l D = (1 + \varepsilon_n') l (2^{-n^2}),
$$
(end of page 4)
where $\varepsilon_n'$ vanishes with growing $n$, i.e.
$$
D = \eta_n 2^{-n^2}
$$
and further
$$
D_{\alpha \beta} = 
\eta_n \left(
\frac{\beta - \alpha}{4}
\right)^{n^2},
$$
where $\eta_n$ approaches the unit with growing $n$.
Now it is, according to a theorem by H. Minkowski [2], always possible,
for a definite quadratic form, to choose the $n$ variables such that
the value of the quadratic form turns out less than the $n$-fold
of the $n^{\tiny \mbox{th}}$ root of its discriminant.
If therefore the positive difference $\beta - \alpha$ is assumed
less than $4$, it follows that it is always possible
to determine a polynomial (lit.: "whole rational function")
$f(x)$ with integer coefficients for which the value of
the integral $I = \int\limits_\alpha^\beta f^2(x)\,dx$ turns
out smaller than 
$n \eta_n' \left( \frac{|\beta - \alpha|}{4} \right)^n$.
But because $\eta_n'=\sqrt[n]{\eta_n}$ approaches the unit too
with growing $n$, we get the result

The integral $\int\limits_\alpha^\beta f^2(x) \, dx$ can
  attain an arbitrary small positive value, if one chooses
  the polynomial (lit.: "whole integer function") $f(x)$
  properly, under the condition that the interval of
  integration $\alpha$ to $\beta$ is smaller than $4$. 

Königsberg in Prussia, March 13th, 1893.
[2] Crelle's Journal, Vol. 107, p. 291. 
(end of page 5)
(Special thanks to Donald E. Knuth and the MathJax team)