After further research, I suspect no source other than Derbyshire's text exists. If so, this would be just a very small slip in an excellent and well written text. I know I have made similar slips.
Here is my reasoning. First, Lebesgue published his first paper in 1898, three in 1899 and then two in 1900 (all most easily found in his collected works). None of these contain any references to the prime numbers:
- Sur l'approximation des fonctions, Bull. Sci. Math. 22 (1898), 278--287.
- Sur la définition de l'aire d'une surface, C. R. Math. Acad. Sci. Paris 129 (1899), 870--873.
- Sur les fonctions de plusieurs variables, C. R. Math. Acad. Sci. Paris 128 (1899), 811--813.
- Sur quelques surfaces non réglées applicables sur le plan, C. R. Math. Acad. Sci. Paris 128 (1899), 1502--1505.
- Sur la définition de certaines intégrales de surface, C. R. Math. Acad. Sci. Paris 131 (1900), 867--870.
- Sur le minimum de certaines intégrales, C. R. Math. Acad. Sci. Paris 131 (1900), 935--937.
So if Lebesgue stated that one was prime in 1899, it appears not to be in a published work of his. (This does not rule out lectures, interviews, works written about him by others...)
Second, all of the Internet references to this that we checked either cite Derbyshire, or were posted well after his work. For example, a friend of mine checked Wikipedia, and this statement about Lebesgue and unity appears in the English and Dutch entry for prime, but not the French, German, Spanish, Italian, Portuguese, Polish, Russian, Czech, or Swedish. In English it was added in 2006, after Derbyshire's 2003 text. In English only it is also found in the Wikipedia page for "Henri Lebesgue":
Is this proof Derbyshire's text is the source, absolutely not. Recall Derbyshire said that he can not recall his source, but that there was one. So if you can find a source that predates his text, I'd like to know; otherwise I think it may just have been a small transcription error while written this popular text. I have done the same myself.
As for the related question: who was the last mathematician of any importance who considered the number 1 to be a prime, my student and I settle on G. H. Hardy in our draft paper (http://arxiv.org/abs/1209.2007). Hardy's A Course of Pure Mathematics, 6th edition, 1933, presented Euclid's proof that there are infinitely many primes with a sequence of primes beginning with 1:
(This was changed in the next edition.) As discussed in our draft article, there is even a remnant of Hardy listing 1 as prime in the revised 10th edition of his text published recently.
We have a list of over 125 references pertinent to the question "is one prime" collected here: http://primes.utm.edu/notes/one.pdf
For $\mathrm{Re}(s)\gt1$,
$$
\zeta(s)=\sum_{n=1}^\infty\frac1{n^s}\tag{1}
$$
For $\mathrm{Re}(s)\le1$, we need to use Analytic Continuation and another formula for $\zeta(s)$ since $(1)$ does not converge for $\mathrm{Re}(s)\le1$.
One formula that can be used is the Functional Equation for $\zeta$ which says
$$
\zeta(s)\frac{\Gamma(s/2)}{\pi^{s/2}}=\zeta(1-s)\frac{\Gamma((1-s)/2)}{\pi^{(1-s)/2}}\tag{2}
$$
The symmetric form in $(2)$ is given as $(14)$ on MathWorld.
Since the recurrence for $\Gamma$ says that for $n\in\mathbb{Z}$ and $n\le0$, $\frac1{\Gamma(n)}=0$, we get that $\zeta(2n)=0$ for $n\in\mathbb{Z}$ and $n\lt0$.
Analytic Continuation of $\boldsymbol{\zeta}$ Using $\boldsymbol{\eta}$ and Integration by Parts
Define $\eta$, the alternating $\zeta$ function, as
$$
\begin{align}
\eta(s)
&=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}\\
&=\sum_{n=1}^\infty\frac1{n^s}-2\sum_{n=1}^\infty\frac1{(2n)^s}\\[6pt]
&=\zeta(s)\left(1-2^{1-s}\right)\tag{3}
\end{align}
$$
Formula $(3)$ follows since the terms of an alternating series are the terms of the non-alternating series minus twice the even terms.
Formula $(3)$ increases the domain to $\operatorname{Re}(s)\gt0$. $\eta$ also comes in handy to define
$$
\eta(s)\,\Gamma(s)=\int_0^\infty\frac{t^{s-1}}{e^t+1}\,\mathrm{d}t\tag{4}
$$
which also converges for $\operatorname{Re}(s)\gt0$. However, we can integrate $(4)$ by parts $k$ times to get
$$
\bbox[5px,border:2px solid #C0A000]{\eta(s)\,\Gamma(s)=\frac{(-1)^k}{s(s+1)\cdots(s+k-1)}\int_0^\infty t^{s+k-1}\frac{\mathrm{d}^k}{\mathrm{d}t^k}\frac1{e^t+1}\,\mathrm{d}t}\tag{5}
$$
$(5)$ agrees with $(4)$ for $\operatorname{Re}(s)\gt0$ and converges for $\operatorname{Re}(s)\gt-k$. Thus, $(5)$ gives an analytic continuation of $\zeta(s)$ for $\operatorname{Re}(s)\gt-k$.
Using this method, $\zeta(0)=-\frac12$ is computed in this answer and $\zeta(-1)=-\frac1{12}$ is computed in this answer.
Proving $\boldsymbol{\zeta(-2n)=0}$
Note that $(5)$ can be rewritten as
$$
\eta(s)\,\Gamma(s+k)=(-1)^k\int_0^\infty t^{s+k-1}\frac{\mathrm{d}^k}{\mathrm{d}t^k}\left(\frac12-\frac12\frac{e^{t/2}-e^{-t/2}}{e^{t/2}+e^{-t/2}}\right)\,\mathrm{d}t\tag{6}
$$
Setting $s=1-k$ in $(6)$ gives
$$
\begin{align}
\eta(1-k)
&=(-1)^k\int_0^\infty\frac{\mathrm{d}^k}{\mathrm{d}t^k}\left(\frac12-\frac12\frac{e^{t/2}-e^{-t/2}}{e^{t/2}+e^{-t/2}}\right)\,\mathrm{d}t\\[6pt]
&=(-1)^{k-1}\frac{\mathrm{d}^{k-1}}{\mathrm{d}t^{k-1}}\left.\left(\frac12-\frac12\frac{e^{t/2}-e^{-t/2}}{e^{t/2}+e^{-t/2}}\right)\right|_{t=0}\tag{7}
\end{align}
$$
Set $k=2n+1$ and we get
$$
\eta(-2n)
=\frac{\mathrm{d}^{2n}}{\mathrm{d}t^{2n}}\left.\left(\frac12-\frac12\frac{e^{t/2}-e^{-t/2}}{e^{t/2}+e^{-t/2}}\right)\right|_{t=0}\tag{8}
$$
For $n\ge1$, the right side of $(8)$ is an odd function, so evaluating at $t=0$, and applying $(3)$, yields
$$
\bbox[5px,border:2px solid #C0A000]{\zeta(-2n)=0}\tag{9}
$$
Analytic Continuation of $\boldsymbol{\zeta}$ Using the Euler-Maclaurin Sum Formula
As described in this answer, this answer, and this answer, for $\operatorname{Re}(s)\gt-2m-1$, $\zeta(s)$ can be represented as
$$
\hspace{-12pt}\bbox[5px,border:2px solid #C0A000]{\zeta(s)=\lim_{n\to\infty}\left[\sum_{k=1}^n\frac{1}{k^s}-\left(\frac{1}{1-s}n^{1-s}+\frac12n^{-s}-\sum_{k=1}^m\frac{B_{2k}}{2k}\binom{s+2k-2}{2k-1}n^{-s-2k+1}\right)\right]}\tag{10}
$$
where the formula in the parentheses is obtained from the Euler-Maclaurin Sum Formula.
Proving $\boldsymbol{\zeta(-2m)=0}$
If we plug $s=-2m$ into $(10)$, for $m\ge1$, the formula in parentheses exactly matches the sum outside the parentheses by Faulhaber's Formula. In fact, the Euler-Maclaurin Sum Formula is one way to prove Faulhaber's Formula. This means that
$$
\bbox[5px,border:2px solid #C0A000]{\zeta(-2m)=0}\tag{11}
$$
We can also plug $s=-2m+1$, for $m\ge1$, into $(10)$ and note that Faulhaber's Formula does not include the constant term. Thus, we get
$$
\zeta(-2m+1)=-\frac{B_{2m}}{2m}\tag{12}
$$
Functional Equation for $\boldsymbol{\zeta}$
The Fourier Transform of $e^{-\pi x^2t}$ is
$$
\begin{align}
\int_{-\infty}^\infty e^{-\pi x^2t}e^{-2\pi ix\xi}\,\mathrm{d}x
&=\int_{-\infty}^\infty e^{-\pi(x-i\xi/t)^2t}e^{-\pi\xi^2/t}\,\mathrm{d}x\\
&=\frac1{\sqrt{t}}e^{-\pi\xi^2/t}\tag{13}
\end{align}
$$
Applying the Poisson Summation Formula to $(13)$ says that
$$
1+2\sum_{n=1}^\infty e^{-\pi n^2t}
=\frac1{\sqrt{t}}+\frac2{\sqrt{t}}\sum_{n=1}^\infty e^{-\pi n^2/t}\tag{14}
$$
Note that for $s\gt1$,
$$
\begin{align}
&\zeta(s)\frac{\Gamma(s/2)}{\pi^{s/2}}\\
&=\sum_{n=1}^\infty\frac1{n^s}\int_0^\infty e^{-\pi t}t^{\frac{\large s}2}\,\frac{\mathrm{d}t}t\\
&=\int_0^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)t^{\frac{\large s}2}\,\frac{\mathrm{d}t}t\\
&=\int_0^1\left(\frac1{2\sqrt{t}}-\frac12+\frac1{\sqrt{t}}\sum_{n=1}^\infty e^{-\pi n^2/t}\right)t^{\frac{\large s}2}\,\frac{\mathrm{d}t}t
+\int_1^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)t^{\frac{\large s}2}\,\frac{\mathrm{d}t}t\\
&=-\frac1{1-s}-\frac1s+\int_1^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)t^{\frac{1-\large s}2}\,\frac{\mathrm{d}t}t
+\int_1^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)t^{\frac{\large s}2}\,\frac{\mathrm{d}t}t\\
&=-\frac1{1-s}-\frac1s+\int_1^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)\left(t^{\frac{1-\large s}2}+t^{\frac{\large s}2}\right)\,\frac{\mathrm{d}t}t\tag{15}
\end{align}
$$
The following integral is increasing in $\alpha$. For $\alpha\ge0$, we have
$$
\begin{align}
\int_1^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)t^\alpha\,\mathrm{d}t
&=\sum_{n=1}^\infty\frac1{\pi^{\alpha+1}n^{2\alpha+2}}\int_{\pi n^2}^\infty e^{-t}\,t^\alpha\,\mathrm{d}t\\
&\le\sum_{n=1}^\infty\frac1{\pi^{\alpha+1}n^{2\alpha+2}}\left(\int_{\pi n^2}^\infty e^{-t}\,\mathrm{d}t\right)^{1/2}\left(\int_{\pi n^2}^\infty e^{-t}t^{2\alpha}\,\mathrm{d}t\right)^{1/2}\\
&\le\frac{\Gamma(2\alpha+1)^{1/2}}{\pi^{\alpha+1}}\sum_{n=1}^\infty\frac{e^{-\pi n^2/2}}{n^{2+2\alpha}}\\
&\le\frac{\Gamma(2\alpha+1)^{1/2}}{\pi^{\alpha+1}}\sum_{n=1}^\infty\frac2{\pi n^{4+2\alpha}}\\
&=\frac{2\,\Gamma(2\alpha+1)^{1/2}}{\pi^{\alpha+2}}\zeta(4+2\alpha)\tag{16}
\end{align}
$$
Thus, the last integral in $(15)$ defines an entire function. Therefore $(15)$ defines $\zeta(s)\frac{\Gamma(s/2)}{\pi^{s/2}}$ for all $s\in\mathbb{C}$. Since $(15)$ is invariant under $s\leftrightarrow1-s$, we have
$$
\bbox[5px,border:2px solid #C0A000]{\zeta(s)\frac{\Gamma(s/2)}{\pi^{s/2}}
=\zeta(1-s)\frac{\Gamma((1-s)/2)}{\pi^{(1-s)/2}}}\tag{17}
$$
Best Answer
From Riemann's Zeta Function, By Harold M. Edwards:
