The answer by so-called friend Don indicates why the existence of logarithmic density implies the existence of Dirichlet density, with the same value. Below is an argument explaining why the set of primes with a specified initial digit has a logarithmic density (of Benford type), and thus a Dirichlet density. In my experience, the number of times authors cite Serre on this point is much greater than the number of references that provide an actual proof, so another write-up of a proof is probably helpful.

The paper by R. E. Whitney mentioned in the OP uses the Mertens formula $\sum_{p \leq x} 1/p = \log \log x + M + O(1/(\log x)^2)$, where the $O$-term is stronger than the remainder term
typically found in treatments of this estimate, which is $O(1/(\log x))$. The argument below avoids this. We start with a useful technical lemma that will be applied a few times.

**Lemma**. *For $0 < c_1 < c_2$ and $b > 1$*,
$$
\sum_{c_1b^k \leq p \leq c_2b^k} \frac{1}{p} \sim \frac{\log_b(c_2/c_1)}{k}
$$
*as $k \to \infty$, where the sum on the left runs over primes*.

*Proof*.
Our argument is based on p. 7 here, which is the paper cited at the end of the Wikipedia page about Dirichlet density. That paper
proves a much more general result than what we need.

Using partial summation, for $x \geq 1$
$$
\sum_{p \leq x} \frac{1}{p} = \frac{\pi(x)}{x} + \int_1^x \frac{\pi(y)}{y^2}\,dy.
$$
Therefore
\begin{align*}
\sum_{c_1b^k \leq p \leq c_2b^k} \frac{1}{p}
& = \sum_{p \leq c_2b^k} \frac{1}{p} - \sum_{p \leq c_1b^k} \frac{1}{p} + O\left(\frac{1}{b^k}\right) \\
& = \frac{\pi(c_2b^k)}{c_2b^k} - \frac{\pi(c_1b^k)}{c_1b^k} + \int_{c_1b^k}^{c_2b^k} \frac{\pi(y)}{y^2}\,dy + O\left(\frac{1}{b^k}\right).
\end{align*}
To show this is asymptotic to $\log_b(c_2/c_1)/k$ as $k \to \infty$,
we will show
$$
\frac{\pi(c_2b^k)}{c_2b^k} - \frac{\pi(c_1b^k)}{c_1b^k} = o\left(\frac{1}{k}\right), \ \ \
\int_{c_1b^k}^{c_2b^k} \frac{\pi(y)}{y^2}\,dy \sim \frac{\log_b(c_2/c_1)}{k}.
$$

By the Prime Number Theorem, $\pi(c_2b^k) \sim c_2b^k/\log(c_2b^k) \sim c_2b^k/(k\log b)$ and
$\pi(c_1b^k) \sim c_1b^k/(k\log b)$ as $k \to \infty$, so
$$
k\left(\frac{\pi(c_2b^k)}{c_2b^k} - \frac{\pi(c_1b^k)}{c_1b^k}\right) \to \frac{1}{\log b} - \frac{1}{\log b} = 0.
$$
Thus $\pi(c_2b^k)/c_2b^k - \pi(c_1b^k)/c_1b^k = o(1/k)$ as $k \to \infty$.

To estimate $\int_{c_1b^k}^{c_2b^k} (\pi(y)/y^2)\,dy$, pick $\varepsilon > 0$. For
large $y$, say $y \geq y_\varepsilon$, we have
$1-\varepsilon \leq \pi(y)/(y/\log y) \leq 1+\varepsilon$,
so $(1-\varepsilon)/(y\log y) \leq \pi(y)/y^2 \leq (1+\varepsilon)/(y\log y)$.
For $k$ large enough that $c_1b^k \geq y_\varepsilon$,
\begin{equation}\label{int-long}
(1-\varepsilon)\int_{c_1b^k}^{c_2b^k}\frac{dy}{y\log y} \leq
\int_{c_1b^k}^{c_2b^k} \frac{\pi(y)}{y^2}\,dy \leq
(1+\varepsilon)\int_{c_1b^k}^{c_2b^k}\frac{dy}{y\log y},
\end{equation}
and as $k \to \infty$,
\begin{align*}
\int_{c_1b^k}^{c_2b^k}\frac{dy}{y\log y} = \log\log(c_2b^k) - \log\log(c_1b^k)
& = \log\left(\frac{k\log b + \log c_2}{k\log b + \log c_1}\right) \\
& = \log\left(1 + \frac{\log_b(c_2/c_1)}{k + \log_b(c_1)}\right) \\
& = \frac{\log_b(c_2/c_1)}{k + \log_b(c_1)} + O\left(\frac{1}{k^2}\right) \\
& = \frac{\log_b(c_2/c_1)}{k} + O\left(\frac{1}{k^2}\right).
\end{align*}
Therefore
$$
1-\varepsilon \leq \varliminf_{k \to \infty} \frac{\int_{c_1b^k}^{c_2b^k} (\pi(y)/y^2)\,dy}{\log_b(c_2/c_1)/k} \leq
\varlimsup_{k \to \infty} \frac{\int_{c_1b^k}^{c_2b^k} (\pi(y)/y^2)\,dy}{\log_b(c_2/c_1)/k} \leq 1 + \varepsilon.
$$
Taking $\varepsilon \to 0^+$ shows
$\int_{c_1b^k}^{c_2b^k} (\pi(y)/y^2)\,dy \sim \log_b(c_2/c_1)/k$, so we're done. QED

For $d \in \{1, \ldots, 9\}$, let $A_d$ be the set of primes with leading digit $d$. For example,
\begin{align*}
A_1 & = \{11, 13, 17, 19, 101, 103, 107, 109, 113, 127, 131, 137, 139, \ldots\}, \\
A_2 & = \{2, 23, 29, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, \ldots\}.
\end{align*}
We can describe the primes in $A_d$ in terms of inequalities:
$$
A_d = \{p : d \cdot 10^k \leq p < (d+1)10^k \ {\rm for \ some } \ k \geq 0\}.
$$

**Theorem**.
*For each $d$, $A_d$ has logarithmic density $\log_{10}((d+1)/d)$*.

Since existence of logarithmic density implies existence of Dirichlet density with the same value, $A_d$ has Dirichlet density $\log_{10}((d+1)/d)$.

*Proof*.
Instead of estimating
$$
\frac{\sum_{p \in A_d, p \leq x} 1/p}{\log \log x}
$$
for large $x$, we want to restrict $x$ to powers of $10$.
We will first check the ratio changes negligibly when $x$ runs between
consecutive powers of $10$. For $n \geq 1$, if $10^n \leq x < 10^{n+1}$
then
$$
\log n + \log \log 10 \leq \log\log x < \log(n+1) + \log \log 10,
$$ so
$\log \log x \sim \log n$ as $x \to \infty$. We will show

$$
\frac{\sum_{p \in A_d, 10^n \leq p < 10^{n+1}} 1/p}{\log n} \to 0 \ {\rm as } \ n \to \infty,
$$
so
$$
\lim_{x \to \infty} \frac{\sum_{p \in A_d, p \leq x} 1/p}{\log \log x} =
\lim_{n \to \infty} \frac{\sum_{p \in A_d, p < 10^n} 1/p}{\log n},
$$
in the sense that if the limit on the right exists then so does the limit on the left and they agree.

If $d \not= 1$ then $\{p \in A_d : 10^n \leq p < 10^{n+1}\} = \emptyset$.

If $d = 1$ then $\{p \in A_d : 10^n \leq p < 10^{n+1}\} = \{p : 10^n \leq p \leq 2 \cdot 10^{n}\}$ and
$$
\sum_{10^n \leq p \leq 2 \cdot 10^{n}} \frac{1}{p} \sim \frac{\log_{10} 2}{n}
$$
by the lemma, so
$$
\frac{\sum_{p \in A_1, 10^n \leq p < 10^{n+1}} 1/p}{\log n} \sim \frac{\log_{10} 2}{n\log n} \to 0.
$$

It remains to show
$$
\lim_{n \to \infty} \frac{\sum_{p \in A_d, p < 10^n} 1/p}{\log n} = \log_{10}\left(\frac{d+1}{d}\right).
$$
Break up the numerator into sums between consecutive powers of $10$:
$$
\sum_{p \in A_d, p < 10^n} \frac{1}{p} =
\sum_{0 \leq k \leq n-1} \left(\sum_{d \cdot 10^k \leq p < (d+1)10^k} \frac{1}{p}\right).
$$
For $k \geq 1$,
$$
\sum_{d \cdot 10^k \leq p < (d+1)10^k} \frac{1}{p} =
\sum_{d \cdot 10^k \leq p \leq (d+1)10^k} \frac{1}{p} \sim \frac{\log_{10}((d+1)/d)}{k}
$$
by the lemma. Therefore
$$
\sum_{p \in A_d, p < 10^n} \frac{1}{p}
\sim \sum_{1 \leq k \leq n-1} \frac{\log_{10}((d+1)/d)}{k} \sim \log_{10}\left(\frac{d+1}{d}\right)\log n,
$$
so $A_d$ has logarithmic density $\log_{10}((d+1)/d)$. QED

**Remark**.
In a similar way, if $m \geq 2$ and $1 \leq a \leq m-1$, then the set of primes whose number of digits (in base $10$, say) is congruent to $a \bmod m$ has logarithmic density $1/m$ (it has no natural density). For example, the set of primes with an odd number of digits has logarithmic density $1/2$.

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