Alternative approaches to showing that $\Gamma'(1/2)=-\sqrt\pi\left(\gamma+\log(4)\right)$
Since $\Gamma'(x)=\Gamma(x)\psi(x)$ the determination of $\Gamma'(1/2)$ immediately boils down to the determination of $\psi(1/2)$. Since $$ \sum_{n\geq 0}\left(\frac{1}{n+a}-\frac{1}{n+b}\right)=\psi(a)-\psi(b)$$ and $\psi(1)=-\gamma$ by the Weierstrass product for the $\Gamma$ function, we may just pick $a=\frac{1}{2}$, $b=1$ and compute $$ \psi(1/2)+\gamma=\sum_{n\geq 0}\left(\frac{2}{2n+1}-\frac{2}{2n+2}\right)=2\sum_{m\geq 1}\frac{(-1)^{m+1}}{m}=-2\log 2 $$ to deduce $$ \Gamma'(1/2) = \Gamma(1/2)\psi(1/2) = \sqrt{\pi}\psi(1/2) = -\sqrt{\pi}(\gamma+\log 4)$$ without invoking Frullani.
Conversely,
$$\begin{eqnarray*} \gamma=\lim_{n\to +\infty}(H_n-\log n) &=& \sum_{n\geq 1}\left(\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right)\\&\stackrel{\text{Frullani}}{=}&\sum_{n\geq 1}\int_{0}^{+\infty}e^{-nx}-\frac{e^{-nx}-e^{-(n+1)x}}{x}\,dx\\&=&\int_{0}^{+\infty}\left(\frac{1}{e^x-1}-\frac{1}{x e^x}\right)\,dx\\&\stackrel{\color{red}{\text{Devil}}}{=}&-\int_{0}^{+\infty}e^{-x}\log(x)\,dx=-\Gamma'(1)\end{eqnarray*} $$ where the marked equality is justified by this:
$$ \int_{0}^{M}\left(\frac{1}{e^x-1}-\frac{1}{x}\right)\,dx = \log(1-e^{-M})-\log M$$ $$ \int_{0}^{M}\frac{1-e^{-x}}{x}\,dx\stackrel{\text{IBP}}{=}(1-e^{-M})\log M-\int_{0}^{M}e^{-x}\log(x)\,dx. $$
At this point we have
$$ \mathcal{L}\log(x) = -\frac{\gamma+\log(s)}{s},\qquad \mathcal{L}^{-1}\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{\pi s}}$$ hence by the self-adjointness of the Laplace transform
$$ \Gamma'(1/2)=\int_{0}^{+\infty}e^{-x}\log(x)\frac{dx}{\sqrt{x}}=-\frac{1}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{\gamma+\log(s+1)}{(s+1)\sqrt{s}}\,ds $$
where
$$ \int_{0}^{+\infty}\frac{ds}{(s+1)\sqrt{s}}=2\int_{0}^{+\infty}\frac{ds}{s^2+1}=\pi $$ and $$ \int_{0}^{+\infty}\frac{\log(s+1)}{(s+1)\sqrt{s}}\,ds = 2\int_{0}^{+\infty}\frac{\log(1+s^2)}{1+s^2}\,ds = -4\int_{0}^{\pi/2}\log\cos\theta\,d\theta =\pi\log 4.$$
We begin with the integral representation of the Gamma function as given by
$$\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}\,dx\tag1$$
for $z>0$.
In the next section, we show that $\Gamma(z)$ as expressed by $(1)$ can be represented by the limit
$$\Gamma(z)= \lim_{n\to\infty}\frac{n^z\,n!}{z(z+1)(z+2)\cdots (z+n)}$$
Limit Definition of Gamma
Let $G_n(z)$ be the sequence of functions given by
$$G_n(z)=\int_0^n x^{z-1}\left(1-\frac{x}{n}\right)^n\,dx$$
I showed in THIS ANSWER, using only Bernoulli's Inequality, that the sequence $\left(1-\frac{x}{n}\right)^n$ monotonically increases for $x\le n$. Therefore, $\left|x^{z-1} \left(1-\frac{x}{n}\right)^n\right|\le x^{z-1}e^{-x}$ for $x\le n$. The Dominated Convergence Theorem guarantees that we can write
$$\begin{align} \lim_{n\to \infty} G_n(z)=&\lim_{n\to \infty}\int_0^n x^{z-1}\left(1-\frac{x}{n}\right)^n\,dx\\\\ &=\lim_{n\to \infty}\int_0^\infty \xi_{[0,n]}\,s^{x-1}\left(1-\frac{s}{n}\right)^n\,ds\\\\ &=\int_0^\infty \lim_{n\to \infty} \left(\xi_{[0,n]}\,\left(1-\frac{x}{n}\right)^n\right)\,x^{z-1}\,\,dx\\\\ &=\int_0^\infty x^{z-1}e^{-x}\,dx\\\\ &=\Gamma(z) \end{align}$$
ALTERNATIVE PROOF: Limit Definition of Gamma
If one is unfamiliar with the Dominated Convergence Theorem, then we can simply show that
$$\lim_{n\to \infty}\int_0^n x^{z-1}e^{-x}\left(1-e^x\left(1-\frac{x}{n}\right)^n\right)=0$$
To do this, we appeal again to the analysis in THIS ANSWER. Proceeding, we have
$$\begin{align} 1-e^x\left(1-\frac{x}{n}\right)^n &\le 1-\left(1+\frac{x}{n}\right)^n\left(1-\frac{x}{n}\right)^n\\\\ &=1-\left(1-\frac{x^2}{n^2}\right)^n\\\\ &\le 1-\left(1-\frac{x^2}{n}\right)\\\\ &=\frac{x^2}{n} \end{align}$$
where Bernoulli's Inequality was used to arrive at the last inequality. Similarly, we see that
$$\begin{align} 1-e^x\left(1-\frac{x}{n}\right)^n &\ge 1-e^xe^{-x}\\\\ &=0 \end{align}$$
Therefore, applying the squeeze theorem yields to coveted limit
$$\lim_{n\to \infty}\int_0^n x^{z-1}e^{-x}\left(1-e^x\left(1-\frac{x}{n}\right)^n\right)=0$$
which implies $\lim_{n\to \infty}G_n(z)=\Gamma(z)$.
Integrating by parts repeatedly the integral representation of $G_n(z)$ reveals
$$G_n(z)=\frac{n^z\,n!}{z(z+1)(z+2)\cdots (z+n)}$$
so that
$$\bbox[5px,border:2px solid #C0A000]{\Gamma(z)=\lim_{n\to \infty}\frac{n^z\,n!}{z(z+1)(z+2)\cdots (z+n)}}\tag2$$
Now, we use $(2)$ to find a limit representation of the derivative of $\Gamma(z)$. To facilitate analysis, we use $(2)$ to find the logarithm of $\Gamma(z)$. Proceeding we have
$$\log\left(\Gamma(z)\right)=\lim_{n\to \infty}\left(z\log(n)+\log(n!)-\sum_{k=0}^n \log(z+k)\right)\tag3$$
Differentiating $(3)$ reveals
$$\begin{align} \frac{\Gamma'(z)}{\Gamma(z)}&=\lim_{n\to\infty}\left(\log(n)-\sum_{k=0}^n \frac1{z+k}\right)\\\\ &=\lim_{n\to\infty}\left(\log(n)-\sum_{k=0}^n\frac1{k+1}-\sum_{k=0}^n \left(\frac1{z+k}-\frac1{k+1}\right)\right)\\\\ &=-\gamma-\sum_{k=0}^\infty \left(\frac1{z+k}-\frac1{k+1}\right)\tag4 \end{align}$$
Setting $z=1/2$ in $(4)$ and using $\Gamma(1/2)=\sqrt \pi$ yields
$$\begin{align} \Gamma'(1/2)&=\sqrt{\pi}\left(-\gamma-\sum_{k=0}^\infty \left(\frac{1}{k+1/2}-\frac1{k+1}\right)\right)\\\\ &=\sqrt{\pi}\left(-\gamma-2\sum_{k=0}^\infty \left(\frac{1}{2k+1}-\frac1{2k+2}\right)\right)\\\\ &=\sqrt{\pi}\left(-\gamma-2\sum_{k=1}^\infty \left(\frac{1}{2k-1}-\frac1{2k}\right)\right)\\\\ &=\sqrt{\pi}\left(-\gamma-2\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\right)\\\\ &=-\sqrt\pi\left(\gamma+\log(4)\right) \end{align}$$
as was to be shown!