Alternative approaches to showing that $\Gamma'(1/2)=-\sqrt\pi\left(\gamma+\log(4)\right)$ Starting from the definition of the Gamma function as expressed by
$$\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}\,dx\tag1$$
we can show that the derivative of $\Gamma(z)$ evaluated at $z=1/2$ is given by
$$\Gamma'(1/2)=-\sqrt{\pi} \left(\gamma+\log(4)\right)\tag2$$



Proof of $(2)$:  Here, I present for completeness the approach that I took.  One can skip this part without losing context.

Differentiating $(1)$ and setting $z=1/2$ reveals
$$\Gamma'(1/2)=\int_0^\infty \frac{e^{-x}}{\sqrt{x}}\log(x)\,dx\tag3$$
Next, we represent the logarithm function in $(3)$ by a Frullani integral to find that
$$\begin{align}
\Gamma'(1/2)&=\int_0^\infty \frac{e^{-x}}{\sqrt{x}}\int_0^\infty \frac{e^{-y}-e^{-xy}}{y}\,dy\,dx\\\\
&=\int_0^\infty \frac1y\int_0^\infty \frac{e^{-x}e^{-y}-e^{-(y+1)x}}{\sqrt{x}}\,dx\,dy\\\\
&=\sqrt\pi\int_0^\infty \frac1y \left(e^{-y}-\frac1{\sqrt{y+1}}\right)\,dy\tag4
\end{align}$$
Integrating by parts the integral on the right-hand side of $(4)$, we obtain
$$\Gamma'(1/2)=-\sqrt\pi(\gamma+\log(4))\tag5$$
as was to be shown.



QUESTION: So, what are alternative approaches to evaluating $\Gamma'(1/2)$ if we begin with $(1)$?

 A: 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
$$ \int_{0}^{+\infty}\frac{\log(s+1)}{(s+1)\sqrt{s}}\,ds = 2\int_{0}^{+\infty}\frac{\log(1+s^2)}{1+s^2}\,ds = -4\int_{0}^{\pi/2}\log\cos\theta\,d\theta =\pi\log 4.$$
A: 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!
A: I fear you might not like this "non rigorous" methode, so just to provoke you. We begin with the fact that the gamma function is a product:
$$\Gamma'(1/2)=\frac{\prod_{n=1}^{-3/2+h}(1+n)-\prod_{n=1}^{-3/2}(1+n)}{h}$$
$$\prod_{n=1}^{-3/2+h}(1+n)=\prod_{n=1}^{-3/2}(1+n)\prod_{n=-1/2}^{-3/2+h}(1+n)$$
$$\prod_{n=1}^{-3/2}(1+n)=(-1/2)!=(\pi)^{1/2}$$
$$=\prod_{n=-1/2}^{-3/2+h}(1+n)=\prod_{n=1}^{h}(-1/2+n)=\prod_{n=1}^{h}(1/2)\prod_{n=1}^{h}(2n-1)$$
$$\prod_{n=1}^{h}(1/2)=(1/2)^h=1-h\ln(2)$$
$$\prod_{n=1}^{h}(2n-1)=\frac{\prod_{n=1}^{2h}(n)}{\prod_{n=1}^{h}(2n)}$$
We know that  $\prod_{n=1}^{h}(n)=1-h\gamma$*
$$\frac{\prod_{n=1}^{2h}(n)}{\prod_{n=1}^{h}(2n)}=(1-h\ln(2))\frac{1-2h\gamma}{1-h\gamma}$$
All together gives: $\prod_{n=1}^{-3/2+h}(1+n)=\pi^{1/2}(1-h\ln(2))^2(1-2h\gamma)(1+h\gamma)=$
$$\pi^{1/2}\big(1-h\big(\gamma+2\ln(2)\big)+O(h^2)$$
Clearly the order h is the derivative
*the fact that it's the Euler–Mascheroni constant is directly visible from either it's polynomal alternatively, it comes instinctively because $\prod_{n=1}^{h}(1+n)$ as h goes to 0, we have it's polynomal form of n^s,  because it's the first derivative we need it's order $n^1$. Thuse we have the $\sum_s \sum_n n^s/s!$, times the last (or first, depends on your view) refined stirling number, which is $(-1)^{s+1}(s-1)!$. The order n in $\sum_n n^s=s\zeta(1-s)$, therefor:
The regularised sum:
$$\prod_{n=1}^{h}(1+n)=1+h\sum_{s=1}^{s\infty} (-1)^{s+1}\zeta(1-s)=1+h(1-\gamma)$$
