Consider the Barnes beta like integral: Where Re$(a)$ ,Re$(b)$ is greater than $0$ and $c$ is not a negative integer, where $|z|<1$ and $|$arg$(-z)$|$<\pi$, $$\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-z)^s\, ds$$
Since there is a simple pole to $0$. I computed it using the right-side semicircle contour with a left-side semicircle $\epsilon$ away from the origin. Traversed clockwise. Call the whole contour $C$
Naming the Bigger arc $\Gamma$, smaller arc $\gamma_\epsilon$,
I have no trouble proving $\frac{1}{2\pi i}\int_{\Gamma}\frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-z)^s\, ds$ tends to zero as $\Gamma$ gets larger and larger.
Using the Residue theorem:(Since the contour is oriented clockwise)
$$\frac{1}{2\pi i}\int_{C}{\underbrace{\frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-z)^s}_{f(s)}}\, ds=-\sum_{n=0}^{\infty}\mathbf{Res}(f,s=n)$$ $$=-\sum_{n=0}^{\infty}\frac{\Gamma(a+n)\Gamma(b+n)}{\Gamma(c+n)}(-z)^n\frac{(-1)^{n+1}}{n!}=\frac{\Gamma(a)\Gamma(b)}{\Gamma(c)}{}_2F_1(a,b;c;z)$$
Now computing $\int_{\gamma_\epsilon}f$, letting $s=\epsilon e^{-i\theta}$ and $s=-u$ gives $$=\int_{-3\pi/2}^{-\pi/2}f(\epsilon e^{-i\theta})(-i\epsilon e^{-i\theta})\, d\theta=-\int_{\pi/2}^{3\pi/2}\frac{\Gamma(a+\epsilon e^{iu})\Gamma(b+\epsilon e^{iu})\Gamma(-\epsilon e^{iu})}{\Gamma(c+\epsilon e^{iu})}(-z)^{\epsilon e^{iu}}i\epsilon e^{iu}\, du$$
Now letting epsilon go to zero: $$\int_{\gamma_{\epsilon}}f(s)\, ds=\pi i\frac{\Gamma(a)\Gamma(b)}{\Gamma(c)}$$
Since integrating on $C-\Gamma-\gamma_\epsilon$ gives the result.
We subtract the integrals to get the answer $$\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-z)^s\, ds=\frac{\Gamma(a)\Gamma(b)}{\Gamma(c)}\Big({}_2F_1(a,b;c;z)-\frac{1}{2}\Big)$$
However, the correct result should be $\frac{\Gamma(a)\Gamma(b)}{\Gamma(c)}{}_2F_1(a,b;c;z)$.
I found many derivations and it seems that they didn't compute $\gamma_\epsilon$.
Why is that? (I feel like I got something wrong conceptually)
Reference: Derivation I found on the internet
It turns out that the statement $$\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-z)^s\, ds$$ incudes the contour $\gamma_\epsilon$. So I did not have to subtract it.
See this MO post