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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 at $0$. 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 $\gamma_\epsilon$.

Q: Why is that?

Note that I've asked this question on this MSE post as well. The question has poorly received attention and the post is still unanswered.

Any help is appreciated.

Reference: Derivation I found on the internet (Go to Page 3 Theorem 1)

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  • $\begingroup$ The contour in the Mellin-Barnes integral is defined to include $\gamma_{\epsilon} $, so you shouldn't subtract it. You just need to evaluate $C-\Gamma $. $\endgroup$
    – Michael Engelhardt
    Commented Jun 11, 2023 at 14:01
  • $\begingroup$ Ah I see, it is written in the condition, my bad I overlooked it. Still I find it kind of weird to define it this way because if we consider the contour $\gamma_\epsilon$ traversed right hand side instead. The way we write the integral is still $\int_{-i\infty}^{i\infty}$. Yet in this way the integral would yield a different answer $\endgroup$
    – Dqrksun
    Commented Jun 11, 2023 at 23:15
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    $\begingroup$ Certainly, the notation $\int_{-i\infty }^{i\infty } $ on its own is ambiguous, and one can define a left integral version of it, or a right integral version, or a principal value version (which I think your evaluation is related to) ... that's why this is usually accompanied by an explanation of which contour is meant. My understanding is that the standard definition is chosen the way it is in order to properly construct the inverse Mellin transform. $\endgroup$
    – Michael Engelhardt
    Commented Jun 11, 2023 at 23:47

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