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In the following paper and in order to prove the Barnes integral

$$\frac{1}{2\pi i}\int^{i\infty}_{-i\infty}\frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-z)^s\,ds=\frac{\Gamma(a)\Gamma(b)}{\Gamma(c)}{}_2F_1(a,b;c;z)$$

The author uses the asymptotic approximation

$$\frac{\Gamma(a+s)\Gamma(b+s)}{\Gamma(c+s)\Gamma(1+s)}\cdot \frac{\pi }{\sin(\pi s)}(-z)^s\sim \frac{\pi(-z)^s}{\sin(\pi s)}\text{exp} \left[\left(a+b-c-1 \right) \log(s)\right]$$

But I have no idea how to conclude that.

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First note that the asymptotic relation can be simplified to

$$\frac{\Gamma(a+s)\Gamma(b+s)}{\Gamma(c+s)\Gamma(1+s)} \sim \exp[(a+b-c-1)\log(s)]$$

Taking the logarithm of the left-hand side gives

$$\log\frac{\Gamma(a+s)\Gamma(b+s)}{\Gamma(c+s)\Gamma(1+s)}=\log\Gamma(a+s)+\log\Gamma(b+s)-\log\Gamma(c+s)-\log\Gamma(1+s)$$

Now, the authors use the fact that

$$\log \Gamma(s+a)=\left(s+a-\frac12\right)\log s-s+\frac12\log(2\pi)+o(1) \qquad (*)$$

Most of the terms in this expression cancel out, so that $$\log\Gamma(a+s)+\log\Gamma(b+s)-\log\Gamma(c+s)-\log\Gamma(1+s)=\log s(a+b-c-1)$$

Now, exponentiating both sides gives you the desired asymptotic relation.


$(*)$

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More on this relation can be found on Wikipedia, according to which it was derived by Gauss and Rocktaeschel.

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