Another beta integral due to Cauchy.

Question

I have the following identity which I want to prove:

$$C(x,y):= \int_{-\infty}^{\infty} \frac{dt}{(1+it)^x(1-it)^y} = \frac{\pi \cdot 2^{2-x-y}\Gamma(x+y-1)}{\Gamma(x)\Gamma(y)}$$ where $\Re(x+y)>1$. I proved so far that $C(x,y) = \frac{2^{2n} (x)_n (y)_n }{(x+y-1)_{2n}}C(x+n,y+n)$ and that: $C(x+n,y+n) = \int_{-\infty}^{\infty} \frac{dt}{(1+t^2)^n(1+it)^x(1-it)^y}$; now they ask me to set $t\to t/\sqrt{n}$ and then let $n\to \infty$, I don't see what does this integral converge to?

P.S Here I use the following notation: $(a)_n = a(a+1)(a+2)\ldots (a+n-1)$.

Thanks in advance.

Answer 1

\begin{align*} C(x+n,y+n) &= \int_{-\infty}^\infty \frac{dt}{(1+t^2)^n(1+it)^x(1-it)^y} \\ &= \int_{-\infty}^\infty \frac{du/\sqrt{n}}{(1+u^2/n)^n(1+iu/\sqrt{n})^x(1-iu/\sqrt{n})^y} \qquad (\textrm{let }u=t/\sqrt{n}) \\ &= \int_{-\infty}^\infty \frac{1}{\sqrt{n}} e^{-u^2} + O(1/n) \qquad (\textrm{standard limit for }e) \\ &= \sqrt{\frac{\pi}{n}} + O(1/n) \qquad (\textrm{standard integral}) \end{align*}

Recall that $\lim_{n\rightarrow\infty}(1+1/n)^n = e$. Similarly, $\lim_{n\rightarrow\infty}(1+x/n)^n = e^x$. Thus, we find $$\lim_{n\rightarrow\infty}\frac{1}{(1+u^2/n)^n} = e^{-u^2}.$$ Also $$\lim_{n\rightarrow\infty}\frac{1}{(1+iu/\sqrt{n})^x} = \lim_{n\rightarrow\infty}\frac{1}{(1-iu/\sqrt{n})^y}=1.$$