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Let $c>1$ be a real number, let $\alpha$ be a complex number with $0<\mathrm{Re}(\alpha)\leq 1$ and $T$ a large positive real number. Now consider the integral $$\frac{1}{2\pi i}\int_{c-iT}^{c+iT}\frac{\Gamma\left(w\right)}{\Gamma\left(w+\alpha\right)}dw.$$ Recalling the definition of Beta function $$B\left(a,b\right)=\int_{0}^{1}\left(1-t\right)^{a-1}t^{b-1}dt,\,\textrm{Re}\left(a\right)>0,\,\textrm{Re}\left(b\right)>0$$ from the inverse Mellin transform we get $$f\left(t\right)=\frac{\Gamma\left(a\right)}{2\pi i}\int_{c-i\infty}^{c+i\infty}t^{-b}\frac{\Gamma\left(b\right)}{\Gamma\left(a+b\right)}db,\,c>0,\,\textrm{Re}\left(a\right)>0$$ where $$f\left(t\right)=\begin{cases} \left(1-t\right)^{a-1} & 0<t<1\\ 0, & t\geq1. \end{cases}$$ Then $$\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\Gamma\left(w\right)}{\Gamma\left(w+\alpha\right)}dw=0$$ so it means that $$\frac{1}{2\pi i}\int_{c-iT}^{c+iT}\frac{\Gamma\left(w\right)}{\Gamma\left(w+\alpha\right)}dw\rightarrow 0$$ as $T\rightarrow \infty.$

Question: Is it possible to find a function $f(\alpha,T)$ such that $$\frac{1}{2\pi i}\int_{c-iT}^{c+iT}\frac{\Gamma\left(w\right)}{\Gamma\left(w+\alpha\right)}dw=O(f(\alpha,T))$$ and $$\lim_{T\rightarrow\infty}f\left(\alpha,T\right)=0? $$

I tried to use the residue theorem but the limitation $0<\mathrm{Re}(\alpha)\leq 1$ is annoying. Thank you.

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  • $\begingroup$ Plugging $t = 1$ in $f(t) = (1-t)^{a-1} 1_{t \in (0,1)}$ is allowed only if $\Re(a) > 1$. For $\Re(a) < 1$, $f(1) = \infty$ so you'll hardly find the bound you want. $\endgroup$ Commented Sep 16, 2017 at 21:27
  • $\begingroup$ @reuns So the inversion is not valid if $0<\mathrm{Re}\left(\alpha\right)<1$? $\endgroup$ Commented Sep 16, 2017 at 21:29
  • $\begingroup$ It is valid in the sense of distributions or $L^2$. And it is valid pointwise away from $t=1$. $\endgroup$ Commented Sep 16, 2017 at 21:31

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About the Fourier/Laplace/Mellin inversion theorem.

Let $\phi_a(x) = \int_{-a}^a e^{ikx}dk = \frac{e^{iax}-e^{-iax}}{ix} = \frac{2 \sin(ax)}{x} = a \, \Phi'(ax)$ where $\Phi(x) = \int_{-\infty}^x \frac{2\sin(y)}{y} dy, \Phi(-\infty) = 0, \Phi(+\infty) = 2\pi$. If $F,F' \in L^1$ then integrating by parts $$\int_{-\infty}^\infty \hat{F}(k) dk = \lim_{a \to \infty} \int_{-a}^a \int_{-\infty}^\infty F(x)e^{-ikx}dx dk = \lim_{a \to\infty}\int_{-\infty}^\infty \phi_a(x) F(x)dx \\= \lim_{a \to\infty}-\int_{-\infty}^\infty \Phi(ax) F'(x)dx = -\int_{-\infty}^\infty 2 \pi 1_{x > 0} F'(x)dx =2\pi F(0)$$

Here $F(x) = f(e^x)e^{\Re(b) x}, f(t) = (1-t)^{a-1}$ so $F' \not \in L^1$ for $\Re(a) \in (0,1)$ but you can look at $F \ast n e^{-\pi n^2 x^2}$, obtaining that

$$\lim_{a \to \infty}\int_{-a}^a \hat{F}(k) dk = \lim_{n \to \infty}\int_{-\infty}^\infty \hat{F}(k) e^{-\pi k^2/n^2} dk= \lim_{n \to \infty} F \ast n e^{-\pi n^2 x^2}|_{x=0} = \infty$$

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