# Non-normalized sinc function

In looking for an efficient way to show the relation

$$\int_{-\infty}^\infty \frac{e^{inx}}{\Gamma(\alpha+x)\Gamma(\beta-x)} \, dx = \frac{(2 \cos n/2)^{\alpha+\beta-2}}{\Gamma(\alpha+\beta-1)} e^{ \frac{in(\beta-\alpha)}{2}} \text{ or }~~ 0$$

acc. as $|n|< \pi$ or $|n|>\pi,$ respectively, and $\operatorname{Re}(\alpha+\beta)> 1,$ it occurred to me that this looks to be the Fourier transform (or inverse transform ) of a sinc function, and then I noticed in a Wiki article that

$$\frac{\sin(\pi x)}{ \pi x} = \frac{1}{\Gamma(1+x)\Gamma(1-x)}.$$

I am familiar with the basic properties like linearity and scaling for transform pairs and have found some obvious points but seem far from finding a simple way to modify the normalized sinc function above to give the desired rectangular function.

Can someone tell me what the correct form of the sinc function should be?

Thanks for any help.

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Well, I have a little different result, but still hope this will help. I think the application of beta-function will be fruitful enough.
Just remember, that $$\mathrm{B}(\alpha+x,\beta-x)=\frac{\Gamma(\alpha+x)\Gamma(\beta-x)}{\Gamma(\alpha+\beta)}$$ $$\frac{1}{\Gamma(\alpha+x)\Gamma(\beta-x)}=\frac{1}{\Gamma(\alpha+\beta)}\frac{1}{\mathrm{B}(\alpha+x,\beta-x)}$$ So: $$\int_{-\infty}^\infty \frac{e^{inx}}{\Gamma(\alpha+x)\Gamma(\beta-x)} \, dx=\frac{1}{\Gamma(\alpha+\beta)}\int_{-\infty}^\infty \frac{e^{inx}}{\mathrm{B}(\alpha+x,\beta-x)} \, dx$$ Then one can use the definition of the beta-function (5.12.6): $$\int _{0}^{\pi}(\mathop{\sin}\nolimits t)^{{a-1}}e^{{ibt}}dt=\frac{\pi}{2^{{a-1}}}\frac{e^{{i\pi b/2}}}{a\mathop{\mathrm{B}}\nolimits\!\left(\frac{1}{2}(a+b+1),\frac{1}{2}(a-b+1)\right)}$$ Setting $a=\alpha+\beta-1, \ b=\alpha-\beta+2x$, after simplifications one will obtain: $$\frac{1}{\mathrm{B}(\alpha+x,\beta-x)}=\frac{2^{\alpha+\beta-2}e^{i\frac{\pi}{2}(\beta-\alpha)}}{\pi}(\alpha+\beta-1)\int _{0}^{\pi}(\mathop{\sin}\nolimits t)^{{\alpha+\beta-2}}e^{{i(\alpha-\beta+2x)t}}dt,$$ which is true, when $\mathrm{Re}(a)>0$, or $\mathrm{Re}(\alpha+\beta)>1$.
So the initial integral will look like (here you already can see the terms as in the answer ;)): $$\int_{-\infty}^\infty \!\frac{e^{inx}}{\Gamma(\alpha+x)\Gamma(\beta-x)} \!\,\! dx= \frac{2^{\alpha+\beta-2}e^{i\frac{\pi}{2}(\beta-\alpha)}}{\pi}\frac{(\alpha+\beta-1)}{\Gamma(\alpha+\beta)} \int_{-\infty}^\infty \!e^{inx} \!\!\int _{0}^{\pi}(\mathop{\sin}\nolimits t)^{{\alpha+\beta-2}}\!e^{{i(\alpha-\beta+2x)t}}\!dt \ dx$$

$$\int_{-\infty}^\infty \!\frac{e^{inx}}{\Gamma(\alpha+x)\Gamma(\beta-x)} \!\,\! dx= \frac{2^{\alpha+\beta-2}e^{i\frac{\pi}{2}(\beta-\alpha)}}{\pi \ \Gamma(\alpha+\beta-1)} \int _{0}^{\pi} \!(\mathop{\sin}\nolimits t)^{{\alpha+\beta-2}} e^{{i(\alpha-\beta)t}} \int_{-\infty}^\infty \!e^{i(n+2t)x} \!dx \ dt$$ The inside integral is the definition of the Dirac delta function: $$\int_{-\infty}^\infty \!e^{i(n+2t)x} \!dx= 2\pi \delta(2t+n)=\pi \delta\bigg(t+\frac{n}{2}\bigg)$$ And if $0\leq -\frac{n}{2}\leq \pi$: $$\int _{0}^{\pi} \!(\mathop{\sin}\nolimits t)^{{\alpha+\beta-2}} e^{{i(\alpha-\beta)t}} \delta\bigg(t+\frac{n}{2}\bigg) dt=\theta (-n) e^{-\frac{1}{2} i n (\alpha -\beta )} \left(-\sin\bigg(\frac{n}{2}\bigg)\right)^{\alpha +\beta -2}$$ where is $\theta (-n)$ the Heaviside theta function.
So finally the result is: $$\int_{-\infty}^\infty \frac{e^{inx}}{\Gamma(\alpha+x)\Gamma(\beta-x)} \, dx= \frac{2^{\alpha+\beta-2}e^{i\frac{\pi}{2}(\beta-\alpha)}}{ \Gamma(\alpha+\beta-1)}\theta (-n) e^{-\frac{1}{2} i n (\alpha -\beta )} \left(-\sin\bigg(\frac{n}{2}\bigg)\right)^{\alpha +\beta -2}$$ Or after some simplification: $$\int_{-\infty}^\infty \frac{e^{inx}}{\Gamma(\alpha+x)\Gamma(\beta-x)} \, dx= \frac{(2 \sin(\frac{n}{2}))^{\alpha+\beta-2}}{ \Gamma(\alpha+\beta-1)}e^{i\frac{n}{2}(\beta-\alpha)}\theta (-n)$$

OR

One can use the definition of the beta-function (5.12.5): $$\int _{0}^{\frac{\pi}{2}}(\mathop{\cos}\nolimits t)^{a-1}\cos(bt)dt=\frac{\pi}{2^{{a}}}\frac{1}{a\mathop{\mathrm{B}}\nolimits\!\left(\frac{1}{2}(a+b+1),\frac{1}{2}(a-b+1)\right)}$$ using the same substitutions for $a$ and $b$ and expanding $cos(bt)$ in terms of complex exponents one will get:

$$\int_{-\infty}^\infty \!\!\frac{e^{inx}}{\Gamma(\!\alpha+x)\!\Gamma(\beta-x)} \!\,\! dx= \frac{2^{\alpha+\beta-2}e^{i\frac{\pi}{2}(\beta-\alpha)}}{\pi}\frac{(\alpha+\beta-1)}{\Gamma(\alpha+\beta)} \int_{-\infty}^\infty \!e^{inx} \!\!\int _{0}^{\frac{\pi}{2}}(\mathop{\cos}\nolimits t)^{{\alpha+\beta-2}}\!e^{{i(\alpha-\beta+2x)t}}\!dt \ dx \! + \\ +\frac{2^{\alpha+\beta-2}e^{i\frac{\pi}{2}(\beta-\alpha)}}{\pi}\frac{(\alpha+\beta-1)}{\Gamma(\alpha+\beta)} \int_{-\infty}^\infty \!e^{inx} \!\!\int _{0}^{\frac{\pi}{2}}(\mathop{\cos}\nolimits t)^{{\alpha+\beta-2}}\!e^{{-i(\alpha-\beta+2x)t}}\!dt \ dx$$ Using precisely the same reasoning one will come up with the answer: $$\int_{-\infty}^\infty \frac{e^{inx}}{\Gamma(\alpha+x)\Gamma(\beta-x)} \, dx= \frac{(2 \cos(\frac{n}{2}))^{\alpha+\beta-2}}{ \Gamma(\alpha+\beta-1)}e^{i\frac{n}{2}(\beta-\alpha)}(\theta (n,\pi -n)+\theta (-n,n+\pi ))$$ and $$\theta (-n,n+\pi )+\theta (n,\pi -n)= \begin{cases} 1, & -\pi <n<\pi \\ 0, & \mathrm{otherwise} \end{cases}$$

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@daniel First, you should prove that the result is correct by showing the reference, where it is taken from, or by showing you own effort. I have shown my solution and the obtained result, so one can easily track whether there is a fault or not. Second, you start with In looking for an efficient way to show the relation and end it occurred to me that this looks to be the Fourier transform (or inverse transform ) of a sinc function, and then I noticed in a Wiki article that, don't you see that there can be a more efficient way to solve the problem? –  Caran-d'Ache Apr 15 '13 at 4:06
@daniel When I was answering your question was Does the approach work and if so can someone tell me what the correct form of the sinc function should be? I personally think (I may be wrong but by now there is no other opinion) that this approach will not work and showed the approach that gave the answer. I edited the answer, adding one more transrormation and now the answer is exactly the same. The derivations in both cases are similar and correct, but the answers are different, and for now I see no reasons to make favour for one or another. –  Caran-d'Ache Apr 15 '13 at 10:46
@daniel Speaking why I think the approach with sinc functions will not work I can comment, that the equality that you found will help only if $\alpha=\beta$. It is because $\frac{\sin(\pi (x-\gamma))}{ \pi (x-\gamma)} = \frac{1}{\Gamma(1+(x-\gamma))\Gamma(1-(x-\gamma))}$, so scaling sinc you scale both gamma functions simultaneously. In this case everything gets a little easier, but you didn't set them equal, so my opinion that this approach will not work. –  Caran-d'Ache Apr 15 '13 at 10:55
I had hoped there was a faster way but now that I have worked through your proof I doubt it--anything that works might be about this involved. This is a really nice answer and I hope you get more votes for it. –  daniel Apr 18 '13 at 1:11
@daniel Thank you. Indeed it is a very interesting problem. –  Caran-d'Ache Apr 18 '13 at 16:23