I would like to evaluate this integral: $$\mathcal F(a)=\int_{-\infty}^\infty\frac{\Gamma(x)\,\sin(\pi x)}{\Gamma\left(x+a\right)}\,dx,\quad a>0.\tag1$$ For all $a>0$ the integrand is a smooth oscillating function decaying for $x\to\pm\infty$. The poles of the gamma function in the numerator are cancelled by the sine factor.

For $a\in\mathbb N$, the ratio of the gamma functions simplifies to a polynomial in the denominator, and in each case the integral can be pretty easily evaluated in a closed form, e.g. $$\mathcal F(3)=\int_{-\infty}^\infty\frac{\sin(\pi x)}{x\,(x+1)\,(x+2)}\,dx=2\pi.\tag2$$ Can we find a general formula for $\mathcal F(a)$ valid both for integer and non-integers positive values of $a$?

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    $\begingroup$ Interesting question. By using the reflection formula, your integral can be written as: $$ I(a) = \pi \int_{-\infty}^{+\infty}\frac{dx}{\Gamma(x+a)\Gamma(1-x)}\tag{1}$$ $\endgroup$ Jun 9, 2016 at 16:12
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    $\begingroup$ @JackD'Aurizio, which would be then equal to: $$I(a) = \frac{\pi}{\Gamma(a+1)} \int_{-\infty}^{+\infty}\frac{dx}{B(x+a,1-x)}\tag{2}$$, right? $\endgroup$
    – Yuriy S
    Jun 9, 2016 at 16:39

3 Answers 3


Interesting question.
By using the reflection formula, your integral can be written as a convolution integral: $$ I(a) = \pi \int_{-\infty}^{+\infty}\frac{dx}{\Gamma(x+a)\Gamma(1-x)}.\tag{1}$$ We may also notice that when $n\in\mathbb{N}$ we have: $$ \int_{-\infty}^{+\infty}\frac{\sin(\pi x)\,dx}{x(x+1)\cdot\ldots\cdot(x+n)}=\pi\sum_{k=0}^{n}(-1)^k \text{Res}\left(\frac{1}{x(x+1)\cdot\ldots\cdot(x+n)},x=-k\right)\tag{2}$$ where the $RHS$ of $(2)$ equals: $$ \frac{\pi}{n!}\sum_{k=0}^{n}\binom{n}{k} = \color{red}{\frac{\pi\, 2^n}{n!}}\tag{3} $$ so the most reasonable conjecture is:

$$ I(a) = \pi \int_{-\infty}^{+\infty}\frac{dx}{\Gamma(x+a)\Gamma(1-x)}\stackrel{!}{=}\color{red}{\frac{\pi\, 2^{a-1}}{\Gamma(a)}}.\tag{4}$$

Numerical simulations support $(4)$. Probably it is enough to exploit the log-convexity of the $\Gamma$ function to extend the validity of $(4)$ from $a\in\mathbb{N}$ to $a\in\mathbb{R}^+$.

Update: As a matter of fact, this Sangchul Lee's blogpost proves that the partial fraction decomposition through the residue theorem still works for $a\in\mathbb{R}^+\setminus\mathbb{N}$, setting $(4)$ as an identity. Credits to him for another great piece of cooperative math.

Following Yuriy S' approach, $$ \int_{-\infty}^{+\infty}B(a,x)\sin(\pi x)\,dx \\=\int_{0}^{+\infty}\int_{0}^{1}\left[(1-t)^{a-1}t^{x-1}\sin(\pi x)+t^{x-a}(1-t)^{a-1}\sin(\pi(a-x))\right]\,dt\,dx$$ leads to an integral that can be easily evaluated through the residue theorem, giving an alterntive proof of $(4)$.

  • $\begingroup$ Interestingly, if we return to the original expression for $I(a)$ and multiply through by $\Gamma(a)$ we can write (4) in terms of the beta function as $$\int_{-\infty}^\infty \text{B}(x,a)\sin(\pi x)\,dx=\pi 2^{a-1}.$$ This seems rather cute, if not particularly revealing. $\endgroup$ Jun 9, 2016 at 16:26
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    $\begingroup$ I guess my postig is related to this. $\endgroup$ Jun 9, 2016 at 16:35
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    $\begingroup$ @SangchulLee: that is exactly the missing part, thank you. $\endgroup$ Jun 9, 2016 at 16:37
  • $\begingroup$ Following upon my previous comment, is it evident how one would calculate $$\int_{-\infty}^\infty \text{B}(x,a)e^{i \pi x}\,dx?$$ The imaginary part is evidently $\pi 2^{a-1}$. $\endgroup$ Jun 9, 2016 at 16:48
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    $\begingroup$ @Semiclassical: it is not evident at first sight, but Sangchul Lee gave a convincing argument :D $\endgroup$ Jun 9, 2016 at 16:51


$$g(x) = \frac{\Gamma(x)\,\exp(i\pi x)}{\Gamma\left(x+a\right)},\quad a>0$$


$$\mathcal G(a)=\int_{-\infty}^\infty g(x)\,dx.$$

The residues of $g$ are along the real line at $z\in-\mathbb N$:

\begin{align} r_n&=\operatorname*{Res}_{z = -n}g(z) \\ &= \lim_{z\to-n} (z+n) \frac{\Gamma(z+n+1)}{(z+n)(z+n-1)\cdots(z+1)} \frac{e^{i\pi z}}{\Gamma\left(z+a\right)}\\ &= \frac{e^{-i\pi n}}{\Gamma(a-n)} \frac{(-1)^n}{n!} \\ &= \frac{1}{n! \Gamma(a-n)} \end{align}

Then, considering a semicircular contour along $\mathbb R$ and $R e^{i\theta}$, indented around each pole, we have

$$\mathcal G(a) = \pi i\sum_{n=0}^\infty r_n = \pi i \frac{1}{\Gamma(a)}\sum_{n=0}^\infty {a-1\choose n} = \pi i \frac{2^{a-1}}{\Gamma(a)}.$$

It is clear that the outer part of the contour makes no contribution since for large $|z|$, $|g(z)| \le \frac{R e^{-\pi y}}{|P_{\lfloor a \rfloor}(z)|}$.

Now take the complex part to conclude

$$\boxed{\mathcal F(a) = \pi\frac{2^{a-1}}{\Gamma(a)}}$$

which is valid for all $a>0$.


To me the most obvious action would be:

$$\int_{-\infty}^\infty\frac{\Gamma(x)\,\sin(\pi x)}{\Gamma\left(x+a\right)} dx=\frac{1}{\Gamma(a)}\int_{-\infty}^\infty B(x,a)\,\sin(\pi x) dx$$

$$\int_{-\infty}^\infty B(x,a)\,\sin(\pi x) dx=\int_{-\infty}^\infty \int_0^1 t^{x-1}(1-t)^{a-1} \sin(\pi x) ~dt~ dx$$

However the integral for $x$ diverges, we can only find the part for $x>0$ (because $\ln t<0$):

$$\int_{0}^\infty t^{x} \sin(\pi x) ~dx=\int_{0}^\infty e^{(\ln t)x} \sin(\pi x) ~dx= \frac{\pi}{\pi^2+\ln^2 t}$$

So we have for the part with $x>0$:

$$\int_{0}^\infty\frac{\Gamma(x)\,\sin(\pi x)}{\Gamma\left(x+a\right)} dx=\frac{\pi}{\Gamma(a)} \int_0^1 \frac{(1-t)^{a-1}}{t(\pi^2+\ln^2 t)} ~dt$$

We can make an obvious substitution:

$$ \int_0^1 \frac{(1-t)^{a-1}}{t(\pi^2+\ln^2 t)} ~dt=\int_0^\infty \frac{(1-e^{-u})^{a-1}}{\pi^2+u^2} du$$

Or we can use the integral representation of the Gregory coefficients somehow (thanks, Jack D'Aurizio!):

$$\displaystyle G_n=(-1)^{n-1}\!\int\limits_0^{\infty}\frac{dx}{(1+x)^n\left(\ln^2 x + \pi^2\right)}$$

If this leads anywhere, I'll edit my post.

  • $\begingroup$ You may use the integral representation for Gregory coefficients: en.wikipedia.org/wiki/Gregory_coefficients $\endgroup$ Jun 9, 2016 at 16:54
  • $\begingroup$ @JackD'Aurizio, nice idea, thanks $\endgroup$
    – Yuriy S
    Jun 9, 2016 at 17:00
  • $\begingroup$ Or, you may simply put together the integral over x>0 and the integral over x<0 to get an integral straightforward to compute through the residue theorem. $\endgroup$ Jun 9, 2016 at 17:29
  • $\begingroup$ @JackD'Aurizio, I'm bad at residues, still didn't take the time to brush up on my complex analysis $\endgroup$
    – Yuriy S
    Jun 9, 2016 at 17:30

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