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 $a$?

  • 2
    $\begingroup$ 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 '16 at 16:12
  • 1
    $\begingroup$ @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 '16 at 16:39

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 '16 at 16:26
  • 3
    $\begingroup$ I guess my postig is related to this. $\endgroup$ Jun 9 '16 at 16:35
  • 1
    $\begingroup$ @SangchulLee: that is exactly the missing part, thank you. $\endgroup$ Jun 9 '16 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 '16 at 16:48
  • 1
    $\begingroup$ @Semiclassical: it is not evident at first sight, but Sangchul Lee gave a convincing argument :D $\endgroup$ Jun 9 '16 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 '16 at 16:54
  • $\begingroup$ @JackD'Aurizio, nice idea, thanks $\endgroup$
    – Yuriy S
    Jun 9 '16 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 '16 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 '16 at 17:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.