Is $\frac{1}{\pi}\int_{0}^{\infty} \Gamma(\sigma +xi)^2\,\Gamma(\sigma-xi)^2 \,dx = \frac{\Gamma(2\,\sigma)^4}{\Gamma(4\,\sigma)}$?

Question

Using the approach from the answer to this question, it can be shown that for $\sigma \in \mathbb{C}, x\in \mathbb{R}$:

$$\frac{1}{\pi}\int_{0}^{\infty} \Gamma(\sigma +xi)\,\Gamma(\sigma-xi) \,dx =\frac{\Gamma(2\sigma)}{2^{2\sigma}} \qquad \Re(\sigma)>0$$

In a nutshell:

$$\Gamma(s) = \int_0^\infty \frac{x^{s-1}}{e^{x}}dx \overset{(x \, = \,e^u)}= \int_{-\infty}^\infty \frac{e^{su}}{e^{e^u}}du = \int_{-\infty}^\infty \frac{e^{-su}}{e^{e^{-u}}}du$$

$\Gamma(s)$ is the Laplace transform of $\dfrac{1}{e^{e^{-u}}}$ and $F_\sigma(\xi) = \Gamma(\sigma+2i \pi \xi)$ is the Fourier transform of $f_\sigma(u) = \dfrac{e^{-\sigma u}}{e^{e^{-u}}}$. Apply the Parseval theorem and get: $$\int_{-\infty}^\infty |F_\sigma(\xi)|^2d\xi = \int_{-\infty}^\infty |f_\sigma(u)|^2du = \int_{-\infty}^\infty \frac{e^{-2\sigma u}}{(e^{e^{-u}})^2}du $$

$$=\int_{-\infty}^\infty \frac{e^{2\sigma u}}{(e^{e^{u}})^2}du \overset{(e^u \, = \,x)}= \int_0^\infty \frac{x^{2\sigma-1}}{e^{2x}}dx = \frac{\Gamma(2\sigma)}{2^{2\sigma}}$$

For the slightly altered function, I derived from numerical testing that:

$$\frac{1}{\pi}\int_{0}^{\infty} \Gamma(\sigma +xi)^2\,\Gamma(\sigma-xi)^2 \,dx = \frac{\Gamma(2\,\sigma)^4}{\Gamma(4\,\sigma)}\ \qquad \Re(\sigma)>0$$

Could this be proven? (I tried a similar approach as above, however the squares got me stuck).

Thanks!

Answer 1

Notice that it suffices to prove the following result:

Barnes' beta integral. For $a, b, c, d \in \mathbb{H}_{\to} := \{ z \in \Bbb{C} : \Re(z) > 0 \}$, we have $$ \begin{split} &\frac{1}{2\pi} \int_{-\infty}^{\infty} \Gamma(a+it)\Gamma(b+it)\Gamma(c-it)\Gamma(d-it) \, \mathrm{d}t \\ &\hspace{12em} = \frac{\Gamma(a+c)\Gamma(a+d)\Gamma(b+c)\Gamma(b+d)}{\Gamma(a+b+c+d)}. \end{split} \tag{1} $$

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NEW ANSWER. We begin with the following generalized beta integral:

$$ \int_{[0,\infty)^k} \frac{x_1^{a_1-1} x_2^{a_2-1} \cdots x_k^{a_k-1} \, \mathrm{d}x_1\mathrm{d}x_2 \cdots \mathrm{d}x_k}{(x_1+x_2+\cdots+x_k+1)^{a_0 + a_1 + \cdots + a_k}} = \frac{\Gamma(a_0)\Gamma(a_1)\cdots\Gamma(a_k)}{\Gamma(a_0 + a_1 + \cdots + a_k)}, $$

valid for any $k \geq 1$ and $a_0, a_1, \ldots, a_k \in \mathbb{H}_{\to}$. (This can be proved by induction on $k$, starting from the beta function identity.) Then, starting from the gaussian-regularized integral,

$$ I_{\varepsilon} := \frac{1}{2\pi} \int_{-\infty}^{\infty} \Gamma(a+it)\Gamma(b+it)\Gamma(c-it)\Gamma(d-it)e^{-\frac{\varepsilon}{2}t^2} \, \mathrm{d}t, $$

we have

$$\begin{align*} I_{\varepsilon} &= \frac{\Gamma(a+b+c+d)}{2\pi} \int_{[0,\infty)^3} \int_{-\infty}^{\infty} \frac{x^{a+it-1} y^{b+it-1} z^{c-it-1}}{(x+y+z+1)^{a+b+c+d}}e^{-\frac{\varepsilon}{2}t^2} \, \mathrm{d}t \mathrm{d}x\mathrm{d}y\mathrm{d}z \\ &= \Gamma(a+b+c+d) \int_{[0,\infty)^3} \frac{x^{a-1} y^{b-1} z^{c-1}}{(x+y+z+1)^{a+b+c+d}} \delta_{\varepsilon}\left(\log\frac{z}{xy}\right) \, \mathrm{d}x\mathrm{d}y\mathrm{d}z, \end{align*}$$

where $\delta_{\varepsilon}(x) := \frac{1}{\sqrt{2\pi\varepsilon}}e^{-x^2/2\varepsilon}$ and we invoked the gaussian integral to compute the integral w.r.t. $t$. Substituting $u = \log\frac{z}{xy}$ for the integral w.r.t. $z$, we have $\mathrm{d}z = xy e^u \, \mathrm{d}u$ and

$$\begin{align*} I_{\varepsilon} &= \Gamma(a+b+c+d) \int_{[0,\infty)^2}\int_{-\infty}^{\infty} \frac{x^{a+c-1} y^{b+c-1} e^{cu}}{(x+y+xy e^u+1)^{a+b+c+d}} \delta_{\varepsilon}(u) \, \mathrm{d}u\mathrm{d}x\mathrm{d}y, \end{align*}$$

We can check that $\delta_{\varepsilon}(u)$ indeed behaves like the Dirac delta as $\varepsilon \to 0^+$, yielding:

$$\begin{align*} I_0 &= \Gamma(a+b+c+d) \int_{[0,\infty)^2} \left[ \frac{x^{a+c-1} y^{b+c-1} e^{cu}}{(x+y+xy e^u+1)^{a+b+c+d}} \right]_{u=0} \, \mathrm{d}x\mathrm{d}y \\ &= \Gamma(a+b+c+d) \int_{[0,\infty)^2} \frac{x^{a+c-1}}{(x+1)^{a+b+c+d}} \frac{y^{b+c-1}}{(1+y)^{a+b+c+d}} \, \mathrm{d}x\mathrm{d}y \end{align*}$$

Using the beta integral, this reduces to the RHS of $(1)$ as desired. $\blacksquare$

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OLD ANSWER. In view of the principle of analytic continuation, it suffices to consider the case when $a > b > 0$ and $c > d > 0$. If we denote by $B(a,b,c,d)$ the LHS of $\text{(1)}$, then it is easy to check that $B(a,b,c,d) \in \Bbb{R}$:

$$ \overline{B(a,b,c,d)} = \frac{1}{2\pi} \int_{-\infty}^{\infty} \Gamma(a-it)\Gamma(b-it)\Gamma(c+it)\Gamma(d+it) \, \mathrm{d}t = B(a,b,c,d) $$

by applying the substitution $t \mapsto -t$. Now in order to compute $B(a,b,c,d)$, we utilize the following formula formula for the beta function:

$$ \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} = \int_{0}^{\infty} \frac{u^{a-1}}{(1+u)^{a+b}} \, \mathrm{d}u. $$

Then we have

$$ B(a,b,c,d) = \frac{1}{2\pi} \Gamma(a+c)\Gamma(b+d)\int_{-\infty}^{\infty} \mathrm{d}t \int_{0}^{\infty} \mathrm{d}x \int_{0}^{\infty} \mathrm{d}y \, \frac{x^{a-1+it}}{(1+x)^{a+c}} \frac{y^{b-1+it}}{(1+y)^{b+d}}. \tag{2} $$

In order to compute this nasty triple integral, we want to interchange the order of integral. Even if we pretend that the Fubini's theorem works, it turns out that the resulting integral fails to be improperly integrable. So we introduce a small trick using the following theorem:

Theorem. (Abel). If $f(x)e^{-\epsilon x} \in L^1([0,\infty))$ for all $\epsilon > 0$ and $I := \lim_{R\to\infty} \int_{0}^{R} f(x) \, \mathrm{d}x$ exists, then we have the following convergence: $$ \lim_{\epsilon \to 0^+} \int_{0}^{\infty} f(x) e^{-\epsilon x} \, dx = I. $$

Using this, the triple integral in the RHS of $\text{(2)}$ can be computed as

\begin{align*} &\int_{-\infty}^{\infty} \mathrm{d}t \int_{0}^{\infty} \mathrm{d}x \int_{0}^{\infty} \mathrm{d}y \, \frac{x^{a-1+it}}{(1+x)^{a+c}} \frac{y^{b-1+it}}{(1+y)^{b+d}} \\ &\hspace{3em} \stackrel{\text{(3)}}{=} \int_{-\infty}^{\infty} \mathrm{d}t \int_{0}^{\infty} \mathrm{d}x \int_{0}^{\infty} \mathrm{d}y \, \frac{x^{a-1}}{(1+x)^{a+c}} \frac{y^{b-1}}{(1+y)^{b+d}} \, \cos(t\log(xy)) \\ &\hspace{3em} \stackrel{\text{(4)}}{=} \lim_{\epsilon \to 0^+} \int_{-\infty}^{\infty} \mathrm{d}t \int_{0}^{\infty} \mathrm{d}x \int_{0}^{\infty} \mathrm{d}y \, \frac{x^{a-1}}{(1+x)^{a+c}} \frac{y^{b-1}}{(1+y)^{b+d}} \, \cos(t\log(xy)) e^{-\epsilon |t|} \\ &\hspace{3em} \stackrel{\text{(5)}}{=} \lim_{\epsilon \to 0^+} \int_{0}^{\infty} \mathrm{d}x \int_{0}^{\infty} \mathrm{d}y \, \frac{x^{a-1}}{(1+x)^{a+c}} \frac{y^{b-1}}{(1+y)^{b+d}} \frac{2\epsilon}{\epsilon^2 + \log^2(xy)}. \end{align*}

Here are some explanation to each step:

• (3) is obtained by taking the real part of the integrand, since we know that the the integral is real.
• (4) is an application of the theorem above.
• (5) follows from the Fubini's theorem and the Laplace transform of the cosine function.

Now we apply the substitution $\log(xy) = \epsilon u$ to the inner integral. Then

\begin{align*} &\int_{-\infty}^{\infty} \mathrm{d}t \int_{0}^{\infty} \mathrm{d}x \int_{0}^{\infty} \mathrm{d}y \, \frac{x^{a-1+it}}{(1+x)^{a+c}} \frac{y^{b-1+it}}{(1+y)^{b+d}} \\ &\hspace{3em} \stackrel{\text{(6)}}{=} \lim_{\epsilon \to 0^+} \int_{0}^{\infty} \mathrm{d}x \int_{-\infty}^{\infty} \mathrm{d}u \, \frac{x^{a+d-1}}{(1+x)^{a+c}} \frac{e^{\epsilon b u}}{(e^{\epsilon u} + x)^{b+d}} \frac{2}{1+u^2}. \end{align*}

Now it is easy to check that

$$\frac{e^{\epsilon b u}}{(e^{\epsilon u} + x)^{b+d}} \leq 1+ x^{-b-d}.$$

Thus by the assumptions on the magnitude of $a, b, c, d$, the integrand of the RHS of $\text{(6)}$ is dominated by an integrable function. Then by the dominated convergence theorem, we can put the limit inside the integral and we have

\begin{align*} &\int_{-\infty}^{\infty} \mathrm{d}t \int_{0}^{\infty} \mathrm{d}x \int_{0}^{\infty} \mathrm{d}y \, \frac{x^{a-1+it}}{(1+x)^{a+c}} \frac{y^{b-1+it}}{(1+y)^{b+d}} \\ &\hspace{3em} = \int_{0}^{\infty} \mathrm{d}x \int_{-\infty}^{\infty} \mathrm{d}u \, \frac{x^{a+d-1}}{(1+x)^{a+b+c+d}} \frac{2}{1+u^2} \\ &\hspace{3em} = 2\pi \frac{\Gamma(a+d)\Gamma(b+c)}{\Gamma(a+b+c+d)}. \end{align*}

This completes the proof of $\text{(1)}$. $\blacksquare$

Answer 2

UPDATE:

I just realized $5$ years later that Barnes' beta integral $$ \begin{split} & \int_{-\infty}^{\infty} \Gamma(a+it)\Gamma(b+it)\Gamma(c-it)\Gamma(d-it) \, \mathrm d t \\ &\hspace{12em} = 2 \pi \, \frac{\Gamma(a+c)\Gamma(a+d)\Gamma(b+c)\Gamma(b+d)}{\Gamma(a+b+c+d)} \end{split} $$ is just a particular case of the generalized Parseval/Plancherel theorem $$\int_{-\infty}^{\infty} F(it) \overline{G(it)} \, \mathrm dt = 2 \pi \int_{0}^{\infty} \frac{f(x) \overline{g(x)}}{x} \, \mathrm dx \, ,$$ where $F(s)$ is the Mellin transform of $f(x)$ and $G(s)$ is the Mellin transform of $g(x)$.

I added a proof at the end.

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If you're still interested in evaluating the integral using Parseval's/Plancherel's theorem for the Mellin transform, we just have to show that $$2\int_{0}^{\infty} x^{s-1} K_{0}(2 \sqrt{x}) \, \mathrm dx = \Gamma^{2}(s) \; , \quad \Re(s) > 0 \, , \tag{1}$$

where $K_{0}(z)$ is the modified Bessel function of the second kind of order zero.

Then for $\sigma >0$,

$$ \int_{-\infty}^{\infty} \Gamma^{2} (\sigma +it) \Gamma^{2} (\sigma -it) \, \mathrm dt = \int_{-\infty}^{\infty} \lvert \Gamma^{2}(\sigma +it) \rvert ^{2} \, \mathrm dt = 8 \pi\int_{0}^{\infty} x^{2\sigma -1} \, K_{0}^{2}(2 \sqrt{x}) \, \mathrm dx .$$

To show $(1)$, we can use the integral representation $$K_{0}(z) = \int_{0}^{\infty} e^{-z \cosh t} \, \mathrm dt \, , \quad \operatorname{Re}(z) >0 $$

and then switch the order of integration (which is justified since the integrand is nonnegative).

$$ \begin{align} 2\int_{0}^{\infty} x^{s-1} K_{0}(2 \sqrt{x}) \, \mathrm dx &= 4 \int_{0}^{\infty} u^{2s-1} K_{0} (2u) \, \mathrm du \\ &= 4 \int_{0}^{\infty} u^{2s-1} \int_{0}^{\infty} e^{-2 u \cosh t} \, \mathrm dt \,\mathrm du \\ &=4 \int_{0}^{\infty} \int_{0}^{\infty} u^{2s-1} e^{-2u \cosh t} \, \mathrm du \, \mathrm dt \\ &=\frac{4 \, \Gamma(2s)}{2^{2s}}\int_{0}^{\infty} (\cosh t)^{-2s} \, \, \mathrm dt \\ &= \frac{2 \, \Gamma(2s)}{2^{2s-1}} \frac{\sqrt{\pi}}{2} \frac{\Gamma(s)}{\Gamma(s+ \frac{1}{2})} \tag{2} \\ &= \Gamma^{2}(s) \tag{3} \end{align}$$

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$(2)$ Closed form of $\int_0^\infty \frac{1}{\left(a+\cosh x\right)^{1/n}} \, dx$ for $a=0,1$

$(3)$ Legendre Duplication Formula

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If we then use the integral representation$$K_{0}^{2}(z) = 2 \int_{0}^{\infty} K_{0}(2z \cosh t) \, \mathrm dt \, , \quad \operatorname{Re}(z) >0, \tag{4}$$ we get

$$\begin{align}\int_{-\infty}^{\infty} \Gamma^{2} (\sigma +it) \Gamma^{2} (\sigma -it) \, \mathrm dt &=8 \pi\int_{0}^{\infty} x^{2\sigma -1} \, K_{0}^{2}(2 \sqrt{x}) \, \mathrm dx \\ &= 16 \pi\int_{0}^{\infty} u^{4\sigma -1} \, K_{0}^{2}(2 u) \, \mathrm du \\ & =32 \pi \int_{0}^{\infty} u^{4 \sigma-1} \int_{0}^{\infty} K_{0}(4u \cosh v) \, \mathrm dv \, \mathrm du \\ &= 32 \pi \int_{0}^{\infty} u^{4 \sigma -1} \int_{0}^{\infty} \int_{0}^{\infty} e^{-4u \cosh v \cosh w} \, \mathrm dw \, \mathrm dv \, \mathrm du \\ &= 32 \pi \int_{0}^{\infty} \int_{0}^{\infty} \int_{0}^{\infty} u ^{4 \sigma-1} e^{-4u \cosh v \cosh w} \, \mathrm du \, \mathrm dw \, \mathrm dv \\ &= 32 \pi \int_{0}^{\infty} \int_{0}^{\infty} (4 \cosh v \cosh w)^{-4 \sigma} \, \Gamma(4 \sigma) \, \mathrm dw \, \mathrm dv \\ &= \frac{32 \pi \, \Gamma(4 \sigma)}{2^{8 \sigma}} \left(\int_{0}^{\infty} (\cosh v)^{- 4 \sigma} \, \mathrm dv \right)^{2} \\ &= \frac{32 \pi \, \Gamma(4 \sigma)}{2^{8 \sigma}} \frac{\pi}{4} \frac{\Gamma^{2}(2 \sigma)}{\Gamma^{2}(2 \sigma + \frac{1}{2})} \\ &= \frac{32 \pi \, \Gamma(4 \sigma)}{2^{8 \sigma}} \frac{\pi}{4} \, \Gamma^{2}(2 \sigma) \, \frac{2^{8 \sigma-2} \, \Gamma^{2}(2 \sigma)}{\pi \, \Gamma^{2}(4 \sigma)} \\ &= \frac{2 \pi \, \Gamma^{4}(2 \sigma)}{\Gamma(4 \sigma)}. \end{align}$$

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Proof of Barnes' beta integral using Parseval's theorem:

Assume that the parameters $a$, $b$, $c$, and $d$ are all greater than zero.

Let $F(s) = \Gamma(a+s)\Gamma(b+s)$ and $G(s) = \Gamma(c+s) \Gamma(d+s)$.

For $\Re(s \pm \frac{v}{2}) >0$, the Mellin transform of $2 K_{\nu }(2 \sqrt{x})$ is $$ 2 \int_{0}^{\infty} t^{s-1} K_{\nu}(2 \sqrt{x}) \,\mathrm dx = 2^{2(1-s)} \int_{0}^{\infty} u^{2s-1} K_{ \nu}(u) \, \mathrm du = \Gamma \left(\frac{\nu}{2}+s \right) \Gamma \left(-\frac{\nu}{2}+s\right).$$

See this answer.

Therefore, the Mellin transform of $2x^{(a+b)/2}K_{a-b}(2 \sqrt{x})$ is $\Gamma(a+s) \Gamma(b+s)$, and the Mellin transform of $2x^{(c+d)/2}K_{c-d}(2 \sqrt{x})$ is $\Gamma(c+s) \Gamma(d+s)$.

And from this answer we have that the Mellin transform of the product $K_{\nu}(x) K_{ \mu}(x) $ is $$\small \int_{0}^{\infty} x^{s-1} K_{\nu}(x) K_{\mu}(x) \, dx = \frac{2^{s-3}}{\Gamma(s)} \, \Gamma \left(\frac{s+ \nu + \mu}{2} \right) \Gamma \left(\frac{s+\nu - \mu}{2} \right) \Gamma \left(\frac{s- \nu + \mu}{2} \right) \Gamma \left(\frac{s- \nu - \mu}{2}\right), $$ which holds for $ \Re(s \pm \nu \pm \mu) >0$.

Therefore, since $\overline{\Gamma(z)} = \Gamma(\overline{z})$, Parseval's theorem says

$$ \begin{align} &\int_{-\infty}^{\infty} \Gamma(a+it)\Gamma(b+it)\Gamma(c-it)\Gamma(d-it) \, \mathrm d t \\& = 8 \pi \int_{0}^{\infty} x^{(a+b+c+d)/2-1}K_{a-b}(2 \sqrt{x}) K_{c-d}(2 \sqrt{x}) \, \mathrm dx \\& = \frac{2 \pi }{2^{a+b+c+d-3}} \int_{0}^{\infty} u^{a+b+c+d-1} K_{a-b}(u) K_{c-d}(u) \, \mathrm du \\ &= \frac{2 \pi }{2^{a+b+c+d-3}} \frac{2^{a+b+c+d-3}}{\Gamma(a+b+c+d)} \, \Gamma \left(\frac{(a+b+c+d)+(a-b)+(c-d)}{2} \right) \\ &\times \Gamma \left(\frac{(a+b+c+d)+(a-b)-(c-d)}{2} \right) \Gamma \left(\frac{(a+b+c+d)-(a-b)+(c-d)}{2} \right) \\& \times \Gamma \left( \frac{(a+b+c+d)-(a-b)-(c-d)}{2}\right) \\ &= 2 \pi \, \frac{\Gamma(a+c) \Gamma(a+d) \Gamma(b+c) \Gamma(b+d)}{\Gamma(a+b+c+d)}. \end{align}$$

Answer 3

We have that $\Gamma(\sigma+ix)\,\Gamma(\sigma-ix)$ is an even function and since $$\text{(Weierstrass product)}\qquad \Gamma(z)=e^{-\gamma z}\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{z/n} \tag{1}$$ we also have $$ \Gamma(\sigma+ix)\,\Gamma(\sigma-ix) = e^{-2\gamma \sigma}\prod_{n\geq 1}\frac{\exp\left(\frac{2\sigma}{n}\right)}{\left(1+\frac{\sigma+ix}{n}\right)\left(1+\frac{\sigma-ix}{n}\right)}\tag{2}$$ and we may compute such integrals through the residue theorem and Legendre's duplication formula.