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

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!

• duplication formula Jan 1, 2017 at 14:29
• You have rediscovered a particular instance of the Barnes' beta integral. Let me see if I can find a reference. Jan 1, 2017 at 18:12

Notice that it suffices to prove the following result:

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

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:

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

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 $\text{(1)}$.

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}

$$(2)$$ https://math.stackexchange.com/a/1379526

$$(3)$$ Legendre Duplication Forumula

If we then use the integral representation$$K_{0}^{2}(z) = 2 \int_{0}^{\infty} K_{0}(2z \cosh t) \, \mathrm dt \, , \quad \operatorname{Re}(z) >0, \tag{4}$$ (which I'll talk about at the end), 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}

Integral $$(4)$$ can be derived by first expressing $$K_{0}^{2}(z)$$ as the iterated integral $$\frac{1}{4} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{- z \cosh t} e^{-z \cosh u} \, \mathrm dt \, \mathrm du.$$ Then interpret the iterated integral as a double integral and make the transformation $$v = \frac{t+u}{2}$$, $$w= \frac{t-u}{2}$$ (which does not change the limits of integration).

Finally use the identity $$\cosh (v+w) + \cosh(v-w) = 2 \cosh(v) \cosh(w)$$ and then integrate.

G.N. Watson uses this approach in A Treatise on the Theory of Bessel Functions to derive the generalization of $$(4)$$ that I linked to above.

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.