Evaluate
$$\int_{-\infty}^{\infty}|\Gamma(r+xi)|dx$$
as a function of $r$, where $r\notin\{0,-1,-2,...\}$ and $x\in\mathbb{R}$.
For some values of $r$, these visually look like bell curves (similar to $e^{-x^2}$). See this picture for reference.
But I don't believe they are exactly the same. I'm not sure. How would you evaluate the integral as a function of $r$? Can it be written in a way that is more familiar?
On
Partial solution
We can get a closed form for integral $$J(r)=\int_{-\infty }^{\infty } |\Gamma (r+i x)|^2 \, dx\,, \,\,r>0$$ $$J(r)=\int_{-\infty }^{\infty } \Gamma (r+i x) \Gamma (r-i x)\, dx=\int_{-\infty }^{\infty }dx\int_0^{\infty}\int_0^{\infty}du dv \,e^{-u-v}(uv)^{r-1}\Bigl(\frac{u}{v}\Bigr)^{ix}$$ $$=\int_{-\infty }^{\infty }dx\int_0^{\infty}\int_0^{\infty}du dv \,e^{-(u+v)}(uv)^{r-1}e^{ix\log\frac{u}{v}}$$ Making change $u=st$ and $v=s(1-t)$ and evaluating Jacobian $=\frac{\partial (u,v)}{\partial(t,s)}=s\,$ we get $$J(r)=\int_{-\infty }^{\infty }dx\int_0^{\infty}ds\int_0^{1}dt\,e^{-s}s^{2r-1}\bigl(t(1-t)\bigr)^{r-1}e^{ix\log{\frac{t}{1-t}}}$$ $$=\Gamma(2r)\int_0^{1}dt\bigl(t(1-t)\bigr)^{r-1}\int_{-\infty }^{\infty }e^{ix\log{\frac{t}{1-t}}}dx$$ Integral over $x$ gives delta-function for $t\to\frac{1}{2}+\epsilon$ $$\log{\frac{t}{1-t}}\to\log\frac{1/2+\epsilon}{1/2-\epsilon}\to 4\,\epsilon$$ $$J(r)=\Gamma(2r)\int_0^{1}dt\bigl(t(1-t)\bigr)^{r-1}\int_{-\infty }^{\infty }e^{4ix(t-1/2)}dx$$ $$=\frac{2\pi}{4}\Gamma(2r)\bigl(t(1-t)\bigr)^{r-1}|_{t=1/2}$$ $$J(r)=2\pi\frac{\Gamma(2r)}{2^{2r}}$$
Check: at $r=1$ $$\Gamma(1+ix)\Gamma(1-ix)=ix\Gamma(ix)\Gamma(1-ix)=ix\frac{\pi}{\sin\pi ix}=\frac{\pi x}{\sinh\pi x}$$ $$\int_{-\infty }^{\infty }|\Gamma(1+ix)|^2dx=\int_{-\infty }^{\infty }\frac{\pi x}{\sinh\pi x}dx=\frac{\pi}{2}$$
For $$I(r)=\int_{-\infty }^{\infty } |\Gamma (r+i x)| \, dx$$ assuming $r>0$, I am not sure that we can obtain any anlytical result.
Numerically, a quick and dirty regression $$\log[I(r)]=a + b\, r^c$$ gives, with $R^2=0.999861$, $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & -3.79271 & 0.2560 & \{-4.30087,-3.28455\} \\ b & +0.99496 & 0.0121 & \{+0.97088,+1.01903\} \\ c & +1.28364 & 0.0026 & \{+1.27847,+1.28880\} \\ \end{array}$$ which probably hides a logarithmic contribution.