On the Laplace transform of Beta function

Question

While trying to evaluate the Laplace transform below $$I = \int_{0}^{\infty}e^{-st}B(\frac{1}{2}-it,\frac{3}{2}+it)\mathrm{d}t,$$

invoking

LaplaceTransform[Beta[0.5-it, 1.5+it], t, s],

yields

$$\frac{B(\frac{1}{2}-it,\frac{3}{2}+it)}{s},$$

which is quite weird since the output of a Laplace transform has to be an exclusive function of $s$, but this output still includes $t$. What am I missing here?

Answer 1

$Version

(* "14.0.0 for Mac OS X ARM (64-bit) (December 13, 2023)" *)

Clear["Global`*"]

Mathematica appears to be able to handle imaginary values for i

LaplaceTransform[Beta[1/2 - I*t, 3/2 + I*t] //
   FunctionExpand, t, s] // FullSimplify

(* (1/(8 π))(-2 π PolyGamma[0, (π + s)/(4 π)] + 
  2 π PolyGamma[0, 1/4 (3 + s/π)] + 
  I (PolyGamma[1, (π + s)/(4 π)] - PolyGamma[1, 1/4 (3 + s/π)])) *)

Letting i == a*I,

f[i_, s_] = ((LaplaceTransform[(Beta[1/2 - i*t, 3/2 + i*t] /. i -> a*I) //
        FunctionExpand, t, s] // FullSimplify) /. a -> i/I) // FullSimplify

(* (1/(8 i π))(-2 I π (PolyGamma[0, 1/4 + (I s)/(4 i π)] - 
     PolyGamma[0, 3/4 + (I s)/(4 i π)]) - 
  PolyGamma[1, 1/4 (1 + (I s)/(i π))] + 
  PolyGamma[1, 1/4 (3 + (I s)/(i π))]) *)