# On the Laplace transform of Beta function

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?

• Looks like a bug, you shouldn't have s in output..I came across similar bugs, the solution was to rewrite integral using Feynam trick Commented Feb 6 at 1:39
• @YaroslavBulatov: $s$ is OK. One should not have $t$ in the output.
– User
Commented Feb 6 at 1:42
• There were couple of bugs I ran across in laplace.transform, my workaround is using ilaplaceFeynmann and ilaplaceMellin helpers from mathematica.stackexchange.com/q/286895/217 Commented Feb 6 at 1:43
• @YaroslavBulatov: The definitions of these transforms are apparently not the same as that ofLaplaceTransform. So, how do their results help here?
– User
Commented Feb 6 at 1:46
• The Beta function does not include t, only the variable it. What did you intend? i*t or I*t? Commented Feb 6 at 2:27

\$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))]) *)
`