# Calculate an inverse Laplace transform?

Posted 1 month ago
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 Hello, I need an inverse Laplace transform of the expression: Log[s+a]/Sqrt[s] where s is the Laplace variable and a is a nonnegative real constant. I recall that the older version (5) of MATHEMATICA returned some formula containing a generalized hypergeometric series but the formula was clearly wrong (did not agree with the numerically inverted transform) and also it was problematic to obtain the values of the series for arguments larger than about 100. I would be happy to get the result in terms of more widely known and computable functions, if possible. Is MATHEMATICA currently able to derive this result? Lesław
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Posted 1 month ago
 We can express Log[s+a] by series: SeriesCoefficient[Log[s + a], {a, 0, n}] We can check: FullSimplify[Log[s] + Sum[((-1)^(1 + n) s^-n)/n*a^n, {n, 1, Infinity}], Assumptions -> {a > 0, s > 0}] (*Log[a + s]*) Then InverseLaplaceTransform is: InverseLaplaceTransform[(Log[s] + ((-1)^(1 + n) s^-n)/n*a^n)/Sqrt[s], s, t] (*((-1)^(1 + n) a^n t^(-(1/2) + n))/( n Gamma[1/2 + n]) + (-Log[t] + PolyGamma[0, 1/2])/( Sqrt[\[Pi]] Sqrt[t])*) and Sum[((-1)^(1 + n) a^n t^(-(1/2) + n))/( n Gamma[1/2 + n]), {n, 1, Infinity}] + (-Log[t] + PolyGamma[0, 1/2])/(Sqrt[\[Pi]] Sqrt[t]) (*(2 a Sqrt[t] HypergeometricPFQ[{1, 1}, {3/2, 2}, -a t])/Sqrt[\[Pi]] + (-Log[t] + PolyGamma[0, 1/2])/(Sqrt[\[Pi]] Sqrt[t])*) $$\mathcal{L}_s^{-1}\left[\frac{\log (s+a)}{\sqrt{s}}\right](t)=\frac{2 a t \, _2F_2\left(1,1;\frac{3}{2},2;-a t\right)-\log (t)+\psi ^{(0)}\left(\frac{1}{2}\right)}{\sqrt{\pi } \sqrt{t}}$$Regards M.I.
 Typically, increasing $MaxExtraPrecision will resolve the problem:$Version (*"12.0.0 for Microsoft Windows (64-bit) (April 6, 2019)"*) Block[{$MaxExtraPrecision = Infinity}, N[HypergeometricPFQ[{1, 1}, {3/2, 2}, -10^12], 20]] // AbsoluteTiming (*Calculating time is very very long *) Better is asymptotic expansion hypergeometric function calculating time is very short: An example: Block[{$MaxExtraPrecision = Infinity}, N[HypergeometricPFQ[{1, 1}, {3/2, 2}, -10^4], 50]] (*0.00055869001971234905256956762831658880400079883301504*) sol = Series[ HypergeometricPFQ[{1, 1}, {3/2, 2}, -a*t], {a, Infinity, 20}] // Normal; Block[{$MaxExtraPrecision = Infinity}, N[sol /. a -> 10^4 /. t -> 1, 50]] (*0.00055869001971234905256956762831658880400079883301504 + 0.*10^-4349 I*) For *at =10^12:** Block[{$MaxExtraPrecision = Infinity}, N[sol /. a -> 10^12 /. t -> 1, 50]] (*1.4797265570974735843828436707105563028890688734102*10^-11 + 4.9626916249394098003201139691172499219204358091276*10^-434294481922 \ I*) Regards M.I.