How to realize the inverse Laplace Transform?

Questions
1

This expression came from the Laplace Transform,

(Sqrt[Dc] E^((
 Sqrt[s] (lh - x))/Sqrt[Dc]) F1)/(Sqrt[s] (s - \[Theta]c))

where s is the unique argument.
Now it is needed to make the inverse Laplace Transform and it can be realized by the convolution theorem.
But it will be simpler to do it with the Residue Theorem.
Would you like to help me with this issue?
Thanks.

2

Hello,with convolution theorem I have:

L1 = InverseLaplaceTransform[(
  Sqrt[Dc] E^((Sqrt[s] (lh - x))/Sqrt[Dc]) F1)/Sqrt[s] , s, t]
L2 = InverseLaplaceTransform[  1/(s - \[Theta]c), s, t]

L12 = Integrate[(L1[[1]] /. t -> (t - \[Tau]))*(L2 /. 
     t -> \[Tau]), {\[Tau], 0, t}, 
  Assumptions -> {t > 0, F1 > 0, Dc > 0, \[Theta]c > 0, lh > 0, 
    x > 0}]
 (*(Dc E^(t \[Theta]c - 
   Sqrt[\[Theta]c/Dc] Abs[lh - x]) F1 (1 + 
    Erf[(2 t Sqrt[Dc \[Theta]c] - Abs[lh - x])/(2 Sqrt[Dc t])] + 
    E^(2 Sqrt[\[Theta]c/Dc]
       Abs[lh - x]) (-1 + 
       Erf[(2 t Sqrt[Dc \[Theta]c] + Abs[lh - x])/(
        2 Sqrt[Dc t])])))/(2 Sqrt[Dc \[Theta]c])*)

\mathcal{L}_s^{-1}\left[\frac{\sqrt{\text{Dc}} e^{\frac{\sqrt{s} (\text{lh}-x)}{\sqrt{\text{Dc}}}} \text{F1}}{\sqrt{s} (s-\text{\theta c})}\right](t)=\frac{\text{Dc} e^{t \text{\theta c}-\sqrt{\frac{\text{\theta
c}}{\text{Dc}}} | \text{lh}-x| } \text{F1} \left(1+\text{erf}\left(\frac{2 t \sqrt{\text{Dc} \text{\theta c}}-| \text{lh}-x| }{2 \sqrt{\text{Dc} t}}\right)+e^{2 \sqrt{\frac{\text{\theta $c}}{\text{Dc}}} |
\text{lh}-x| } \left(-1+\text{erf}\left(\frac{2 t \sqrt{\text{Dc} \text{$\theta c}}+| \text{lh}-x| }{2 \sqrt{\text{Dc} t}}\right)\right)\right)}{2 \sqrt{\text{Dc} \text{\theta c}}}

Regards M.I.

Untitled-1 ver1.nb (15 KB)

3

The ILT by convolution has already done. I am seeking for the ILT by the residue theorem which seems more capable.

In[88]:= (*ILT of eq1 which came from LT.*)

In[89]:= eq1 = (Sqrt[Dc] E^((Sqrt[s] (lh - x))/Sqrt[Dc]) F1)/(
 Sqrt[s] (s - \[Theta]c))

Out[89]= (Sqrt[Dc] E^((
 Sqrt[s] (lh - x))/Sqrt[Dc]) F1)/(Sqrt[s] (s - \[Theta]c))

In[90]:= (*The numerator of eq1 is the first part.*)

In[91]:= eq2 = Numerator[eq1]/Sqrt[s]

Out[91]= (Sqrt[Dc] E^((Sqrt[s] (lh - x))/Sqrt[Dc]) F1)/Sqrt[s]

In[92]:= (*The denominator of eq1 is the second part.*)

In[93]:= eq3 = Sqrt[s]/Denominator[eq1]

Out[93]= 1/(s - \[Theta]c)

In[94]:= (*The first part.*)

In[95]:= eq4 = InverseLaplaceTransform[eq2, s, t] // Simplify // Normal

Out[95]= (E^(-((lh - x)^2/(4 Dc t))) F1 Sqrt[Dc/t])/Sqrt[\[Pi]]

In[96]:= (*The second parts.*)

In[97]:= eq5 = InverseLaplaceTransform[eq3, s, t] // Simplify // Normal

Out[97]= E^(t \[Theta]c)

In[98]:= (*convolution of eq4 and eq5.*)

In[99]:= eq6 = Integrate[eq4*(eq5 /. {t -> (-t + T) }), {t, 0, T}]

Out[99]= \!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(T\)]\(
\*FractionBox[\(
\*SuperscriptBox[\(E\), \(\(-
\*FractionBox[
SuperscriptBox[\((lh - x)\), \(2\)], \(4\ Dc\ t\)]\) + \((\(-t\) + 
         T)\)\ \[Theta]c\)]\ F1\ 
\*SqrtBox[
FractionBox[\(Dc\), \(t\)]]\), 
SqrtBox[\(\[Pi]\)]] \[DifferentialD]t\)\)

In[100]:= (*change the variable's symbol of eq6. Solution after \
InverseLaplaceTransform.*)

In[101]:= eq7 = eq6 /. {t -> \[Tau]} /. {T -> t}

Out[101]= \!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(t\)]\(
\*FractionBox[\(
\*SuperscriptBox[\(E\), \(\[Theta]c\ \((t - \[Tau])\) - 
\*FractionBox[
SuperscriptBox[\((lh - x)\), \(2\)], \(4\ Dc\ \[Tau]\)]\)]\ F1\ 
\*SqrtBox[
FractionBox[\(Dc\), \(\[Tau]\)]]\), 
SqrtBox[\(\[Pi]\)]] \[DifferentialD]\[Tau]\)\)
4

Thanks.
I have read your codes intensively.

5

Your version of Mathematica is higher.