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]\)\)