Limit[Integrate[Sin[\[Omega]*t]*Exp[-s*t],{t,0,x},
Assumptions->s>0],x->\[Infinity]]
to answer \[Omega]/(\[Omega]^2+s^2) ?
Perhaps by telling Mathematica that omega is real... since Mathematica
works over the field of complex numbers, by default.
In[9]:= Assuming[s > 0 && Im[\[Omega]] == 0,
Limit[Integrate[Sin[\[Omega]*t]*Exp[-s*t], {t, 0, x}],
x -> \[Infinity]]]
Out[9]= \[Omega]/(s^2 + \[Omega]^2)
Regards,
-- Jean-Marc
Keeping in mind that
Integrate[f[x],{x,x0,Infinity}] is by def.
Limit[Integrate[f[x],{x,x0,x1},x1->Infinity],
Integrate[Sin[\[Omega]*t]*Exp[-s*t],{t,0,Infinity},Assumptions -> s>0]
or
LaplaceTransform[Sin[\[Omega]*t],t,s]
should (don't have Mathematica installed on this system) give the desired result.
Integrate[Sin[\[Omega]*t]*Exp[(-t)*s], {t, 0, Infinity},
Assumptions -> Inequality[s, Greater, 0, Equal, Im[\[Omega]]]]
or - because you'll get s^2+omega^2 in the denominator:
Integrate[Sin[\[Omega]*t]/E^(t*s), {t, 0, Infinity},
Assumptions -> s > Abs[Im[\[Omega]]]]
Peter
Limit[Integrate[Sin[\[Omega]*t]*Exp[-s*t], {t, 0, x}, Assumptions -> s
> 0], x -> \[Infinity], Assumptions -> \[Omega] > 0 && s > 0]
Cheers
You have to tell the assumptions to Limit too:
Limit[Integrate[Sin[\[Omega] t] E^(-s t), {t, 0, x},
Assumptions -> s > 0], x -> \[Infinity],
Assumptions -> s > 0 && \[Omega] \[Element] Reals]
To make things simpler and avoid duplication we could use Assuming:
Assuming[s > 0 && \[Omega] \[Element] Reals,
Limit[Integrate[Sin[\[Omega] t] E^(-s t), {t, 0, x}],
x -> \[Infinity]]]
In[2]:= Assuming[s > 0 && Element[\[Omega], Reals],
Limit[Integrate[Sin[\[Omega]*t]*Exp[(-s)*t], {t, 0, x}], x -> Infinity]]
Out[2]= \[Omega]/(s^2 + \[Omega]^2)
David
Correct. The limit comes from a Laplace Transform homework exercise
for sophomores. Mathematica was allowed, along with three other CAS.
Students check their solution with tables.