Re: How to interpret this integral?

*To*: mathgroup at smc.vnet.net*Subject*: [mg112610] Re: How to interpret this integral?*From*: Leonid Shifrin <lshifr at gmail.com>*Date*: Thu, 23 Sep 2010 04:20:30 -0400 (EDT)

Hi Julian, Here is your expression: In[137]:= expr = Integrate[(aV*t+v0)*Cos[(aW*t^2)/2+th0+t*w0],{t,0,Ts}] Out[137]= (1/(aW^(3/2)))(Sqrt[\[Pi]] (-aW v0+aV w0) Cos[th0-w0^2/(2 aW)] FresnelC[w0/(Sqrt[aW] Sqrt[\[Pi]])]+Sqrt[\[Pi]] (aW v0-aV w0) Cos[th0-w0^2/(2 aW)] FresnelC[(aW Ts+w0)/(Sqrt[aW] Sqrt[\[Pi]])]+aV Sqrt[aW] (-Sin[th0]+Sin[th0+(aW Ts^2)/2+Ts w0])+Sqrt[\[Pi]] (aW v0-aV w0) (FresnelS[w0/(Sqrt[aW] Sqrt[\[Pi]])]-FresnelS[(aW Ts+w0)/(Sqrt[aW] Sqrt[\[Pi]])]) Sin[th0-w0^2/(2 aW)]) If you want to treat the point aW = 0 specially, the simplest is to put it to zero right when the integral is computed: In[138]:= exprAt0 = Integrate[(aV*t+v0)*Cos[th0+t*w0],{t,0,Ts}] Out[138]= (-aV Cos[th0]+aV Cos[th0+Ts w0]-v0 w0 Sin[th0]+(aV Ts+v0) w0 Sin[th0+Ts w0])/w0^2 Alternatively, you can extract it from the full expression, but it st a little more involved. One way is to change variable as t = 1/aW, then expand in t around t=Infinity, then take the limit t->Infinity of the resulting expression. One subtlety here that you will need to expand to the second order in t, to get the correct result: In[140]:= exprAt0Alt = Limit[Normal@Series [expr/.aW->1/t,{t,Infinity,2}],t->Infinity] Out[140]= (-aV Cos[th0]+aV Cos[th0+Ts w0]+w0 (-v0 Sin[th0]+aV Ts Sin[th0+Ts w0]+v0 Sin[th0+Ts w0]))/w0^2 You can check that the results are the same: In[141]:= FullSimplify[exprAt0-exprAt0Alt] Out[141]= 0 Hope this helps. Regards, Leonid On Wed, Sep 22, 2010 at 9:57 AM, Julian <julian.stoev at gmail.com> wrote: > Hello All, > > After long interruption with symbolic computing, I am struggling how > to make a useful result out of this integral. > > Integrate[(aV*t + v0)*Cos[(aW*t^2)/2 + th0 + t*w0, {t, 0, Ts}] > > It comes from the differential equations describing the motion of a > wheeled robot. aV and aW are accelerations and v0, w0, th0 are initial > conditions. The numerical solution clearly exists for different real > accelerations, including positive, negative and zero. > > However the symbolic solution of Mathematica is: > (Sqrt[Pi]*(-(aW*v0) + aV*w0)*Cos[th0 - w0^2/(2*aW)]* > FresnelC[w0/(Sqrt[aW]*Sqrt[Pi])] + Sqrt[Pi]*(aW*v0 - aV*w0)* > Cos[th0 - w0^2/(2*aW)]*FresnelC[(aW*Ts + w0)/(Sqrt[aW]*Sqrt[Pi])] > + > aV*Sqrt[aW]*(-Sin[th0] + Sin[th0 + (aW*Ts^2)/2 + Ts*w0]) + > Sqrt[Pi]*(aW*v0 - aV*w0)*(FresnelS[w0/(Sqrt[aW]*Sqrt[Pi])] - > FresnelS[(aW*Ts + w0)/(Sqrt[aW]*Sqrt[Pi])])*Sin[th0 - w0^2/ > (2*aW)])/ > aW^(3/2) > > Note that aW can be found inside a Sqrt function and also in the > denominator. While I can see that FresnelS[Infinity] is well defined, > so aW==0 should not be a real problem, it is still very problematic > how to use this result at this particular point. The case of negative > aW is also interesting, because it will require a complex FresnelS and > FresnelC. I somehow managed to remove this problem using > ComplexExpand, but I am not sure this is the good solution. > > I tried to give Assumptions -> Element[{aW, aV, w0, v0, Ts, th0}, > Reals] to the integral, but there is no change in the solution. > > While I understand that the solution of Mathematica is correct in the > strict mathematical sense, I have to use the result to generate a C- > code, which will be evaluated numerically and the solution I have now > is not working for this. > > Can some experienced user give a good advice how use get a solution of > the problem? > > Thank you in advance! > > --JS > >