Re: ideas needed!

*To*: mathgroup at smc.vnet.net*Subject*: [mg76876] Re: ideas needed!*From*: dimitris <dimmechan at yahoo.com>*Date*: Tue, 29 May 2007 04:56:52 -0400 (EDT)*References*: <f38sf0$ije$1@smc.vnet.net><f3bgnq$34c$1@smc.vnet.net>

After a private response I was informed that the integral appears (slightly modificated) at gradshteyn-ryzhik and it is equal to x * y /(m^2 * (x^2+y^2)) * BesselK(2,sqrt(x^2+y^2)/m). More specifically it is the formula (3.914) and someone has to differentiate wrt b. I use Mathematica 5.2 and I don't know what version 6 does but it seems not quite good (at least to me) that a definite integral like this cannot be done by Mathematica. Of course I wait for opposite opinions. And someone would say if CASs could do all the integrals who would need tables of integrals? Anyway, the integral can be done with contour integration but my question was getting the integral with Mathematica. The following is based on some workaround of my colleague Panagiotis Gourgiotis. It should be noted that he got the result without knowing the presence of the GR-formula! Here is the integrand In[5]:= integrand = (u*Sin[u*x])/E^((Sqrt[1 + m^2*u^2]*y)/m); Integarate wrt x and y, take the series expansion of the reulting funtion and integrate the result wrt u from 0 to infinity. In[6]:= Integrate[integrand, x, y] Normal[Series[%, {y, 0, 3}]] Intser = (Integrate[#1, {u, 0, Infinity}, GenerateConditions -> False] & ) /@ % Out[6]= (m*Cos[u*x])/(E^((Sqrt[1 + m^2*u^2]*y)/m)*Sqrt[1 + m^2*u^2]) Out[7]= (m*Cos[u*x])/Sqrt[1 + m^2*u^2] - y*Cos[u*x] + (Sqrt[1 + m^2*u^2]*y^2*Cos[u*x])/(2*m) + y^3*(-(Cos[u*x]/(6*m^2)) - (1/6)*u^2*Cos[u*x]) Out[8]= (m*BesselK[0, Sqrt[x^2]/Sqrt[m^2]])/Sqrt[m^2] - (y^2*BesselK[1, Sqrt[x^2]/Sqrt[m^2]])/(2*m*Sqrt[x^2]) Take the series of BesselK[0, Sqrt[x^2 + y^2]/m] In[11]:= serBK1 = FullSimplify[Normal[Series[BesselK[0, Sqrt[x^2 + y^2]/m], {y, 0, 3}]]] Out[11]= BesselK[0, Sqrt[x^2]/m] - (y^2*BesselK[1, Sqrt[x^2]/m])/ (2*m*Sqrt[x^2]) Intser and serBK1 are the same. Taking more terms in above series expansion shows that the coincedence continues. So the result of the integration Integrate[integrand, x, y] is BesselK[0, Sqrt[x^2 + y^2]/m] Differentiate wrt x and y, BesselK[0, Sqrt[x^2 + y^2]/m], we get In[13]:= FullSimplify[D[BesselK[0, Sqrt[x^2 + y^2]/m], x, y]] Out[13]= (x*y*BesselK[2, Sqrt[x^2 + y^2]/m])/(m^2*(x^2 + y^2)) A quick check follows now In[16]:= NIntegrate[(u*Sin[u*x])/E^((Sqrt[1 + m^2*u^2]*y)/m) /. {x -> 1/2, m -> 1, y -> 2}, {u, 0, Infinity}] Out[16]= 0.05430810820969097 In[16]:= N[x*(y/(m^2*(x^2 + y^2)))*BesselK[2, Sqrt[x^2 + y^2]/m] /. {x -> 1/2, m -> 1, y -> 2}] Out[16]= 0.05430810820909097 Dimitris / dimitris : > Not care for assumptions, convergence. > Even a result in Hadamard sense is ok! > > Dimitris > > > / dimitris : > > Integrate[u*Sin[u*x]*Exp[(-y)*(Sqrt[1 + m^2*u^2]/ m)], {u, 0, > > Infinity}] > > > > Any ideas for getting a closed form expression? > > > > Dimitris