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