Re: two integrals
- To: mathgroup at smc.vnet.net
- Subject: [mg78381] Re: two integrals
- From: dimitris <dimmechan at yahoo.com>
- Date: Fri, 29 Jun 2007 05:53:52 -0400 (EDT)
- References: <f5vraa$jj1$1@smc.vnet.net>
dimitris :
> Any ideas about getting a closed form expression?
> No need fo convegrence; a result valid in the Hadamard sense is ok!
>
> Integrate[(Sqrt[1 + m^2*u^2]/u)*Exp[(-y)*(Sqrt[1 + m^2*u^2]/m)]*
> Sin[u*x], {x, 0, Infinity}]
>
> Integrate[Exp[(-y)*(Sqrt[1 + m^2*u^2]/m)]*Cos[u*x], {x, 0, Infinity}]
>
> Dimitris
The second integral I am interested in is
>Integrate[Exp[(-y)*(Sqrt[1 + m^2*u^2]/m)]*Cos[u*x], {x, 0, Infinity}]
However, having in hand the result
> Integrate[u*Sin[u*x]*Exp[(-y)*(Sqrt[1 + m^2*u^2]/m)],{u,0,Infinity}]=BesselK[0, Sqrt[x^2 + y^2]/m]
(which I prove in a post called "a definite integral and a
workaround")
and considering that
In[86]:=
-D[Exp[(-y)*(Sqrt[1 + m^2*u^2]/m)]*Cos[u*x], x]
Out[86]=
(u*Sin[u*x])/E^((Sqrt[1 + m^2*u^2]*y)/m)
we finally have
(*Integrate[Exp[(-y)*(Sqrt[1 + m^2*u^2]/m)]*Cos[u*x], {x, 0,
Infinity}] =)
In[83]:=
-Integrate[(x*y*BesselK[2, Sqrt[x^2 + y^2]/m])/(m^2*(x^2 + y^2)), x]
Out[83]=
(y*BesselK[1, Sqrt[x^2 + y^2]/m])/(m*Sqrt[x^2 + y^2])
Voila!
But the quest for the first integral keeps on!
Dimitris