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MathGroup Archive 2007

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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



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